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The linear interpolation (assuming that observations are uniformly distributed within a bucket) is a solid and simple assumption in lack of any other information. However, the exponential bucketing used by standard schemas of native histograms has been chosen to cover the whole range of observations in a way that bucket populations are spread out over buckets in a reasonably way for typical distributions encountered in real-world scenarios. This is the origin of the idea implemented here: If we divide a given bucket into two (or more) smaller exponential buckets, we "most naturally" expect that the samples in the original buckets will split among those smaller buckets in a more or less uniform fashion. With this assumption, we end up with an "exponential interpolation", which therefore appears to be a better match for histograms with exponential bucketing. This commit leaves the linear interpolation in place for NHCB, but changes the interpolation for exponential native histograms to exponential. This affects `histogram_quantile` and `histogram_fraction` (because the latter is more or less the inverse of the former). The zero bucket has to be treated specially because the assumption above would lead to an "interpolation to zero" (the bucket density approaches infinity around zero, and with the postulated uniform usage of buckets, we would end up with an estimate of zero for all quantiles ending up in the zero bucket). We simply fall back to linear interpolation within the zero bucket. At the same time, this commit makes the call to stick with the assumption that the zero bucket only contains positive observations for native histograms without negative buckets (and vice versa). (This is an assumption relevant for interpolation. It is a mostly academic point, as the zero bucket is supposed to be very small anyway. However, in cases where it _is_ relevantly broad, the assumption helps a lot in practice.) This commit also updates and completes the documentation to match both details about interpolation. As a more high level note: The approach here attempts to strike a balance between a more simplistic approach without any assumption, and a more involved approach with more sophisticated assumptions. I will shortly describe both for reference: The "zero assumption" approach would be to not interpolate at all, but _always_ return the harmonic mean of the bucket boundaries of the bucket the quantile ends up in. This has the advantage of minimizing the maximum possible relative error of the quantile estimation. (Depending on the exact definition of the relative error of an estimation, there is also an argument to return the arithmetic mean of the bucket boundaries.) While limiting the maximum possible relative error is a good property, this approach would throw away the information if a quantile is closer to the upper or lower end of the population within a bucket. This can be valuable trending information in a dashboard. With any kind of interpolation, the maximum possible error of a quantile estimation increases to the full width of a bucket (i.e. it more than doubles for the harmonic mean approach, and precisely doubles for the arithmetic mean approach). However, in return the _expectation value_ of the error decreases. The increase of the theoretical maximum only has practical relevance for pathologic distributions. For example, if there are thousand observations within a bucket, they could _all_ be at the upper bound of the bucket. If the quantile calculation picks the 1st observation in the bucket as the relevant one, an interpolation will yield a value close to the lower bucket boundary, while the true quantile value is close to the upper boundary. The "fancy interpolation" approach would be one that analyses the _actual_ distribution of samples in the histogram. A lot of statistics could be applied based on the information we have available in the histogram. This would include the population of neighboring (or even all) buckets in the histogram. In general, the resolution of a native histogram should be quite high, and therefore, those "fancy" approaches would increase the computational cost quite a bit with very little practical benefits (i.e. just tiny corrections of the estimated quantile value). The results are also much harder to reason with. Signed-off-by: beorn7 <beorn@grafana.com>
1165 lines
42 KiB
Plaintext
1165 lines
42 KiB
Plaintext
# Minimal valid case: an empty histogram.
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load 5m
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empty_histogram {{}}
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eval instant at 1m empty_histogram
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{__name__="empty_histogram"} {{}}
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eval instant at 1m histogram_count(empty_histogram)
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{} 0
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eval instant at 1m histogram_sum(empty_histogram)
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{} 0
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eval instant at 1m histogram_avg(empty_histogram)
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{} NaN
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eval instant at 1m histogram_fraction(-Inf, +Inf, empty_histogram)
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{} NaN
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eval instant at 1m histogram_fraction(0, 8, empty_histogram)
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{} NaN
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clear
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# buckets:[1 2 1] means 1 observation in the 1st bucket, 2 observations in the 2nd and 1 observation in the 3rd (total 4).
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load 5m
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single_histogram {{schema:0 sum:5 count:4 buckets:[1 2 1]}}
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# histogram_count extracts the count property from the histogram.
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eval instant at 1m histogram_count(single_histogram)
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{} 4
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# histogram_sum extracts the sum property from the histogram.
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eval instant at 1m histogram_sum(single_histogram)
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{} 5
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# histogram_avg calculates the average from sum and count properties.
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eval instant at 1m histogram_avg(single_histogram)
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{} 1.25
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# We expect half of the values to fall in the range 1 < x <= 2.
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eval instant at 1m histogram_fraction(1, 2, single_histogram)
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{} 0.5
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# We expect all values to fall in the range 0 < x <= 8.
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eval instant at 1m histogram_fraction(0, 8, single_histogram)
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{} 1
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# Median is 1.414213562373095 (2**2**-1, or sqrt(2)) due to
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# exponential interpolation, i.e. the "midpoint" within range 1 < x <=
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# 2 is assumed where the bucket boundary would be if we increased the
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# resolution of the histogram by one step.
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eval instant at 1m histogram_quantile(0.5, single_histogram)
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{} 1.414213562373095
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clear
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# Repeat the same histogram 10 times.
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load 5m
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multi_histogram {{schema:0 sum:5 count:4 buckets:[1 2 1]}}x10
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eval instant at 5m histogram_count(multi_histogram)
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{} 4
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eval instant at 5m histogram_sum(multi_histogram)
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{} 5
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eval instant at 5m histogram_avg(multi_histogram)
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{} 1.25
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eval instant at 5m histogram_fraction(1, 2, multi_histogram)
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{} 0.5
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# See explanation for exponential interpolation above.
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eval instant at 5m histogram_quantile(0.5, multi_histogram)
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{} 1.414213562373095
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# Each entry should look the same as the first.
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eval instant at 50m histogram_count(multi_histogram)
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{} 4
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eval instant at 50m histogram_sum(multi_histogram)
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{} 5
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eval instant at 50m histogram_avg(multi_histogram)
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{} 1.25
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eval instant at 50m histogram_fraction(1, 2, multi_histogram)
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{} 0.5
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# See explanation for exponential interpolation above.
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eval instant at 50m histogram_quantile(0.5, multi_histogram)
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{} 1.414213562373095
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clear
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# Accumulate the histogram addition for 10 iterations, offset is a bucket position where offset:0 is always the bucket
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# with an upper limit of 1 and offset:1 is the bucket which follows to the right. Negative offsets represent bucket
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# positions for upper limits <1 (tending toward zero), where offset:-1 is the bucket to the left of offset:0.
