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162612ea86
Oversight during review of #12525. Signed-off-by: beorn7 <beorn@grafana.com>
410 lines
12 KiB
Go
410 lines
12 KiB
Go
// Copyright 2015 The Prometheus Authors
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// Licensed under the Apache License, Version 2.0 (the "License");
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// you may not use this file except in compliance with the License.
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// You may obtain a copy of the License at
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//
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// http://www.apache.org/licenses/LICENSE-2.0
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//
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// Unless required by applicable law or agreed to in writing, software
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// distributed under the License is distributed on an "AS IS" BASIS,
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// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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// See the License for the specific language governing permissions and
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// limitations under the License.
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package promql
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import (
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"math"
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"sort"
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"golang.org/x/exp/slices"
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"github.com/prometheus/prometheus/model/histogram"
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"github.com/prometheus/prometheus/model/labels"
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)
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// Helpers to calculate quantiles.
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// excludedLabels are the labels to exclude from signature calculation for
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// quantiles.
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var excludedLabels = []string{
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labels.MetricName,
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labels.BucketLabel,
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}
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type bucket struct {
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upperBound float64
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count float64
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}
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// buckets implements sort.Interface.
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type buckets []bucket
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type metricWithBuckets struct {
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metric labels.Labels
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buckets buckets
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}
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// bucketQuantile calculates the quantile 'q' based on the given buckets. The
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// buckets will be sorted by upperBound by this function (i.e. no sorting
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// needed before calling this function). The quantile value is interpolated
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// assuming a linear distribution within a bucket. However, if the quantile
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// falls into the highest bucket, the upper bound of the 2nd highest bucket is
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// returned. A natural lower bound of 0 is assumed if the upper bound of the
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// lowest bucket is greater 0. In that case, interpolation in the lowest bucket
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// happens linearly between 0 and the upper bound of the lowest bucket.
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// However, if the lowest bucket has an upper bound less or equal 0, this upper
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// bound is returned if the quantile falls into the lowest bucket.
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//
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// There are a number of special cases (once we have a way to report errors
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// happening during evaluations of AST functions, we should report those
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// explicitly):
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//
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// If 'buckets' has 0 observations, NaN is returned.
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//
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// If 'buckets' has fewer than 2 elements, NaN is returned.
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//
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// If the highest bucket is not +Inf, NaN is returned.
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//
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// If q==NaN, NaN is returned.
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//
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// If q<0, -Inf is returned.
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//
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// If q>1, +Inf is returned.
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func bucketQuantile(q float64, buckets buckets) float64 {
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if math.IsNaN(q) {
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return math.NaN()
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}
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if q < 0 {
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return math.Inf(-1)
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}
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if q > 1 {
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return math.Inf(+1)
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}
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slices.SortFunc(buckets, func(a, b bucket) bool {
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return a.upperBound < b.upperBound
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})
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if !math.IsInf(buckets[len(buckets)-1].upperBound, +1) {
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return math.NaN()
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}
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buckets = coalesceBuckets(buckets)
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ensureMonotonic(buckets)
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if len(buckets) < 2 {
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return math.NaN()
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}
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observations := buckets[len(buckets)-1].count
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if observations == 0 {
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return math.NaN()
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}
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rank := q * observations
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b := sort.Search(len(buckets)-1, func(i int) bool { return buckets[i].count >= rank })
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if b == len(buckets)-1 {
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return buckets[len(buckets)-2].upperBound
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}
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if b == 0 && buckets[0].upperBound <= 0 {
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return buckets[0].upperBound
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}
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var (
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bucketStart float64
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bucketEnd = buckets[b].upperBound
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count = buckets[b].count
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)
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if b > 0 {
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bucketStart = buckets[b-1].upperBound
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count -= buckets[b-1].count
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rank -= buckets[b-1].count
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}
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return bucketStart + (bucketEnd-bucketStart)*(rank/count)
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}
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// histogramQuantile calculates the quantile 'q' based on the given histogram.
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//
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// The quantile value is interpolated assuming a linear distribution within a
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// bucket.
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// TODO(beorn7): Find an interpolation method that is a better fit for
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// exponential buckets (and think about configurable interpolation).
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//
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// A natural lower bound of 0 is assumed if the histogram has only positive
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// buckets. Likewise, a natural upper bound of 0 is assumed if the histogram has
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// only negative buckets.
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// TODO(beorn7): Come to terms if we want that.
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//
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// There are a number of special cases (once we have a way to report errors
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// happening during evaluations of AST functions, we should report those
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// explicitly):
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//
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// If the histogram has 0 observations, NaN is returned.
