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106e20cde5
For conventional histograms, we need to gather all the individual bucket timeseries at a data point to do the quantile calculation. The code so far mirrored this behavior for the new native histograms. However, since a single data point contains all the buckets alreade, that's actually not needed. This PR simplifies the code while still detecting a mix of conventional and native histograms. The weird signature calculation for the conventional histograms is getting even weirder because of that. If this PR turns out to do the right thing, I will implement a proper fix for the signature calculation upstream. Signed-off-by: beorn7 <beorn@grafana.com>
283 lines
8.6 KiB
Go
283 lines
8.6 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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"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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func (b buckets) Len() int { return len(b) }
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func (b buckets) Swap(i, j int) { b[i], b[j] = b[j], b[i] }
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func (b buckets) Less(i, j int) bool { return b[i].upperBound < b[j].upperBound }
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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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sort.Sort(buckets)
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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 no negative buckets.
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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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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 {
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return math.NaN()
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}
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var (
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bucket histogram.FloatBucket
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count float64
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it = h.AllBucketIterator()
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rank = 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 && len(h.NegativeBuckets) == 0 {
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// The result is in the zero bucket and the histogram has no
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// negative buckets. So we consider 0 to be the lower bound.
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bucket.Lower = 0
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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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// (observations not counted in any bucket are considered "overflow"
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// observations above the highest bucket), in which case we simple
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// return the upper limit of the 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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rank -= count - bucket.Count
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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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// 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)].V*(1-weight) + values[int(upperIndex)].V*weight
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}
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