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load 5m
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incr_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}}+{{sum:2 count:1 buckets:[1] offset:1}}x10
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eval instant at 5m histogram_count(incr_histogram)
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{} 5
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eval instant at 5m histogram_sum(incr_histogram)
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{} 6
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eval instant at 5m histogram_avg(incr_histogram)
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{} 1.2
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# We expect 3/5ths of the values to fall in the range 1 < x <= 2.
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eval instant at 5m histogram_fraction(1, 2, incr_histogram)
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{} 0.6
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# See explanation for exponential interpolation above.
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eval instant at 5m histogram_quantile(0.5, incr_histogram)
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{} 1.414213562373095
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eval instant at 50m incr_histogram
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{__name__="incr_histogram"} {{count:14 sum:24 buckets:[1 12 1]}}
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eval instant at 50m histogram_count(incr_histogram)
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{} 14
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eval instant at 50m histogram_sum(incr_histogram)
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{} 24
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eval instant at 50m histogram_avg(incr_histogram)
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{} 1.7142857142857142
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# We expect 12/14ths of the values to fall in the range 1 < x <= 2.
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eval instant at 50m histogram_fraction(1, 2, incr_histogram)
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{} 0.8571428571428571
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# See explanation for exponential interpolation above.
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eval instant at 50m histogram_quantile(0.5, incr_histogram)
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{} 1.414213562373095
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# Per-second average rate of increase should be 1/(5*60) for count and buckets, then 2/(5*60) for sum.
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eval instant at 50m rate(incr_histogram[10m])
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{} {{count:0.0033333333333333335 sum:0.006666666666666667 offset:1 buckets:[0.0033333333333333335]}}
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# Calculate the 50th percentile of observations over the last 10m.
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# See explanation for exponential interpolation above.
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eval instant at 50m histogram_quantile(0.5, rate(incr_histogram[10m]))
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{} 1.414213562373095
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clear
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# Schema represents the histogram resolution, different schema have compatible bucket boundaries, e.g.:
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# 0: 1 2 4 8 16 32 64 (higher resolution)
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# -1: 1 4 16 64 (lower resolution)
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#
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# Histograms can be merged as long as the histogram to the right is same resolution or higher.
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load 5m
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low_res_histogram {{schema:-1 sum:4 count:1 buckets:[1] offset:1}}+{{schema:0 sum:4 count:4 buckets:[2 2] offset:1}}x1
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eval instant at 5m low_res_histogram
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{__name__="low_res_histogram"} {{schema:-1 count:5 sum:8 offset:1 buckets:[5]}}
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eval instant at 5m histogram_count(low_res_histogram)
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{} 5
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eval instant at 5m histogram_sum(low_res_histogram)
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{} 8
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eval instant at 5m histogram_avg(low_res_histogram)
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{} 1.6
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# We expect all values to fall into the lower-resolution bucket with the range 1 < x <= 4.
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eval instant at 5m histogram_fraction(1, 4, low_res_histogram)
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{} 1
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clear
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# z_bucket:1 means there is one observation in the zero bucket and z_bucket_w:0.5 means the zero bucket has the range
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# 0 < x <= 0.5. Sum and count are expected to represent all observations in the histogram, including those in the zero bucket.
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load 5m
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single_zero_histogram {{schema:0 z_bucket:1 z_bucket_w:0.5 sum:0.25 count:1}}
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eval instant at 1m histogram_count(single_zero_histogram)
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{} 1
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eval instant at 1m histogram_sum(single_zero_histogram)
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{} 0.25
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eval instant at 1m histogram_avg(single_zero_histogram)
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{} 0.25
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# When only the zero bucket is populated, or there are negative buckets, the distribution is assumed to be equally
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# distributed around zero; i.e. that there are an equal number of positive and negative observations. Therefore the
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# entire distribution must lie within the full range of the zero bucket, in this case: -0.5 < x <= +0.5.
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eval instant at 1m histogram_fraction(-0.5, 0.5, single_zero_histogram)
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{} 1
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# Half of the observations are estimated to be zero, as this is the midpoint between -0.5 and +0.5.
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eval instant at 1m histogram_quantile(0.5, single_zero_histogram)
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{} 0
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clear
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# Let's turn single_histogram upside-down.
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load 5m
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negative_histogram {{schema:0 sum:-5 count:4 n_buckets:[1 2 1]}}
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eval instant at 1m histogram_count(negative_histogram)
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{} 4
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eval instant at 1m histogram_sum(negative_histogram)
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{} -5
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eval instant at 1m histogram_avg(negative_histogram)
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{} -1.25
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# We expect half of the values to fall in the range -2 < x <= -1.
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eval instant at 1m histogram_fraction(-2, -1, negative_histogram)
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{} 0.5
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# Exponential interpolation works the same as for positive buckets, just mirrored.
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eval instant at 1m histogram_quantile(0.5, negative_histogram)
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{} -1.414213562373095
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clear
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# Two histogram samples.
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load 5m
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two_samples_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}} {{schema:0 sum:-4 count:4 n_buckets:[1 2 1]}}
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# We expect to see the newest sample.
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eval instant at 5m histogram_count(two_samples_histogram)
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{} 4
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eval instant at 5m histogram_sum(two_samples_histogram)
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{} -4
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eval instant at 5m histogram_avg(two_samples_histogram)
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{} -1
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eval instant at 5m histogram_fraction(-2, -1, two_samples_histogram)
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{} 0.5
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# See explanation for exponential interpolation above.
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eval instant at 5m histogram_quantile(0.5, two_samples_histogram)
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{} -1.414213562373095
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clear
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# Add two histograms with negated data.
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load 5m
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balanced_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}}+{{schema:0 sum:-4 count:4 n_buckets:[1 2 1]}}x1
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eval instant at 5m histogram_count(balanced_histogram)
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{} 8
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eval instant at 5m histogram_sum(balanced_histogram)
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{} 0
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eval instant at 5m histogram_avg(balanced_histogram)
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{} 0
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eval instant at 5m histogram_fraction(0, 4, balanced_histogram)
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{} 0.5
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# If the quantile happens to be located in a span of empty buckets, the actually returned value is the lower bound of
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# the first populated bucket after the span of empty buckets.
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eval instant at 5m histogram_quantile(0.5, balanced_histogram)
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{} 0.5
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clear
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# Add histogram to test sum(last_over_time) regression
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load 5m
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incr_sum_histogram{number="1"} {{schema:0 sum:0 count:0 buckets:[1]}}+{{schema:0 sum:1 count:1 buckets:[1]}}x10
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incr_sum_histogram{number="2"} {{schema:0 sum:0 count:0 buckets:[1]}}+{{schema:0 sum:2 count:1 buckets:[1]}}x10
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eval instant at 50m histogram_sum(sum(incr_sum_histogram))
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{} 30
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eval instant at 50m histogram_sum(sum(last_over_time(incr_sum_histogram[5m])))
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{} 30
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clear
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# Apply rate function to histogram.