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//
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// If q<0, -Inf is returned.
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//
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// If q>1, +Inf is returned.
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//
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// If q is NaN, NaN is returned.
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func histogramQuantile(q float64, h *histogram.FloatHistogram) float64 {
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if q < 0 {
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return math.Inf(-1)
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}
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if q > 1 {
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return math.Inf(+1)
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}
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if h.Count == 0 || math.IsNaN(q) {
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return math.NaN()
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}
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var (
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bucket histogram.Bucket[float64]
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count float64
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it histogram.BucketIterator[float64]
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rank float64
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)
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// if there are NaN observations in the histogram (h.Sum is NaN), use the forward iterator
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// if the q < 0.5, use the forward iterator
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// if the q >= 0.5, use the reverse iterator
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if math.IsNaN(h.Sum) || q < 0.5 {
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it = h.AllBucketIterator()
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rank = q * h.Count
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} else {
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it = h.AllReverseBucketIterator()
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rank = (1 - q) * h.Count
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}
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for it.Next() {
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bucket = it.At()
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count += bucket.Count
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if count >= rank {
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break
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}
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}
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if bucket.Lower < 0 && bucket.Upper > 0 {
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switch {
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case len(h.NegativeBuckets) == 0 && len(h.PositiveBuckets) > 0:
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// The result is in the zero bucket and the histogram has only
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// positive buckets. So we consider 0 to be the lower bound.
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bucket.Lower = 0
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case len(h.PositiveBuckets) == 0 && len(h.NegativeBuckets) > 0:
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// The result is in the zero bucket and the histogram has only
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// negative buckets. So we consider 0 to be the upper bound.
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bucket.Upper = 0
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}
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}
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// Due to numerical inaccuracies, we could end up with a higher count
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// than h.Count. Thus, make sure count is never higher than h.Count.
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if count > h.Count {
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count = h.Count
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}
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// We could have hit the highest bucket without even reaching the rank
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// (this should only happen if the histogram contains observations of
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// the value NaN), in which case we simply return the upper limit of the
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// highest explicit bucket.
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if count < rank {
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return bucket.Upper
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}
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// NaN observations increase h.Count but not the total number of
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// observations in the buckets. Therefore, we have to use the forward
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// iterator to find percentiles. We recognize histograms containing NaN
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// observations by checking if their h.Sum is NaN.
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if math.IsNaN(h.Sum) || q < 0.5 {
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rank -= count - bucket.Count
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} else {
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rank = count - rank
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}
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// TODO(codesome): Use a better estimation than linear.
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return bucket.Lower + (bucket.Upper-bucket.Lower)*(rank/bucket.Count)
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}
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// histogramFraction calculates the fraction of observations between the
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// provided lower and upper bounds, based on the provided histogram.
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//
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// histogramFraction is in a certain way the inverse of histogramQuantile. If
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// histogramQuantile(0.9, h) returns 123.4, then histogramFraction(-Inf, 123.4, h)
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// returns 0.9.
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//
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// The same notes (and TODOs) with regard to interpolation and assumptions about
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// the zero bucket boundaries apply as for histogramQuantile.
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//
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// Whether either boundary is inclusive or exclusive doesn’t actually matter as
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// long as interpolation has to be performed anyway. In the case of a boundary
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// coinciding with a bucket boundary, the inclusive or exclusive nature of the
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// boundary determines the exact behavior of the threshold. With the current
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// implementation, that means that lower is exclusive for positive values and
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// inclusive for negative values, while upper is inclusive for positive values
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// and exclusive for negative values.
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//
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// Special cases:
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//
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// If the histogram has 0 observations, NaN is returned.
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//
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// Use a lower bound of -Inf to get the fraction of all observations below the
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// upper bound.
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//
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// Use an upper bound of +Inf to get the fraction of all observations above the
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// lower bound.
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//
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// If lower or upper is NaN, NaN is returned.
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//
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// If lower >= upper and the histogram has at least 1 observation, zero is returned.
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func histogramFraction(lower, upper float64, h *histogram.FloatHistogram) float64 {
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if h.Count == 0 || math.IsNaN(lower) || math.IsNaN(upper) {
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return math.NaN()
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}
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if lower >= upper {
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return 0
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}
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var (
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rank, lowerRank, upperRank float64
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lowerSet, upperSet bool
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it = h.AllBucketIterator()
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)
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for it.Next() {
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b := it.At()
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if b.Lower < 0 && b.Upper > 0 {
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switch {
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case len(h.NegativeBuckets) == 0 && len(h.PositiveBuckets) > 0:
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// This is the zero bucket and the histogram has only
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// positive buckets. So we consider 0 to be the lower
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// bound.
