prometheus/promql/quantile.go

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// Copyright 2015 The Prometheus Authors
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
package promql
import (
"math"
"slices"
"sort"
"github.com/prometheus/prometheus/model/histogram"
"github.com/prometheus/prometheus/model/labels"
)
// smallDeltaTolerance is the threshold for relative deltas between classic
// histogram buckets that will be ignored by the histogram_quantile function
// because they are most likely artifacts of floating point precision issues.
// Testing on 2 sets of real data with bugs arising from small deltas,
// the safe ranges were from:
// - 1e-05 to 1e-15
// - 1e-06 to 1e-15
// Anything to the left of that would cause non-query-sharded data to have
// small deltas ignored (unnecessary and we should avoid this), and anything
// to the right of that would cause query-sharded data to not have its small
// deltas ignored (so the problem won't be fixed).
// For context, query sharding triggers these float precision errors in Mimir.
// To illustrate, with a relative deviation of 1e-12, we need to have 1e12
// observations in the bucket so that the change of one observation is small
// enough to get ignored. With the usual observation rate even of very busy
// services, this will hardly be reached in timeframes that matters for
// monitoring.
const smallDeltaTolerance = 1e-12
// Helpers to calculate quantiles.
// excludedLabels are the labels to exclude from signature calculation for
// quantiles.
var excludedLabels = []string{
labels.MetricName,
labels.BucketLabel,
}
type bucket struct {
upperBound float64
count float64
}
// buckets implements sort.Interface.
type buckets []bucket
type metricWithBuckets struct {
metric labels.Labels
buckets buckets
}
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// bucketQuantile calculates the quantile 'q' based on the given buckets. The
// buckets will be sorted by upperBound by this function (i.e. no sorting
// needed before calling this function). The quantile value is interpolated
// assuming a linear distribution within a bucket. However, if the quantile
// falls into the highest bucket, the upper bound of the 2nd highest bucket is
// returned. A natural lower bound of 0 is assumed if the upper bound of the
// lowest bucket is greater 0. In that case, interpolation in the lowest bucket
// happens linearly between 0 and the upper bound of the lowest bucket.
// However, if the lowest bucket has an upper bound less or equal 0, this upper
// bound is returned if the quantile falls into the lowest bucket.
//
// There are a number of special cases (once we have a way to report errors
// happening during evaluations of AST functions, we should report those
// explicitly):
//
// If 'buckets' has 0 observations, NaN is returned.
//
// If 'buckets' has fewer than 2 elements, NaN is returned.
//
// If the highest bucket is not +Inf, NaN is returned.
//
// If q==NaN, NaN is returned.
//
// If q<0, -Inf is returned.
//
// If q>1, +Inf is returned.
//
// We also return a bool to indicate if monotonicity needed to be forced,
// and another bool to indicate if small differences between buckets (that
// are likely artifacts of floating point precision issues) have been
// ignored.
func bucketQuantile(q float64, buckets buckets) (float64, bool, bool) {
if math.IsNaN(q) {
return math.NaN(), false, false
}
if q < 0 {
return math.Inf(-1), false, false
}
if q > 1 {
return math.Inf(+1), false, false
}
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slices.SortFunc(buckets, func(a, b bucket) int {
// We don't expect the bucket boundary to be a NaN.
if a.upperBound < b.upperBound {
return -1
}
if a.upperBound > b.upperBound {
return +1
}
return 0
})
if !math.IsInf(buckets[len(buckets)-1].upperBound, +1) {
return math.NaN(), false, false
}
buckets = coalesceBuckets(buckets)
forcedMonotonic, fixedPrecision := ensureMonotonicAndIgnoreSmallDeltas(buckets, smallDeltaTolerance)
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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if len(buckets) < 2 {
return math.NaN(), false, false
}
observations := buckets[len(buckets)-1].count
if observations == 0 {
return math.NaN(), false, false
}
rank := q * observations
b := sort.Search(len(buckets)-1, func(i int) bool { return buckets[i].count >= rank })
if b == len(buckets)-1 {
return buckets[len(buckets)-2].upperBound, forcedMonotonic, fixedPrecision
}
if b == 0 && buckets[0].upperBound <= 0 {
return buckets[0].upperBound, forcedMonotonic, fixedPrecision
}
var (
bucketStart float64
bucketEnd = buckets[b].upperBound
count = buckets[b].count
)
if b > 0 {
bucketStart = buckets[b-1].upperBound
count -= buckets[b-1].count
rank -= buckets[b-1].count
}
return bucketStart + (bucketEnd-bucketStart)*(rank/count), forcedMonotonic, fixedPrecision
}
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// histogramQuantile calculates the quantile 'q' based on the given histogram.
//
// The quantile value is interpolated assuming a linear distribution within a
// bucket.