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load 15s
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histogram_rate {{schema:1 count:12 sum:18.4 z_bucket:2 z_bucket_w:0.001 buckets:[1 2 0 1 1] n_buckets:[1 2 0 1 1]}}+{{schema:1 count:9 sum:18.4 z_bucket:1 z_bucket_w:0.001 buckets:[1 1 0 1 1] n_buckets:[1 1 0 1 1]}}x100
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eval instant at 5m rate(histogram_rate[45s])
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{} {{schema:1 count:0.6 sum:1.2266666666666652 z_bucket:0.06666666666666667 z_bucket_w:0.001 buckets:[0.06666666666666667 0.06666666666666667 0 0.06666666666666667 0.06666666666666667] n_buckets:[0.06666666666666667 0.06666666666666667 0 0.06666666666666667 0.06666666666666667]}}
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eval range from 5m to 5m30s step 30s rate(histogram_rate[45s])
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{} {{schema:1 count:0.6 sum:1.2266666666666652 z_bucket:0.06666666666666667 z_bucket_w:0.001 buckets:[0.06666666666666667 0.06666666666666667 0 0.06666666666666667 0.06666666666666667] n_buckets:[0.06666666666666667 0.06666666666666667 0 0.06666666666666667 0.06666666666666667]}}x1
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clear
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# Apply count and sum function to histogram.
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load 10m
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histogram_count_sum_2 {{schema:0 count:24 sum:100 z_bucket:4 z_bucket_w:0.001 buckets:[2 3 0 1 4] n_buckets:[2 3 0 1 4]}}x1
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eval instant at 10m histogram_count(histogram_count_sum_2)
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{} 24
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eval instant at 10m histogram_sum(histogram_count_sum_2)
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{} 100
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clear
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# Apply stddev and stdvar function to histogram with {1, 2, 3, 4} (low res).
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load 10m
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histogram_stddev_stdvar_1 {{schema:2 count:4 sum:10 buckets:[1 0 0 0 1 0 0 1 1]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_1)
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{} 1.0787993180043811
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_1)
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{} 1.163807968526718
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clear
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# Apply stddev and stdvar function to histogram with {1, 1, 1, 1} (high res).
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load 10m
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histogram_stddev_stdvar_2 {{schema:8 count:10 sum:10 buckets:[1 2 3 4]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_2)
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{} 0.0048960313898237465
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_2)
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{} 2.3971123370139447e-05
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clear
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# Apply stddev and stdvar function to histogram with {-50, -8, 0, 3, 8, 9}.
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load 10m
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histogram_stddev_stdvar_3 {{schema:3 count:7 sum:62 z_bucket:1 buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] n_buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_3)
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{} 42.947236400258
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_3)
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{} 1844.4651144196398
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clear
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# Apply stddev and stdvar function to histogram with {-100000, -10000, -1000, -888, -888, -100, -50, -9, -8, -3}.
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load 10m
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histogram_stddev_stdvar_4 {{schema:0 count:10 sum:-112946 z_bucket:0 n_buckets:[0 0 1 1 1 0 1 1 0 0 3 0 0 0 1 0 0 1]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_4)
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{} 27556.344499842
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_4)
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{} 759352122.1939945
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clear
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# Apply stddev and stdvar function to histogram with {-10x10}.
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load 10m
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histogram_stddev_stdvar_5 {{schema:0 count:10 sum:-100 z_bucket:0 n_buckets:[0 0 0 0 10]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_5)
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{} 1.3137084989848
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_5)
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{} 1.725830020304794
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clear
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# Apply stddev and stdvar function to histogram with {-50, -8, 0, 3, 8, 9, NaN}.
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load 10m
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histogram_stddev_stdvar_6 {{schema:3 count:7 sum:NaN z_bucket:1 buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] n_buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ]}}x1
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eval instant at 10m histogram_stddev(histogram_stddev_stdvar_6)
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{} NaN
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eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_6)
|
|
{} NaN
|
|
|
|
clear
|
|
|
|
# Apply stddev and stdvar function to histogram with {-50, -8, 0, 3, 8, 9, Inf}.
|
|
load 10m
|
|
histogram_stddev_stdvar_7 {{schema:3 count:7 sum:Inf z_bucket:1 buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ] n_buckets:[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ]}}x1
|
|
|
|
eval instant at 10m histogram_stddev(histogram_stddev_stdvar_7)
|
|
{} Inf
|
|
|
|
eval instant at 10m histogram_stdvar(histogram_stddev_stdvar_7)
|
|
{} Inf
|
|
|
|
clear
|
|
|
|
# Apply quantile function to histogram with all positive buckets with zero bucket.
|
|
load 10m
|
|
histogram_quantile_1 {{schema:0 count:12 sum:100 z_bucket:2 z_bucket_w:0.001 buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval_warn instant at 10m histogram_quantile(1.001, histogram_quantile_1)
|
|
{} Inf
|
|
|
|
eval instant at 10m histogram_quantile(1, histogram_quantile_1)
|
|
{} 16
|
|
|
|
# The following quantiles are within a bucket. Exponential
|
|
# interpolation is applied (rather than linear, as it is done for
|
|
# classic histograms), leading to slightly different quantile values.
|
|
eval instant at 10m histogram_quantile(0.99, histogram_quantile_1)
|
|
{} 15.67072476139083
|
|
|
|
eval instant at 10m histogram_quantile(0.9, histogram_quantile_1)
|
|
{} 12.99603834169977
|
|
|
|
eval instant at 10m histogram_quantile(0.6, histogram_quantile_1)
|
|
{} 4.594793419988138
|
|
|
|
eval instant at 10m histogram_quantile(0.5, histogram_quantile_1)
|
|
{} 1.5874010519681994
|
|
|
|
# Linear interpolation within the zero bucket after all.
|
|
eval instant at 10m histogram_quantile(0.1, histogram_quantile_1)
|
|
{} 0.0006
|
|
|
|
eval instant at 10m histogram_quantile(0, histogram_quantile_1)
|
|
{} 0
|
|
|
|
eval_warn instant at 10m histogram_quantile(-1, histogram_quantile_1)
|
|
{} -Inf
|
|
|
|
clear
|
|
|
|
# Apply quantile function to histogram with all negative buckets with zero bucket.
|
|
load 10m
|
|
histogram_quantile_2 {{schema:0 count:12 sum:100 z_bucket:2 z_bucket_w:0.001 n_buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval_warn instant at 10m histogram_quantile(1.001, histogram_quantile_2)
|
|
{} Inf
|
|
|
|
eval instant at 10m histogram_quantile(1, histogram_quantile_2)
