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"
"github.com/prometheus/prometheus/util/almost"
)
// 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
}
// 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
}
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)
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
}
// 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()
if bucket.Count == 0 {
continue
}
count += bucket.Count
if count >= rank {
break
}
}
if !h.UsesCustomBuckets() && 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
}
} else if h.UsesCustomBuckets() {
if bucket.Lower == math.Inf(-1) {
// first bucket, with lower bound -Inf
if bucket.Upper <= 0 {
return bucket.Upper
}
bucket.Lower = 0
} else if bucket.Upper == math.Inf(1) {
// last bucket, with upper bound +Inf
return bucket.Lower
}
}
// 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]
}
// 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.
//
// 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.
//
// 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 almost.Equal(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
}
prev = curr
}
return forcedMonotonic, fixedPrecision
}
// quantile calculates the given quantile of a vector of samples.
//
// The Vector will be sorted.
// If 'values' has zero elements, NaN is returned.
// If q==NaN, NaN is returned.
// If q<0, -Inf is returned.
// If q>1, +Inf is returned.
func quantile(q float64, values vectorByValueHeap) float64 {
if len(values) == 0 || math.IsNaN(q) {
return math.NaN()
}
if q < 0 {
return math.Inf(-1)
}
if q > 1 {
return math.Inf(+1)
}
sort.Sort(values)
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)
return values[int(lowerIndex)].F*(1-weight) + values[int(upperIndex)].F*weight
}