// TODO(beorn7): Find an interpolation method that is a better fit for
// exponential buckets (and think about configurable interpolation).
//
// A natural lower bound of 0 is assumed if the histogram has only positive
// buckets. Likewise, a natural upper bound of 0 is assumed if the histogram has
// only negative buckets.
// TODO(beorn7): Come to terms if we want that.
//
// There are a number of special cases (once we have a way to report errors
// happening during evaluations of AST functions, we should report those
// explicitly):
//
// If the histogram has 0 observations, NaN is returned.
//
// If q<0, -Inf is returned.
//
// If q>1, +Inf is returned.
//
// If q is NaN, NaN is returned.
func histogramQuantile(q float64, h *histogram.FloatHistogram) float64 {
if q < 0 {
return math.Inf(-1)
}
if q > 1 {
return math.Inf(+1)
}
if h.Count == 0 || math.IsNaN(q) {
return math.NaN()
}
var (
bucket histogram.Bucket[float64]
count float64
it histogram.BucketIterator[float64]
rank float64
)
// if there are NaN observations in the histogram (h.Sum is NaN), use the forward iterator
// if the q < 0.5, use the forward iterator
// if the q >= 0.5, use the reverse iterator
if math.IsNaN(h.Sum) || q < 0.5 {
it = h.AllBucketIterator()
rank = q * h.Count
} else {
it = h.AllReverseBucketIterator()
rank = (1 - q) * h.Count
}
for it.Next() {
bucket = it.At()
count += bucket.Count
if count >= rank {
break
}
}
if bucket.Lower < 0 && bucket.Upper > 0 {
switch {
case len(h.NegativeBuckets) == 0 && len(h.PositiveBuckets) > 0:
// The result is in the zero bucket and the histogram has only
// positive buckets. So we consider 0 to be the lower bound.
bucket.Lower = 0
case len(h.PositiveBuckets) == 0 && len(h.NegativeBuckets) > 0:
// The result is in the zero bucket and the histogram has only
// negative buckets. So we consider 0 to be the upper bound.
bucket.Upper = 0
}
}
// Due to numerical inaccuracies, we could end up with a higher count
// than h.Count. Thus, make sure count is never higher than h.Count.
if count > h.Count {
count = h.Count
}
// We could have hit the highest bucket without even reaching the rank
// (this should only happen if the histogram contains observations of
// the value NaN), in which case we simply return the upper limit of the
// highest explicit bucket.
if count < rank {
return bucket.Upper
}
// NaN observations increase h.Count but not the total number of
// observations in the buckets. Therefore, we have to use the forward
// iterator to find percentiles. We recognize histograms containing NaN
// observations by checking if their h.Sum is NaN.
if math.IsNaN(h.Sum) || q < 0.5 {
rank -= count - bucket.Count
} else {
rank = count - rank
}
// TODO(codesome): Use a better estimation than linear.
return bucket.Lower + (bucket.Upper-bucket.Lower)*(rank/bucket.Count)
}
// histogramFraction calculates the fraction of observations between the
// provided lower and upper bounds, based on the provided histogram.
//
// histogramFraction is in a certain way the inverse of histogramQuantile. If
// histogramQuantile(0.9, h) returns 123.4, then histogramFraction(-Inf, 123.4, h)
// returns 0.9.
//
// The same notes (and TODOs) with regard to interpolation and assumptions about
// the zero bucket boundaries apply as for histogramQuantile.
//
// Whether either boundary is inclusive or exclusive doesnt actually matter as
// long as interpolation has to be performed anyway. In the case of a boundary
// coinciding with a bucket boundary, the inclusive or exclusive nature of the
// boundary determines the exact behavior of the threshold. With the current
// implementation, that means that lower is exclusive for positive values and
// inclusive for negative values, while upper is inclusive for positive values
// and exclusive for negative values.
//
// Special cases:
//
// If the histogram has 0 observations, NaN is returned.
//
// Use a lower bound of -Inf to get the fraction of all observations below the
// upper bound.
//
// Use an upper bound of +Inf to get the fraction of all observations above the
// lower bound.
//
// If lower or upper is NaN, NaN is returned.