|
|
{} 0
|
|
|
|
# Again, the quantile values here are slightly different from what
|
|
# they would be with linear interpolation. Note that quantiles
|
|
# ending up in the zero bucket are linearly interpolated after all.
|
|
eval instant at 10m histogram_quantile(0.99, histogram_quantile_2)
|
|
{} -0.00006
|
|
|
|
eval instant at 10m histogram_quantile(0.9, histogram_quantile_2)
|
|
{} -0.0006
|
|
|
|
eval instant at 10m histogram_quantile(0.5, histogram_quantile_2)
|
|
{} -1.5874010519681996
|
|
|
|
eval instant at 10m histogram_quantile(0.1, histogram_quantile_2)
|
|
{} -12.996038341699768
|
|
|
|
eval instant at 10m histogram_quantile(0, histogram_quantile_2)
|
|
{} -16
|
|
|
|
eval_warn instant at 10m histogram_quantile(-1, histogram_quantile_2)
|
|
{} -Inf
|
|
|
|
clear
|
|
|
|
# Apply quantile function to histogram with both positive and negative
|
|
# buckets with zero bucket.
|
|
# First positive buckets with exponential interpolation.
|
|
load 10m
|
|
histogram_quantile_3 {{schema:0 count:24 sum:100 z_bucket:4 z_bucket_w:0.001 buckets:[2 3 0 1 4] n_buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval_warn instant at 10m histogram_quantile(1.001, histogram_quantile_3)
|
|
{} Inf
|
|
|
|
eval instant at 10m histogram_quantile(1, histogram_quantile_3)
|
|
{} 16
|
|
|
|
eval instant at 10m histogram_quantile(0.99, histogram_quantile_3)
|
|
{} 15.34822590920423
|
|
|
|
eval instant at 10m histogram_quantile(0.9, histogram_quantile_3)
|
|
{} 10.556063286183155
|
|
|
|
eval instant at 10m histogram_quantile(0.7, histogram_quantile_3)
|
|
{} 1.2030250360821164
|
|
|
|
# Linear interpolation in the zero bucket, symmetrically centered around
|
|
# the zero point.
|
|
eval instant at 10m histogram_quantile(0.55, histogram_quantile_3)
|
|
{} 0.0006
|
|
|
|
eval instant at 10m histogram_quantile(0.5, histogram_quantile_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_quantile(0.45, histogram_quantile_3)
|
|
{} -0.0006
|
|
|
|
# Finally negative buckets with mirrored exponential interpolation.
|
|
eval instant at 10m histogram_quantile(0.3, histogram_quantile_3)
|
|
{} -1.2030250360821169
|
|
|
|
eval instant at 10m histogram_quantile(0.1, histogram_quantile_3)
|
|
{} -10.556063286183155
|
|
|
|
eval instant at 10m histogram_quantile(0.01, histogram_quantile_3)
|
|
{} -15.34822590920423
|
|
|
|
eval instant at 10m histogram_quantile(0, histogram_quantile_3)
|
|
{} -16
|
|
|
|
eval_warn instant at 10m histogram_quantile(-1, histogram_quantile_3)
|
|
{} -Inf
|
|
|
|
clear
|
|
|
|
# Try different schemas. (The interpolation logic must not depend on the schema.)
|
|
clear
|
|
load 1m
|
|
var_res_histogram{schema="-1"} {{schema:-1 sum:6 count:5 buckets:[0 5]}}
|
|
var_res_histogram{schema="0"} {{schema:0 sum:4 count:5 buckets:[0 5]}}
|
|
var_res_histogram{schema="+1"} {{schema:1 sum:4 count:5 buckets:[0 5]}}
|
|
|
|
eval instant at 1m histogram_quantile(0.5, var_res_histogram)
|
|
{schema="-1"} 2.0
|
|
{schema="0"} 1.4142135623730951
|
|
{schema="+1"} 1.189207
|
|
|
|
eval instant at 1m histogram_fraction(0, 2, var_res_histogram{schema="-1"})
|
|
{schema="-1"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(0, 1.4142135623730951, var_res_histogram{schema="0"})
|
|
{schema="0"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(0, 1.189207, var_res_histogram{schema="+1"})
|
|
{schema="+1"} 0.5
|
|
|
|
# The same as above, but one bucket "further to the right".
|
|
clear
|
|
load 1m
|
|
var_res_histogram{schema="-1"} {{schema:-1 sum:6 count:5 buckets:[0 0 5]}}
|
|
var_res_histogram{schema="0"} {{schema:0 sum:4 count:5 buckets:[0 0 5]}}
|
|
var_res_histogram{schema="+1"} {{schema:1 sum:4 count:5 buckets:[0 0 5]}}
|
|
|
|
eval instant at 1m histogram_quantile(0.5, var_res_histogram)
|
|
{schema="-1"} 8.0
|
|
{schema="0"} 2.82842712474619
|
|
{schema="+1"} 1.6817928305074292
|
|
|
|
eval instant at 1m histogram_fraction(0, 8, var_res_histogram{schema="-1"})
|
|
{schema="-1"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(0, 2.82842712474619, var_res_histogram{schema="0"})
|
|
{schema="0"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(0, 1.6817928305074292, var_res_histogram{schema="+1"})
|
|
{schema="+1"} 0.5
|
|
|
|
# And everything again but for negative buckets.
|
|
clear
|
|
load 1m
|
|
var_res_histogram{schema="-1"} {{schema:-1 sum:6 count:5 n_buckets:[0 5]}}
|
|
var_res_histogram{schema="0"} {{schema:0 sum:4 count:5 n_buckets:[0 5]}}
|
|
var_res_histogram{schema="+1"} {{schema:1 sum:4 count:5 n_buckets:[0 5]}}
|
|
|
|
eval instant at 1m histogram_quantile(0.5, var_res_histogram)
|
|
{schema="-1"} -2.0
|
|
{schema="0"} -1.4142135623730951
|
|
{schema="+1"} -1.189207
|
|
|
|
eval instant at 1m histogram_fraction(-2, 0, var_res_histogram{schema="-1"})
|
|
{schema="-1"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(-1.4142135623730951, 0, var_res_histogram{schema="0"})
|
|
{schema="0"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(-1.189207, 0, var_res_histogram{schema="+1"})
|
|
{schema="+1"} 0.5
|
|
|
|
clear
|
|
load 1m
|
|
var_res_histogram{schema="-1"} {{schema:-1 sum:6 count:5 n_buckets:[0 0 5]}}
|
|
var_res_histogram{schema="0"} {{schema:0 sum:4 count:5 n_buckets:[0 0 5]}}
|
|
var_res_histogram{schema="+1"} {{schema:1 sum:4 count:5 n_buckets:[0 0 5]}}
|
|
|
|
eval instant at 1m histogram_quantile(0.5, var_res_histogram)
|
|
{schema="-1"} -8.0
|
|
{schema="0"} -2.82842712474619
|
|
{schema="+1"} -1.6817928305074292
|
|
|
|
eval instant at 1m histogram_fraction(-8, 0, var_res_histogram{schema="-1"})
|
|
{schema="-1"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(-2.82842712474619, 0, var_res_histogram{schema="0"})
|
|
{schema="0"} 0.5
|
|
|
|
eval instant at 1m histogram_fraction(-1.6817928305074292, 0, var_res_histogram{schema="+1"})
