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Update from review comments.

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Signed-off-by: György Krajcsovits <gyorgy.krajcsovits@grafana.com>
This commit is contained in:
György Krajcsovits 2024-08-13 15:26:07 +02:00
parent 6aee5b4b38
commit 386fc8b9f6
2 changed files with 10 additions and 8 deletions
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5
promql/promqltest/testdata/histograms.test vendored
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@ -497,8 +497,7 @@ eval instant at 5m rate(const_histogram_bucket[5m])
{le="2.0"} 0
{le="+Inf"} 0
# 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.
# 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]}}
@ -507,7 +506,5 @@ eval instant at 5m rate(const_histogram[5m])
eval instant at 5m histogram_quantile(1.0, sum by (le) (rate(const_histogram_bucket[5m])))
{} NaN
# 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(rate(const_histogram[5m])))
{} NaN

13
promql/promqltest/testdata/native_histograms.test vendored
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@ -794,7 +794,8 @@ load 1m
eval instant at 5m rate(const_histogram[5m])
{} {{schema:0 sum:0 count:0}}
# Zero buckets mean no observations, so average has no meaningful value.
# 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
@ -802,15 +803,19 @@ eval instant at 5m histogram_avg(rate(const_histogram[5m]))
eval instant at 5m histogram_count(rate(const_histogram[5m]))
{} 0.0
# Zero buckets mean no observations, so the sum should be NaN, However
# we return 0 for compatibility with classic histograms.
# 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
# BUG??? Zero buckets mean no observations, thus any fraction should be 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]))