//
// If lower >= upper and the histogram has at least 1 observation, zero is returned.
func histogramFraction(lower, upper float64, h *histogram.FloatHistogram) float64 {
if h.Count == 0 || math.IsNaN(lower) || math.IsNaN(upper) {
return math.NaN()
}
if lower >= upper {
return 0
}
var (
rank, lowerRank, upperRank float64
lowerSet, upperSet bool
it = h.AllBucketIterator()
)
for it.Next() {
b := it.At()
if b.Lower < 0 && b.Upper > 0 {
switch {
case len(h.NegativeBuckets) == 0 && len(h.PositiveBuckets) > 0:
// This is the zero bucket and the histogram has only
// positive buckets. So we consider 0 to be the lower
// bound.
b.Lower = 0
case len(h.PositiveBuckets) == 0 && len(h.NegativeBuckets) > 0:
// This is in the zero bucket and the histogram has only
// negative buckets. So we consider 0 to be the upper
// bound.
b.Upper = 0
}
}
if !lowerSet && b.Lower >= lower {
lowerRank = rank
lowerSet = true
}
if !upperSet && b.Lower >= upper {
upperRank = rank
upperSet = true
}
if lowerSet && upperSet {
break
}
if !lowerSet && b.Lower < lower && b.Upper > lower {
lowerRank = rank + b.Count*(lower-b.Lower)/(b.Upper-b.Lower)
lowerSet = true
}
if !upperSet && b.Lower < upper && b.Upper > upper {
upperRank = rank + b.Count*(upper-b.Lower)/(b.Upper-b.Lower)
upperSet = true
}
if lowerSet && upperSet {
break
}
rank += b.Count
}
if !lowerSet || lowerRank > h.Count {
lowerRank = h.Count
}
if !upperSet || upperRank > h.Count {
upperRank = h.Count
}
return (upperRank - lowerRank) / h.Count
}
// coalesceBuckets merges buckets with the same upper bound.
//
// The input buckets must be sorted.
func coalesceBuckets(buckets buckets) buckets {
last := buckets[0]
i := 0
for _, b := range buckets[1:] {
if b.upperBound == last.upperBound {
last.count += b.count
} else {
buckets[i] = last
last = b
i++
}
}
buckets[i] = last
return buckets[:i+1]
}
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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// The assumption that bucket counts increase monotonically with increasing
// upperBound may be violated during:
//
// - Circumstances where data is already inconsistent at the target's side.
// - Ingestion via the remote write receiver that Prometheus implements.
// - Optimisation of query execution where precision is sacrificed for other
// benefits, not by Prometheus but by systems built on top of it.
// - Circumstances where floating point precision errors accumulate.
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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//
// 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.
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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//
// bucketQuantile depends on that monotonicity to do a binary search for the
// bucket with the φ-quantile count, so breaking the monotonicity
// guarantee causes bucketQuantile() to return undefined (nonsense) results.
//
// As a somewhat hacky solution, we first silently ignore any numerically
// insignificant (relative delta below the requested tolerance and likely to
// be from floating point precision errors) differences between successive
// buckets regardless of the direction. Then we calculate the "envelope" of
// the histogram buckets, essentially removing any decreases in the count
// between successive buckets.
//
// We return a bool to indicate if this monotonicity was forced or not, and
// another bool to indicate if small deltas were ignored or not.
func ensureMonotonicAndIgnoreSmallDeltas(buckets buckets, tolerance float64) (bool, bool) {
var forcedMonotonic, fixedPrecision bool
prev := buckets[0].count
for i := 1; i < len(buckets); i++ {
curr := buckets[i].count // Assumed always positive.
if curr == prev {
// No correction needed if the counts are identical between buckets.
continue
}
if almostEqual(prev, curr, tolerance) {
// Silently correct numerically insignificant differences from floating
// point precision errors, regardless of direction.
// Do not update the 'prev' value as we are ignoring the difference.
buckets[i].count = prev
fixedPrecision = true
continue
}
if curr < prev {
// Force monotonicity by removing any decreases regardless of magnitude.
// Do not update the 'prev' value as we are ignoring the decrease.
buckets[i].count = prev
forcedMonotonic = true
continue
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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}
prev = curr
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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}
return forcedMonotonic, fixedPrecision
Force buckets in a histogram to be monotonic for quantile estimation (#2610) * 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.
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}
// quantile calculates the given quantile of a vector of samples.
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//
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// The Vector will be sorted.
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// If 'values' has zero elements, NaN is returned.
// If q==NaN, NaN is returned.
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// If q<0, -Inf is returned.
// If q>1, +Inf is returned.
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func quantile(q float64, values vectorByValueHeap) float64 {
if len(values) == 0 || math.IsNaN(q) {
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return math.NaN()
}
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if q < 0 {
return math.Inf(-1)
}
if q > 1 {
return math.Inf(+1)
}
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sort.Sort(values)
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n := float64(len(values))
// When the quantile lies between two samples,
// we use a weighted average of the two samples.