|
|
{schema="+1"} 0.5
|
|
|
|
|
|
# Apply fraction function to empty histogram.
|
|
load 10m
|
|
histogram_fraction_1 {{}}x1
|
|
|
|
eval instant at 10m histogram_fraction(3.1415, 42, histogram_fraction_1)
|
|
{} NaN
|
|
|
|
clear
|
|
|
|
# Apply fraction function to histogram with positive and zero buckets.
|
|
load 10m
|
|
histogram_fraction_2 {{schema:0 count:12 sum:100 z_bucket:2 z_bucket_w:0.001 buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval instant at 10m histogram_fraction(0, +Inf, histogram_fraction_2)
|
|
{} 1
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, 0, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-0.001, 0, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0, 0.001, histogram_fraction_2)
|
|
{} 0.16666666666666666
|
|
|
|
# Note that this result and the one above add up to 1.
|
|
eval instant at 10m histogram_fraction(0.001, inf, histogram_fraction_2)
|
|
{} 0.8333333333333334
|
|
|
|
# We are in the zero bucket, resulting in linear interpolation
|
|
eval instant at 10m histogram_fraction(0, 0.0005, histogram_fraction_2)
|
|
{} 0.08333333333333333
|
|
|
|
# Demonstrate that the inverse operation with histogram_quantile yields
|
|
# the original value with the non-trivial result above.
|
|
eval instant at 10m histogram_quantile(0.08333333333333333, histogram_fraction_2)
|
|
{} 0.0005
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -0.001, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(1, 2, histogram_fraction_2)
|
|
{} 0.25
|
|
|
|
# More non-trivial results with interpolation involved below, including
|
|
# some round-trips via histogram_quantile to prove that the inverse
|
|
# operation leads to the same results.
|
|
|
|
eval instant at 10m histogram_fraction(0, 1.5, histogram_fraction_2)
|
|
{} 0.4795739585136224
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 2, histogram_fraction_2)
|
|
{} 0.10375937481971091
|
|
|
|
eval instant at 10m histogram_fraction(1, 8, histogram_fraction_2)
|
|
{} 0.3333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(0, 6, histogram_fraction_2)
|
|
{} 0.6320802083934297
|
|
|
|
eval instant at 10m histogram_quantile(0.6320802083934297, histogram_fraction_2)
|
|
{} 6
|
|
|
|
eval instant at 10m histogram_fraction(1, 6, histogram_fraction_2)
|
|
{} 0.29874687506009634
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 6, histogram_fraction_2)
|
|
{} 0.15250624987980724
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1.5, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-8, -1, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1.5, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(42, 3.1415, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0, 0, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0.000001, 0.000001, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(42, 42, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-3.1, -3.1, histogram_fraction_2)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(3.1415, NaN, histogram_fraction_2)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, 42, histogram_fraction_2)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, NaN, histogram_fraction_2)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, +Inf, histogram_fraction_2)
|
|
{} 1
|
|
|
|
# Apply fraction function to histogram with negative and zero buckets.
|
|
load 10m
|
|
histogram_fraction_3 {{schema:0 count:12 sum:100 z_bucket:2 z_bucket_w:0.001 n_buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval instant at 10m histogram_fraction(0, +Inf, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, 0, histogram_fraction_3)
|
|
{} 1
|
|
|
|
eval instant at 10m histogram_fraction(-0.001, 0, histogram_fraction_3)
|
|
{} 0.16666666666666666
|
|
|
|
eval instant at 10m histogram_fraction(0, 0.001, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-0.0005, 0, histogram_fraction_3)
|
|
{} 0.08333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -0.0005, histogram_fraction_3)
|
|
{} 0.9166666666666666
|
|
|
|
eval instant at 10m histogram_quantile(0.9166666666666666, histogram_fraction_3)
|
|
{} -0.0005
|
|
|
|
eval instant at 10m histogram_fraction(0.001, inf, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -0.001, histogram_fraction_3)
|
|
{} 0.8333333333333334
|
|
|
|
eval instant at 10m histogram_fraction(1, 2, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 2, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(1, 8, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(1, 6, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 6, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1, histogram_fraction_3)
|
|
{} 0.25
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1.5, histogram_fraction_3)
|
|
{} 0.10375937481971091
|
|
|
|
eval instant at 10m histogram_fraction(-8, -1, histogram_fraction_3)
|
|
{} 0.3333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -6, histogram_fraction_3)
|
|
{} 0.36791979160657035
|
|
|
|
eval instant at 10m histogram_quantile(0.36791979160657035, histogram_fraction_3)
|
|
{} -6
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1, histogram_fraction_3)
|
|
{} 0.29874687506009634
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1.5, histogram_fraction_3)
|
|
{} 0.15250624987980724
|
|
|
|
eval instant at 10m histogram_fraction(42, 3.1415, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0, 0, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0.000001, 0.000001, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(42, 42, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-3.1, -3.1, histogram_fraction_3)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(3.1415, NaN, histogram_fraction_3)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, 42, histogram_fraction_3)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, NaN, histogram_fraction_3)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, +Inf, histogram_fraction_3)