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b.Lower = 0
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case len(h.PositiveBuckets) == 0 && len(h.NegativeBuckets) > 0:
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// This is in the zero bucket and the histogram has only
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// negative buckets. So we consider 0 to be the upper
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// bound.
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b.Upper = 0
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}
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}
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if !lowerSet && b.Lower >= lower {
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lowerRank = rank
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lowerSet = true
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}
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if !upperSet && b.Lower >= upper {
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upperRank = rank
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upperSet = true
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}
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if lowerSet && upperSet {
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break
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}
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if !lowerSet && b.Lower < lower && b.Upper > lower {
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lowerRank = rank + b.Count*(lower-b.Lower)/(b.Upper-b.Lower)
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lowerSet = true
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}
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if !upperSet && b.Lower < upper && b.Upper > upper {
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upperRank = rank + b.Count*(upper-b.Lower)/(b.Upper-b.Lower)
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upperSet = true
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}
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if lowerSet && upperSet {
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break
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}
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rank += b.Count
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}
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if !lowerSet || lowerRank > h.Count {
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lowerRank = h.Count
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}
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if !upperSet || upperRank > h.Count {
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upperRank = h.Count
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}
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return (upperRank - lowerRank) / h.Count
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}
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// coalesceBuckets merges buckets with the same upper bound.
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//
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// The input buckets must be sorted.
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func coalesceBuckets(buckets buckets) buckets {
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last := buckets[0]
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i := 0
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for _, b := range buckets[1:] {
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if b.upperBound == last.upperBound {
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last.count += b.count
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} else {
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buckets[i] = last
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last = b
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i++
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}
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}
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buckets[i] = last
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return buckets[:i+1]
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}
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// The assumption that bucket counts increase monotonically with increasing
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// upperBound may be violated during:
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//
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// * Recording rule evaluation of histogram_quantile, especially when rate()
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// has been applied to the underlying bucket timeseries.
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// * Evaluation of histogram_quantile computed over federated bucket
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// timeseries, especially when rate() has been applied.
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//
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// This is because scraped data is not made available to rule evaluation or
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// federation atomically, so some buckets are computed with data from the
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// most recent scrapes, but the other buckets are missing data from the most
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// recent scrape.
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//
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// Monotonicity is usually guaranteed because if a bucket with upper bound
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// u1 has count c1, then any bucket with a higher upper bound u > u1 must
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// have counted all c1 observations and perhaps more, so that c >= c1.
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//
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// Randomly interspersed partial sampling breaks that guarantee, and rate()
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// exacerbates it. Specifically, suppose bucket le=1000 has a count of 10 from
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// 4 samples but the bucket with le=2000 has a count of 7 from 3 samples. The
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// monotonicity is broken. It is exacerbated by rate() because under normal
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// operation, cumulative counting of buckets will cause the bucket counts to
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// diverge such that small differences from missing samples are not a problem.
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// rate() removes this divergence.)
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//
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// bucketQuantile depends on that monotonicity to do a binary search for the
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// bucket with the φ-quantile count, so breaking the monotonicity
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// guarantee causes bucketQuantile() to return undefined (nonsense) results.
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//
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// As a somewhat hacky solution until ingestion is atomic per scrape, we
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// calculate the "envelope" of the histogram buckets, essentially removing
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// any decreases in the count between successive buckets.
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func ensureMonotonic(buckets buckets) {
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max := buckets[0].count
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for i := 1; i < len(buckets); i++ {
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switch {
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case buckets[i].count > max:
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max = buckets[i].count
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case buckets[i].count < max:
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buckets[i].count = max
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}
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}
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}
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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.
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// If q==NaN, NaN is returned.
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// If q<0, -Inf is returned.
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// If q>1, +Inf is returned.
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func quantile(q float64, values vectorByValueHeap) float64 {
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if len(values) == 0 || math.IsNaN(q) {
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return math.NaN()
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}
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if q < 0 {
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return math.Inf(-1)
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}
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if q > 1 {
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return math.Inf(+1)
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}
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sort.Sort(values)
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n := float64(len(values))
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// When the quantile lies between two samples,
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// we use a weighted average of the two samples.
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rank := q * (n - 1)
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lowerIndex := math.Max(0, math.Floor(rank))
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upperIndex := math.Min(n-1, lowerIndex+1)
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weight := rank - math.Floor(rank)
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return values[int(lowerIndex)].F*(1-weight) + values[int(upperIndex)].F*weight
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}
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