This attempts to close#3973.
Handles cases where the length of the input vector to an aggregate topk
/ bottomk function is less than the K paramater. The change updates
Prometheus to allocate a result vector the same length as the input
vector in these cases.
Previously Prometheus would out-of-memory panic for large K values. This
change makes that unlikely unless the size of the input vector is
equally large.
Signed-off-by: David King <dave@davbo.org>
This means that if there is no stale marker, only the usual staleness
delta (5m) applies.
It has occured to me that there is an oddity in the heurestic. It works
fine as long as you have 2 points within the last 5m, but breaks down
when the time window advances to the point where you have just 1 point.
Consider you had points at t=0 and t=10. With the heurestic it goes stale
at t=51, up until t=300. However from t=301 until t=310 we only
see the t=10 point and the series comes back to life. That is not
desirable.
I don't see a way to keep this form of heurestic working given this
issue, so thus I'm removing it.
To cover the cases where stale markers may not be available,
we need to infer the interval and mark series stale based on that.
As we're lacking stale markers this is less accurate, however
it should be good enough for these cases.
We need 4 intervals as if say we had data at t=0 and t=10,
coming via federation. The next data point should be at t=20 however it
could take up to t=30 for it actually to be ingested, t=40 for it to be
scraped via federation and t=50 for it to be ingested.
We then add 10% on to that for slack, as we do elsewhere.
For instant vectors, if "stale" is the newest sample
ignore the timeseries.
For range vectors, filter out "stale" samples.
Make it possible to inject "stale" samples in promql tests.
Make the timestamp of instant vectors be the timestamp of the sample
rather than the evaluation. We were not using this anywhere, so this is
safe.
Add a function to return the timestamp of samples in an instant vector.
Fixes#1557
* Force buckets in a histogram to be monotonic for quantile estimation
The assumption that bucket counts increase monotonically with increasing
upperBound may be violated during:
* Recording rule evaluation of histogram_quantile, especially when rate()
has been applied to the underlying bucket timeseries.
* Evaluation of histogram_quantile computed over federated bucket
timeseries, especially when rate() has been applied
This is because scraped data is not made available to RR evalution or
federation atomically, so some buckets are computed with data from the N
most recent scrapes, but the other buckets are missing the most recent
observations.
Monotonicity is usually guaranteed because if a bucket with upper bound
u1 has count c1, then any bucket with a higher upper bound u > u1 must
have counted all c1 observations and perhaps more, so that c >= c1.
Randomly interspersed partial sampling breaks that guarantee, and rate()
exacerbates it. Specifically, suppose bucket le=1000 has a count of 10 from
4 samples but the bucket with le=2000 has a count of 7, from 3 samples. The
monotonicity is broken. It is exacerbated by rate() because under normal
operation, cumulative counting of buckets will cause the bucket counts to
diverge such that small differences from missing samples are not a problem.
rate() removes this divergence.)
bucketQuantile depends on that monotonicity to do a binary search for the
bucket with the qth percentile count, so breaking the monotonicity
guarantee causes bucketQuantile() to return undefined (nonsense) results.
As a somewhat hacky solution until the Prometheus project is ready to
accept the changes required to make scrapes atomic, we calculate the
"envelope" of the histogram buckets, essentially removing any decreases
in the count between successive buckets.
* Fix up comment docs for ensureMonotonic
* ensureMonotonic: Use switch statement
Use switch statement rather than if/else for better readability.
Process the most frequent cases first.
This offers new semantics in allowing on() for matching
two single-element vectors with no known common labels.
Previosuly this was often done using on(dummy).
This also allows making it explict that you meant
to do an aggregation without labels via by().
Fixes#1597.
If the label doesn't exist on the one side, it's not copied.
All labels on the many inside are included, this is a breaking change
but likely low impact.
The labels listed in the group_ modifier will be copied from the one
side to the many side. It will be valid to specify no labels.
This is intended to replace the existing ON/GROUP_* support.,
The `unless` set operator can be used to return all vector elements from
the LHS which do not match the elements on the RHS. A use case is to
return all metrics for nodes which do not have a specific role:
node_load1 unless on(instance) chef_role{role="app"}
Fixes https://github.com/prometheus/prometheus/issues/1401
This remove the last (and in fact bogus) use of BoundaryValues.
Thus, a whole lot of unused (and arguably sub-optimal / ugly) code can
be removed here, too.
This has the advantage that the user doesn't need
to list all labels they want to keep (as with "by")
but without having to worry about inconsistent labels
as when there's only one time series (as with "keeping_common").
Almost all aggregation should use this rather than the existing
two options as it's much less error prone and easier to maintain
due to not having to always add in "job" plus whatever other common
job-level labels you have like "region".
The new implementation detects the start and end of a series by
looking at the average sample interval within the range. If the first
(last) sample in the range is more than 1.1*interval distant from the
beginning (end) of the range, it is considered the first (last) sample
of the series as a whole, and extrapolation is limited to half the
interval (rather than all the way to the beginning (end) of the
range). In addition, if the extrapolated starting point of a counter
(where it is zero) is within the range, it is used as the starting
point of the series.
Fixes#581
This change is breaking, use the 'bool' modifier for such comprisons.
After this change all comparisons without 'bool' will filter, and all
comparisons with 'bool' will return 0/1. This makes the language more
consistent and orthogonal, and ultimately easier to learn and use.
If we ever figure out sane semantics for filtering scalar/scalar
comparisons we can add them in, which will most likely come out of how
the new vector() function is used.
This change is breaking, use increase() instead.
I'm not cleaning up the function in this PR, as my solution to #581 will
rewrite and simplify increase/rate/delta.
irate is a rate function that only looks at the most
recent two data points, and calucaltes a per-second value
from that. This produces much more granular graphs for
fast moving data, and works sanely across many scrape intervals.
It doesn't do so well for slowly moving data.
Currently the only way to convert a scalar to a vector is to
use absent(), which isn't very clean. This adds a vector()
function that's the inverse of scalar() and lets your optionally
set labels.
Example usage would be
vector(time() % 86400) < 3600
to filter to only the first hour of the day.
When doing comparison operations on vectors, filtering
sometimes gets in the way and you have to go to a fair bit of
effort to workaround it in order to always return a result.
The 'bool' modifier instead of filtering returns 0/1 depending
on the result of the compairson.
This is also a prerequisite to removing plain scalar/scalar comparisons,
as it maintains the current behaviour under a new syntax.
`keep_common` is more in line with the function name
`drop_common_labels()` terminology-wise, and also more in line with
`group_left`/`group_right` (no `...ing` verb suffix).
We could also go the full way and call it `keep_common_labels`. That
would have the benefit of being even more consistent with the function
`drop_common_labels()` and would be more explanatory, but it also seems
quite long.
This calculates how much a counter increases over
a given period of time, which is the area under the curve
of it's rate.
increase(x[5m]) is equivilent to rate(x[5m]) * 300.