|
|
{} 1
|
|
|
|
clear
|
|
|
|
# Apply fraction function to histogram with both positive, negative and zero buckets.
|
|
load 10m
|
|
histogram_fraction_4 {{schema:0 count:24 sum:100 z_bucket:4 z_bucket_w:0.001 buckets:[2 3 0 1 4] n_buckets:[2 3 0 1 4]}}x1
|
|
|
|
eval instant at 10m histogram_fraction(0, +Inf, histogram_fraction_4)
|
|
{} 0.5
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, 0, histogram_fraction_4)
|
|
{} 0.5
|
|
|
|
eval instant at 10m histogram_fraction(-0.001, 0, histogram_fraction_4)
|
|
{} 0.08333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(0, 0.001, histogram_fraction_4)
|
|
{} 0.08333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(-0.0005, 0.0005, histogram_fraction_4)
|
|
{} 0.08333333333333333
|
|
|
|
eval instant at 10m histogram_fraction(-inf, 0.0005, histogram_fraction_4)
|
|
{} 0.5416666666666666
|
|
|
|
eval instant at 10m histogram_quantile(0.5416666666666666, histogram_fraction_4)
|
|
{} 0.0005
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -0.0005, histogram_fraction_4)
|
|
{} 0.4583333333333333
|
|
|
|
eval instant at 10m histogram_quantile(0.4583333333333333, histogram_fraction_4)
|
|
{} -0.0005
|
|
|
|
eval instant at 10m histogram_fraction(0.001, inf, histogram_fraction_4)
|
|
{} 0.4166666666666667
|
|
|
|
eval instant at 10m histogram_fraction(-inf, -0.001, histogram_fraction_4)
|
|
{} 0.4166666666666667
|
|
|
|
eval instant at 10m histogram_fraction(1, 2, histogram_fraction_4)
|
|
{} 0.125
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 2, histogram_fraction_4)
|
|
{} 0.051879687409855414
|
|
|
|
eval instant at 10m histogram_fraction(1, 8, histogram_fraction_4)
|
|
{} 0.16666666666666666
|
|
|
|
eval instant at 10m histogram_fraction(1, 6, histogram_fraction_4)
|
|
{} 0.14937343753004825
|
|
|
|
eval instant at 10m histogram_fraction(1.5, 6, histogram_fraction_4)
|
|
{} 0.07625312493990366
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1, histogram_fraction_4)
|
|
{} 0.125
|
|
|
|
eval instant at 10m histogram_fraction(-2, -1.5, histogram_fraction_4)
|
|
{} 0.051879687409855456
|
|
|
|
eval instant at 10m histogram_fraction(-8, -1, histogram_fraction_4)
|
|
{} 0.16666666666666666
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1, histogram_fraction_4)
|
|
{} 0.14937343753004817
|
|
|
|
eval instant at 10m histogram_fraction(-6, -1.5, histogram_fraction_4)
|
|
{} 0.07625312493990362
|
|
|
|
eval instant at 10m histogram_fraction(42, 3.1415, histogram_fraction_4)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0, 0, histogram_fraction_4)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(0.000001, 0.000001, histogram_fraction_4)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(42, 42, histogram_fraction_4)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(-3.1, -3.1, histogram_fraction_4)
|
|
{} 0
|
|
|
|
eval instant at 10m histogram_fraction(3.1415, NaN, histogram_fraction_4)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, 42, histogram_fraction_4)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(NaN, NaN, histogram_fraction_4)
|
|
{} NaN
|
|
|
|
eval instant at 10m histogram_fraction(-Inf, +Inf, histogram_fraction_4)
|
|
{} 1
|
|
|
|
eval instant at 10m histogram_sum(scalar(histogram_fraction(-Inf, +Inf, sum(histogram_fraction_4))) * histogram_fraction_4)
|
|
{} 100
|
|
|
|
# Apply multiplication and division operator to histogram.
|
|
load 10m
|
|
histogram_mul_div {{schema:0 count:30 sum:33 z_bucket:3 z_bucket_w:0.001 buckets:[3 3 3] n_buckets:[6 6 6]}}x1
|
|
float_series_3 3+0x1
|
|
float_series_0 0+0x1
|
|
|
|
eval instant at 10m histogram_mul_div*3
|
|
{} {{schema:0 count:90 sum:99 z_bucket:9 z_bucket_w:0.001 buckets:[9 9 9] n_buckets:[18 18 18]}}
|
|
|
|
eval instant at 10m histogram_mul_div*-1
|
|
{} {{schema:0 count:-30 sum:-33 z_bucket:-3 z_bucket_w:0.001 buckets:[-3 -3 -3] n_buckets:[-6 -6 -6]}}
|
|
|
|
eval instant at 10m -histogram_mul_div
|
|
{} {{schema:0 count:-30 sum:-33 z_bucket:-3 z_bucket_w:0.001 buckets:[-3 -3 -3] n_buckets:[-6 -6 -6]}}
|
|
|
|
eval instant at 10m histogram_mul_div*-3
|
|
{} {{schema:0 count:-90 sum:-99 z_bucket:-9 z_bucket_w:0.001 buckets:[-9 -9 -9] n_buckets:[-18 -18 -18]}}
|
|
|
|
eval instant at 10m 3*histogram_mul_div
|
|
{} {{schema:0 count:90 sum:99 z_bucket:9 z_bucket_w:0.001 buckets:[9 9 9] n_buckets:[18 18 18]}}
|
|
|
|
eval instant at 10m histogram_mul_div*float_series_3
|
|
{} {{schema:0 count:90 sum:99 z_bucket:9 z_bucket_w:0.001 buckets:[9 9 9] n_buckets:[18 18 18]}}
|
|
|
|
eval instant at 10m float_series_3*histogram_mul_div
|
|
{} {{schema:0 count:90 sum:99 z_bucket:9 z_bucket_w:0.001 buckets:[9 9 9] n_buckets:[18 18 18]}}
|
|
|
|
eval instant at 10m histogram_mul_div/3
|
|
{} {{schema:0 count:10 sum:11 z_bucket:1 z_bucket_w:0.001 buckets:[1 1 1] n_buckets:[2 2 2]}}
|
|
|
|
eval instant at 10m histogram_mul_div/-3
|
|
{} {{schema:0 count:-10 sum:-11 z_bucket:-1 z_bucket_w:0.001 buckets:[-1 -1 -1] n_buckets:[-2 -2 -2]}}
|
|
|
|
eval instant at 10m histogram_mul_div/float_series_3
|
|
{} {{schema:0 count:10 sum:11 z_bucket:1 z_bucket_w:0.001 buckets:[1 1 1] n_buckets:[2 2 2]}}
|
|
|
|
eval instant at 10m histogram_mul_div*0
|
|
{} {{schema:0 count:0 sum:0 z_bucket:0 z_bucket_w:0.001 buckets:[0 0 0] n_buckets:[0 0 0]}}
|
|
|
|
eval instant at 10m 0*histogram_mul_div
|
|
{} {{schema:0 count:0 sum:0 z_bucket:0 z_bucket_w:0.001 buckets:[0 0 0] n_buckets:[0 0 0]}}
|
|
|
|
eval instant at 10m histogram_mul_div*float_series_0
|
|
{} {{schema:0 count:0 sum:0 z_bucket:0 z_bucket_w:0.001 buckets:[0 0 0] n_buckets:[0 0 0]}}
|
|
|
|
eval instant at 10m float_series_0*histogram_mul_div
|
|
{} {{schema:0 count:0 sum:0 z_bucket:0 z_bucket_w:0.001 buckets:[0 0 0] n_buckets:[0 0 0]}}
|
|
|
|
# TODO: (NeerajGartia21) remove all the histogram buckets in case of division with zero. See: https://github.com/prometheus/prometheus/issues/13934
|
|
eval instant at 10m histogram_mul_div/0
|
|
{} {{schema:0 count:Inf sum:Inf z_bucket:Inf z_bucket_w:0.001 buckets:[Inf Inf Inf] n_buckets:[Inf Inf Inf]}}
|
|
|
|
eval instant at 10m histogram_mul_div/float_series_0
|
|
{} {{schema:0 count:Inf sum:Inf z_bucket:Inf z_bucket_w:0.001 buckets:[Inf Inf Inf] n_buckets:[Inf Inf Inf]}}
|
|
|
|
eval instant at 10m histogram_mul_div*0/0
|
|
{} {{schema:0 count:NaN sum:NaN z_bucket:NaN z_bucket_w:0.001 buckets:[NaN NaN NaN] n_buckets:[NaN NaN NaN]}}
