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Merge pull request #14658 from krajorama/zero-rate-nh-quantile 

Native histograms: define behavior when rate is null.
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George Krajcsovits 2024-08-13 16:05:43 +02:00 committed by GitHub
parent aa4b056ad0 386fc8b9f6
commit fe5354df88
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26
promql/promqltest/testdata/histograms.test vendored
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@ -482,3 +482,29 @@ load_with_nhcb 5m
eval_fail instant at 50m histogram_quantile(0.99, {__name__=~"request_duration_seconds\\d*_bucket"})
eval_fail instant at 50m histogram_quantile(0.99, {__name__=~"request_duration_seconds\\d*"})
# Histogram with constant buckets.
load_with_nhcb 1m
const_histogram_bucket{le="0.0"} 1 1 1 1 1
const_histogram_bucket{le="1.0"} 1 1 1 1 1
const_histogram_bucket{le="2.0"} 1 1 1 1 1
const_histogram_bucket{le="+Inf"} 1 1 1 1 1
# There is no change to the bucket count over time, thus rate is 0 in each bucket.
eval instant at 5m rate(const_histogram_bucket[5m])
{le="0.0"} 0
{le="1.0"} 0
{le="2.0"} 0
{le="+Inf"} 0
# Native histograms do not represent empty buckets, so here the zeros are implicit.
eval instant at 5m rate(const_histogram[5m])
{} {{schema:-53 sum:0 count:0 custom_values:[0.0 1.0 2.0]}}
# Zero buckets mean no observations, so there is no value that observations fall below,
# which means that any quantile is a NaN.
eval instant at 5m histogram_quantile(1.0, sum by (le) (rate(const_histogram_bucket[5m])))
{} NaN
eval instant at 5m histogram_quantile(1.0, sum(rate(const_histogram[5m])))
{} NaN

44
promql/promqltest/testdata/native_histograms.test vendored
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@ -784,3 +784,47 @@ eval_warn instant at 1m rate(some_metric[30s])
# Start with exponential, end with custom.
eval_warn instant at 30s rate(some_metric[30s])
# Should produce no results.
# Histogram with constant buckets.
load 1m
const_histogram {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}}
# There is no change to the bucket count over time, thus rate is 0 in each bucket.
# However native histograms do not represent empty buckets, so here the zeros are implicit.
eval instant at 5m rate(const_histogram[5m])
{} {{schema:0 sum:0 count:0}}
# Zero buckets mean no observations, thus the denominator in the average is 0
# leading to 0/0, which is NaN.
eval instant at 5m histogram_avg(rate(const_histogram[5m]))
{} NaN
# Zero buckets mean no observations, so count is 0.
eval instant at 5m histogram_count(rate(const_histogram[5m]))
{} 0.0
# Zero buckets mean no observations and empty histogram has a sum of 0 by definition.
eval instant at 5m histogram_sum(rate(const_histogram[5m]))
{} 0.0
# Zero buckets mean no observations, thus the denominator in the fraction is 0,
# leading to 0/0, which is NaN.
eval instant at 5m histogram_fraction(0.0, 1.0, rate(const_histogram[5m]))
{} NaN
# Workaround to calculate the observation count corresponding to NaN fraction.
eval instant at 5m histogram_count(rate(const_histogram[5m])) == 0.0 or histogram_fraction(0.0, 1.0, rate(const_histogram[5m])) * histogram_count(rate(const_histogram[5m]))
{} 0.0
# Zero buckets mean no observations, so there is no value that observations fall below,
# which means that any quantile is a NaN.
eval instant at 5m histogram_quantile(1.0, rate(const_histogram[5m]))
{} NaN
# Zero buckets mean no observations, so there is no standard deviation.
eval instant at 5m histogram_stddev(rate(const_histogram[5m]))
{} NaN
# Zero buckets mean no observations, so there is no standard variance.
eval instant at 5m histogram_stdvar(rate(const_histogram[5m]))
{} NaN