// 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, } // Bucket represents a bucket of a classic histogram. It is used internally by the promql // package, but it is nevertheless exported for potential use in other PromQL engines. 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. // // Generically speaking, BucketQuantile is for calculating the // histogram_quantile() of classic histograms. See also: HistogramQuantile // for native histograms. // // BucketQuantile is exported as a useful quantile function over a set of // given buckets. It may be used by other PromQL engine implementations. 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. // // For custom buckets, the result is interpolated linearly, i.e. it is assumed // the observations are uniformly distributed within each bucket. (This is a // quite blunt assumption, but it is consistent with the interpolation method // used for classic histograms so far.) // // For exponential buckets, the interpolation is done under the assumption that // the samples within each bucket are distributed in a way that they would // uniformly populate the buckets in a hypothetical histogram with higher // resolution. For example, if the rank calculation suggests that the requested // quantile is right in the middle of the population of the (1,2] bucket, we // assume the quantile would be right at the bucket boundary between the two // buckets the (1,2] bucket would be divided into if the histogram had double // the resolution, which is 2**2**-1 = 1.4142... We call this exponential // interpolation. // // However, for a quantile that ends up in the zero bucket, this method isn't // very helpful (because there is an infinite number of buckets close to zero, // so we would have to assume zero as the result). Therefore, we return to // linear interpolation in the zero bucket. // // 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. // // There are a number of special cases: // // 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. // // HistogramQuantile is for calculating the histogram_quantile() of native // histograms. See also: BucketQuantile for classic histograms. // // HistogramQuantile is exported as it may be used by other PromQL engine // implementations. 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 q < 0.5, use the forward iterator. // If 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 } // The fraction of how far we are into the current bucket. fraction := rank / bucket.Count // Return linear interpolation for custom buckets and for quantiles that // end up in the zero bucket. if h.UsesCustomBuckets() || (bucket.Lower <= 0 && bucket.Upper >= 0) { return bucket.Lower + (bucket.Upper-bucket.Lower)*fraction } // For exponential buckets, we interpolate on a logarithmic scale. On a // logarithmic scale, the exponential bucket boundaries (for any schema) // become linear (every bucket has the same width). Therefore, after // taking the logarithm of both bucket boundaries, we can use the // calculated fraction in the same way as for linear interpolation (see // above). Finally, we return to the normal scale by applying the // exponential function to the result. logLower := math.Log2(math.Abs(bucket.Lower)) logUpper := math.Log2(math.Abs(bucket.Upper)) if bucket.Lower > 0 { // Positive bucket. return math.Exp2(logLower + (logUpper-logLower)*fraction) } // Otherwise, we are in a negative bucket and have to mirror things. return -math.Exp2(logUpper + (logLower-logUpper)*(1-fraction)) } // 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 with regard to interpolation and assumptions about the zero // bucket boundaries apply as for histogramQuantile. // // Whether either boundary is inclusive or exclusive doesn’t 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. // // HistogramFraction is exported as it may be used by other PromQL engine // implementations. 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() zeroBucket := false // interpolateLinearly is used for custom buckets to be // consistent with the linear interpolation known from classic // histograms. It is also used for the zero bucket. interpolateLinearly := func(v float64) float64 { return rank + b.Count*(v-b.Lower)/(b.Upper-b.Lower) } // interpolateExponentially is using the same exponential // interpolation method as above for histogramQuantile. This // method is a better fit for exponential bucketing. interpolateExponentially := func(v float64) float64 { var ( logLower = math.Log2(math.Abs(b.Lower)) logUpper = math.Log2(math.Abs(b.Upper)) logV = math.Log2(math.Abs(v)) fraction float64 ) if v > 0 { fraction = (logV - logLower) / (logUpper - logLower) } else { fraction = 1 - ((logV - logUpper) / (logLower - logUpper)) } return rank + b.Count*fraction } if b.Lower <= 0 && b.Upper >= 0 { zeroBucket = true 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 { // We have hit the lower value at the lower bucket boundary. lowerRank = rank lowerSet = true } if !upperSet && b.Lower >= upper { // We have hit the upper value at the lower bucket boundary. upperRank = rank upperSet = true } if lowerSet && upperSet { break } if !lowerSet && b.Lower < lower && b.Upper > lower { // The lower value is in this bucket. if h.UsesCustomBuckets() || zeroBucket { lowerRank = interpolateLinearly(lower) } else { lowerRank = interpolateExponentially(lower) } lowerSet = true } if !upperSet && b.Lower < upper && b.Upper > upper { // The upper value is in this bucket. if h.UsesCustomBuckets() || zeroBucket { upperRank = interpolateLinearly(upper) } else { upperRank = interpolateExponentially(upper) } 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 }