|
|
|
|
clear
|
|
|
|
# Counter reset only noticeable in a single bucket.
|
|
load 5m
|
|
reset_in_bucket {{schema:0 count:4 sum:5 buckets:[1 2 1]}} {{schema:0 count:5 sum:6 buckets:[1 1 3]}} {{schema:0 count:6 sum:7 buckets:[1 2 3]}}
|
|
|
|
eval instant at 10m increase(reset_in_bucket[15m])
|
|
{} {{count:9 sum:10.5 buckets:[1.5 3 4.5]}}
|
|
|
|
# The following two test the "fast path" where only sum and count is decoded.
|
|
eval instant at 10m histogram_count(increase(reset_in_bucket[15m]))
|
|
{} 9
|
|
|
|
eval instant at 10m histogram_sum(increase(reset_in_bucket[15m]))
|
|
{} 10.5
|
|
|
|
clear
|
|
|
|
# Test native histograms with custom buckets.
|
|
load 5m
|
|
custom_buckets_histogram {{schema:-53 sum:5 count:4 custom_values:[5 10] buckets:[1 2 1]}}x10
|
|
|
|
eval instant at 5m histogram_fraction(5, 10, custom_buckets_histogram)
|
|
{} 0.5
|
|
|
|
eval instant at 5m histogram_quantile(0.5, custom_buckets_histogram)
|
|
{} 7.5
|
|
|
|
eval instant at 5m sum(custom_buckets_histogram)
|
|
{} {{schema:-53 sum:5 count:4 custom_values:[5 10] buckets:[1 2 1]}}
|
|
|
|
clear
|
|
|
|
# Test 'this native histogram metric is not a gauge' warning for rate
|
|
load 30s
|
|
some_metric {{schema:0 sum:1 count:1 buckets:[1] counter_reset_hint:gauge}} {{schema:0 sum:2 count:2 buckets:[2] counter_reset_hint:gauge}} {{schema:0 sum:3 count:3 buckets:[3] counter_reset_hint:gauge}}
|
|
|
|
# Test the case where we only have two points for rate
|
|
eval_warn instant at 30s rate(some_metric[1m])
|
|
{} {{count:0.03333333333333333 sum:0.03333333333333333 buckets:[0.03333333333333333]}}
|
|
|
|
# Test the case where we have more than two points for rate
|
|
eval_warn instant at 1m rate(some_metric[1m])
|
|
{} {{count:0.03333333333333333 sum:0.03333333333333333 buckets:[0.03333333333333333]}}
|
|
|
|
clear
|
|
|
|
# Test rate() over mixed exponential and custom buckets.
|
|
load 30s
|
|
some_metric {{schema:0 sum:1 count:1 buckets:[1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} {{schema:0 sum:5 count:4 buckets:[1 2 1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
|
|
# Start and end with exponential, with custom in the middle.
|
|
eval_warn instant at 1m rate(some_metric[1m])
|
|
# Should produce no results.
|
|
|
|
# Start and end with custom, with exponential in the middle.
|
|
eval_warn instant at 1m30s rate(some_metric[1m])
|
|
# Should produce no results.
|
|
|
|
# Start with custom, end with exponential.
|
|
eval_warn instant at 1m rate(some_metric[1m])
|
|
# Should produce no results.
|
|
|
|
# Start with exponential, end with custom.
|
|
eval_warn instant at 30s rate(some_metric[1m])
|
|
# Should produce no results.
|
|
|
|
clear
|
|
|
|
# Histogram with constant buckets.
|
|
load 1m
|
|
const_histogram {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}}
|
|
|
|
# There is no change to the bucket count over time, thus rate is 0 in each bucket.
|
|
# However native histograms do not represent empty buckets, so here the zeros are implicit.
|
|
eval instant at 5m rate(const_histogram[5m])
|
|
{} {{schema:0 sum:0 count:0}}
|
|
|
|
# Zero buckets mean no observations, thus the denominator in the average is 0
|
|
# leading to 0/0, which is NaN.
|
|
eval instant at 5m histogram_avg(rate(const_histogram[5m]))
|
|
{} NaN
|
|
|
|
# Zero buckets mean no observations, so count is 0.
|
|
eval instant at 5m histogram_count(rate(const_histogram[5m]))
|
|
{} 0.0
|
|
|
|
# Zero buckets mean no observations and empty histogram has a sum of 0 by definition.
|
|
eval instant at 5m histogram_sum(rate(const_histogram[5m]))
|
|
{} 0.0
|
|
|
|
# Zero buckets mean no observations, thus the denominator in the fraction is 0,
|
|
# leading to 0/0, which is NaN.
|
|
eval instant at 5m histogram_fraction(0.0, 1.0, rate(const_histogram[5m]))
|
|
{} NaN
|
|
|
|
# Workaround to calculate the observation count corresponding to NaN fraction.
|
|
eval instant at 5m histogram_count(rate(const_histogram[5m])) == 0.0 or histogram_fraction(0.0, 1.0, rate(const_histogram[5m])) * histogram_count(rate(const_histogram[5m]))
|
|
{} 0.0
|
|
|
|
# Zero buckets mean no observations, so there is no value that observations fall below,
|
|
# which means that any quantile is a NaN.
|
|
eval instant at 5m histogram_quantile(1.0, rate(const_histogram[5m]))
|
|
{} NaN
|
|
|
|
# Zero buckets mean no observations, so there is no standard deviation.
|
|
eval instant at 5m histogram_stddev(rate(const_histogram[5m]))
|
|
{} NaN
|
|
|
|
# Zero buckets mean no observations, so there is no standard variance.
|
|
eval instant at 5m histogram_stdvar(rate(const_histogram[5m]))
|
|
{} NaN
|
|
|
|
clear
|
|
|
|
# Test mixing exponential and custom buckets.
|
|
load 6m
|
|
metric{series="exponential"} {{sum:4 count:3 buckets:[1 2 1]}} _ {{sum:4 count:3 buckets:[1 2 1]}}
|
|
metric{series="other-exponential"} {{sum:3 count:2 buckets:[1 1 1]}} _ {{sum:3 count:2 buckets:[1 1 1]}}
|
|
metric{series="custom"} _ {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
metric{series="other-custom"} _ {{schema:-53 sum:15 count:2 custom_values:[5 10] buckets:[0 2]}} {{schema:-53 sum:15 count:2 custom_values:[5 10] buckets:[0 2]}}
|
|
|
|
# T=0: only exponential
|
|
# T=6: only custom
|
|
# T=12: mixed, should be ignored and emit a warning
|
|
eval_warn range from 0 to 12m step 6m sum(metric)
|
|
{} {{sum:7 count:5 buckets:[2 3 2]}} {{schema:-53 sum:16 count:3 custom_values:[5 10] buckets:[1 2]}} _
|
|
|
|
eval_warn range from 0 to 12m step 6m avg(metric)
|
|
{} {{sum:3.5 count:2.5 buckets:[1 1.5 1]}} {{schema:-53 sum:8 count:1.5 custom_values:[5 10] buckets:[0.5 1]}} _