rank := q * (n - 1)
lowerIndex := math.Max(0, math.Floor(rank))
upperIndex := math.Min(n-1, lowerIndex+1)
weight := rank - math.Floor(rank)
promql: Separate `Point` into `FPoint` and `HPoint` In other words: Instead of having a “polymorphous” `Point` that can either contain a float value or a histogram value, use an `FPoint` for floats and an `HPoint` for histograms. This seemingly small change has a _lot_ of repercussions throughout the codebase. The idea here is to avoid the increase in size of `Point` arrays that happened after native histograms had been added. The higher-level data structures (`Sample`, `Series`, etc.) are still “polymorphous”. The same idea could be applied to them, but at each step the trade-offs needed to be evaluated. The idea with this change is to do the minimum necessary to get back to pre-histogram performance for functions that do not touch histograms. Here are comparisons for the `changes` function. The test data doesn't include histograms yet. Ideally, there would be no change in the benchmark result at all. First runtime v2.39 compared to directly prior to this commit: ``` name old time/op new time/op delta RangeQuery/expr=changes(a_one[1d]),steps=1-16 391µs ± 2% 542µs ± 1% +38.58% (p=0.000 n=9+8) RangeQuery/expr=changes(a_one[1d]),steps=10-16 452µs ± 2% 617µs ± 2% +36.48% (p=0.000 n=10+10) RangeQuery/expr=changes(a_one[1d]),steps=100-16 1.12ms ± 1% 1.36ms ± 2% +21.58% (p=0.000 n=8+10) RangeQuery/expr=changes(a_one[1d]),steps=1000-16 7.83ms ± 1% 8.94ms ± 1% +14.21% (p=0.000 n=10+10) RangeQuery/expr=changes(a_ten[1d]),steps=1-16 2.98ms ± 0% 3.30ms ± 1% +10.67% (p=0.000 n=9+10) RangeQuery/expr=changes(a_ten[1d]),steps=10-16 3.66ms ± 1% 4.10ms ± 1% +11.82% (p=0.000 n=10+10) RangeQuery/expr=changes(a_ten[1d]),steps=100-16 10.5ms ± 0% 11.8ms ± 1% +12.50% (p=0.000 n=8+10) RangeQuery/expr=changes(a_ten[1d]),steps=1000-16 77.6ms ± 1% 87.4ms ± 1% +12.63% (p=0.000 n=9+9) RangeQuery/expr=changes(a_hundred[1d]),steps=1-16 30.4ms ± 2% 32.8ms ± 1% +8.01% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=10-16 37.1ms ± 2% 40.6ms ± 2% +9.64% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=100-16 105ms ± 1% 117ms ± 1% +11.69% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=1000-16 783ms ± 3% 876ms ± 1% +11.83% (p=0.000 n=9+10) ``` And then runtime v2.39 compared to after this commit: ``` name old time/op new time/op delta RangeQuery/expr=changes(a_one[1d]),steps=1-16 391µs ± 2% 547µs ± 1% +39.84% (p=0.000 n=9+8) RangeQuery/expr=changes(a_one[1d]),steps=10-16 452µs ± 2% 616µs ± 2% +36.15% (p=0.000 n=10+10) RangeQuery/expr=changes(a_one[1d]),steps=100-16 1.12ms ± 1% 1.26ms ± 1% +12.20% (p=0.000 n=8+10) RangeQuery/expr=changes(a_one[1d]),steps=1000-16 7.83ms ± 1% 7.95ms ± 1% +1.59% (p=0.000 n=10+8) RangeQuery/expr=changes(a_ten[1d]),steps=1-16 2.98ms ± 0% 3.38ms ± 2% +13.49% (p=0.000 n=9+10) RangeQuery/expr=changes(a_ten[1d]),steps=10-16 3.66ms ± 1% 4.02ms ± 1% +9.80% (p=0.000 n=10+9) RangeQuery/expr=changes(a_ten[1d]),steps=100-16 10.5ms ± 0% 10.8ms ± 1% +3.08% (p=0.000 n=8+10) RangeQuery/expr=changes(a_ten[1d]),steps=1000-16 77.6ms ± 1% 78.1ms ± 1% +0.58% (p=0.035 n=9+10) RangeQuery/expr=changes(a_hundred[1d]),steps=1-16 30.4ms ± 2% 33.5ms ± 4% +10.18% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=10-16 37.1ms ± 2% 40.0ms ± 1% +7.98% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=100-16 105ms ± 1% 107ms ± 1% +1.92% (p=0.000 n=10+10) RangeQuery/expr=changes(a_hundred[1d]),steps=1000-16 783ms ± 3% 775ms ± 1% -1.02% (p=0.019 n=9+9) ``` In summary, the runtime doesn't really improve with this change for queries with just a few steps. For queries with many steps, this commit essentially reinstates the old performance. This is good because the many-step queries are the one that matter most (longest absolute runtime). In terms of allocations, though, this commit doesn't make a dent at all (numbers not shown). The reason is that most of the allocations happen in the sampleRingIterator (in the storage package), which has to be addressed in a separate commit. Signed-off-by: beorn7 <beorn@grafana.com>
2022-10-28 07:58:40 -07:00
return values[int(lowerIndex)].F*(1-weight) + values[int(upperIndex)].F*weight
2016-07-08 05:33:20 -07:00
}