|
|
|
|
clear
|
|
|
|
# Test incompatible custom bucket schemas.
|
|
load 6m
|
|
metric{series="1"} _ {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
metric{series="2"} {{schema:-53 sum:1 count:1 custom_values:[2] buckets:[1]}} _ {{schema:-53 sum:1 count:1 custom_values:[2] buckets:[1]}}
|
|
metric{series="3"} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
|
|
# T=0: incompatible, should be ignored and emit a warning
|
|
# T=6: compatible
|
|
# T=12: incompatible followed by compatible, should be ignored and emit a warning
|
|
eval_warn range from 0 to 12m step 6m sum(metric)
|
|
{} _ {{schema:-53 sum:2 count:2 custom_values:[5 10] buckets:[2]}} _
|
|
|
|
eval_warn range from 0 to 12m step 6m avg(metric)
|
|
{} _ {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}} _
|
|
|
|
clear
|
|
|
|
load 1m
|
|
metric{group="just-floats", series="1"} 2
|
|
metric{group="just-floats", series="2"} 3
|
|
metric{group="just-exponential-histograms", series="1"} {{sum:3 count:4 buckets:[1 2 1]}}
|
|
metric{group="just-exponential-histograms", series="2"} {{sum:2 count:3 buckets:[1 1 1]}}
|
|
metric{group="just-custom-histograms", series="1"} {{schema:-53 sum:1 count:1 custom_values:[2] buckets:[1]}}
|
|
metric{group="just-custom-histograms", series="2"} {{schema:-53 sum:3 count:4 custom_values:[2] buckets:[7]}}
|
|
metric{group="floats-and-histograms", series="1"} 2
|
|
metric{group="floats-and-histograms", series="2"} {{sum:2 count:3 buckets:[1 1 1]}}
|
|
metric{group="exponential-and-custom-histograms", series="1"} {{sum:2 count:3 buckets:[1 1 1]}}
|
|
metric{group="exponential-and-custom-histograms", series="2"} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
metric{group="incompatible-custom-histograms", series="1"} {{schema:-53 sum:1 count:1 custom_values:[5 10] buckets:[1]}}
|
|
metric{group="incompatible-custom-histograms", series="2"} {{schema:-53 sum:1 count:1 custom_values:[2] buckets:[1]}}
|
|
|
|
eval_warn instant at 0 sum by (group) (metric)
|
|
{group="just-floats"} 5
|
|
{group="just-exponential-histograms"} {{sum:5 count:7 buckets:[2 3 2]}}
|
|
{group="just-custom-histograms"} {{schema:-53 sum:4 count:5 custom_values:[2] buckets:[8]}}
|
|
|
|
clear
|
|
|
|
# Test native histograms with sum, count, avg.
|
|
load 10m
|
|
histogram_sum{idx="0"} {{schema:0 count:25 sum:1234.5 z_bucket:4 z_bucket_w:0.001 buckets:[1 2 0 1 1] n_buckets:[2 4 0 0 1 9]}}x1
|
|
histogram_sum{idx="1"} {{schema:0 count:41 sum:2345.6 z_bucket:5 z_bucket_w:0.001 buckets:[1 3 1 2 1 1 1] n_buckets:[0 1 4 2 7 0 0 0 0 5 5 2]}}x1
|
|
histogram_sum{idx="2"} {{schema:0 count:41 sum:1111.1 z_bucket:5 z_bucket_w:0.001 buckets:[1 3 1 2 1 1 1] n_buckets:[0 1 4 2 7 0 0 0 0 5 5 2]}}x1
|
|
histogram_sum{idx="3"} {{schema:1 count:0}}x1
|
|
histogram_sum_float{idx="0"} 42.0x1
|
|
|
|
eval instant at 10m sum(histogram_sum)
|
|
{} {{schema:0 count:107 sum:4691.2 z_bucket:14 z_bucket_w:0.001 buckets:[3 8 2 5 3 2 2] n_buckets:[2 6 8 4 15 9 0 0 0 10 10 4]}}
|
|
|
|
eval_warn instant at 10m sum({idx="0"})
|
|
|
|
eval instant at 10m sum(histogram_sum{idx="0"} + ignoring(idx) histogram_sum{idx="1"} + ignoring(idx) histogram_sum{idx="2"} + ignoring(idx) histogram_sum{idx="3"})
|
|
{} {{schema:0 count:107 sum:4691.2 z_bucket:14 z_bucket_w:0.001 buckets:[3 8 2 5 3 2 2] n_buckets:[2 6 8 4 15 9 0 0 0 10 10 4]}}
|
|
|
|
eval instant at 10m count(histogram_sum)
|
|
{} 4
|
|
|
|
eval instant at 10m avg(histogram_sum)
|
|
{} {{schema:0 count:26.75 sum:1172.8 z_bucket:3.5 z_bucket_w:0.001 buckets:[0.75 2 0.5 1.25 0.75 0.5 0.5] n_buckets:[0.5 1.5 2 1 3.75 2.25 0 0 0 2.5 2.5 1]}}
|
|
|
|
clear
|
|
|
|
# Test native histograms with sum_over_time, avg_over_time.
|
|
load 1m
|
|
histogram_sum_over_time {{schema:0 count:25 sum:1234.5 z_bucket:4 z_bucket_w:0.001 buckets:[1 2 0 1 1] n_buckets:[2 4 0 0 1 9]}} {{schema:0 count:41 sum:2345.6 z_bucket:5 z_bucket_w:0.001 buckets:[1 3 1 2 1 1 1] n_buckets:[0 1 4 2 7 0 0 0 0 5 5 2]}} {{schema:0 count:41 sum:1111.1 z_bucket:5 z_bucket_w:0.001 buckets:[1 3 1 2 1 1 1] n_buckets:[0 1 4 2 7 0 0 0 0 5 5 2]}} {{schema:1 count:0}}
|
|
|
|
eval instant at 3m sum_over_time(histogram_sum_over_time[4m:1m])
|
|
{} {{schema:0 count:107 sum:4691.2 z_bucket:14 z_bucket_w:0.001 buckets:[3 8 2 5 3 2 2] n_buckets:[2 6 8 4 15 9 0 0 0 10 10 4]}}
|
|
|
|
eval instant at 3m avg_over_time(histogram_sum_over_time[4m:1m])
|
|
{} {{schema:0 count:26.75 sum:1172.8 z_bucket:3.5 z_bucket_w:0.001 buckets:[0.75 2 0.5 1.25 0.75 0.5 0.5] n_buckets:[0.5 1.5 2 1 3.75 2.25 0 0 0 2.5 2.5 1]}}
|