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272 lines
8.0 KiB
272 lines
8.0 KiB
# Minimal valid case: an empty histogram.
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load 5m
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empty_histogram {{}}
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eval instant at 5m empty_histogram
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{__name__="empty_histogram"} {{}}
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eval instant at 5m histogram_count(empty_histogram)
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{} 0
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eval instant at 5m histogram_sum(empty_histogram)
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{} 0
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eval instant at 5m histogram_avg(empty_histogram)
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{} NaN
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eval instant at 5m histogram_fraction(-Inf, +Inf, empty_histogram)
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{} NaN
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eval instant at 5m histogram_fraction(0, 8, empty_histogram)
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{} NaN
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# buckets:[1 2 1] means 1 observation in the 1st bucket, 2 observations in the 2nd and 1 observation in the 3rd (total 4).
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load 5m
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single_histogram {{schema:0 sum:5 count:4 buckets:[1 2 1]}}
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# histogram_count extracts the count property from the histogram.
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eval instant at 5m histogram_count(single_histogram)
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{} 4
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# histogram_sum extracts the sum property from the histogram.
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eval instant at 5m histogram_sum(single_histogram)
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{} 5
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# histogram_avg calculates the average from sum and count properties.
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eval instant at 5m histogram_avg(single_histogram)
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{} 1.25
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# We expect half of the values to fall in the range 1 < x <= 2.
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eval instant at 5m histogram_fraction(1, 2, single_histogram)
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{} 0.5
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# We expect all values to fall in the range 0 < x <= 8.
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eval instant at 5m histogram_fraction(0, 8, single_histogram)
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{} 1
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# Median is 1.5 due to linear estimation of the midpoint of the middle bucket, whose values are within range 1 < x <= 2.
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eval instant at 5m histogram_quantile(0.5, single_histogram)
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{} 1.5
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# Repeat the same histogram 10 times.
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load 5m
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multi_histogram {{schema:0 sum:5 count:4 buckets:[1 2 1]}}x10
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eval instant at 5m histogram_count(multi_histogram)
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{} 4
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eval instant at 5m histogram_sum(multi_histogram)
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{} 5
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eval instant at 5m histogram_avg(multi_histogram)
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{} 1.25
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eval instant at 5m histogram_fraction(1, 2, multi_histogram)
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{} 0.5
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eval instant at 5m histogram_quantile(0.5, multi_histogram)
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{} 1.5
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# Each entry should look the same as the first.
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eval instant at 50m histogram_count(multi_histogram)
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{} 4
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eval instant at 50m histogram_sum(multi_histogram)
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{} 5
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eval instant at 50m histogram_avg(multi_histogram)
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{} 1.25
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eval instant at 50m histogram_fraction(1, 2, multi_histogram)
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{} 0.5
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eval instant at 50m histogram_quantile(0.5, multi_histogram)
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{} 1.5
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# Accumulate the histogram addition for 10 iterations, offset is a bucket position where offset:0 is always the bucket
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# with an upper limit of 1 and offset:1 is the bucket which follows to the right. Negative offsets represent bucket
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# positions for upper limits <1 (tending toward zero), where offset:-1 is the bucket to the left of offset:0.
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load 5m
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incr_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}}+{{sum:2 count:1 buckets:[1] offset:1}}x10
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eval instant at 5m histogram_count(incr_histogram)
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{} 5
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eval instant at 5m histogram_sum(incr_histogram)
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{} 6
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eval instant at 5m histogram_avg(incr_histogram)
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{} 1.2
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# We expect 3/5ths of the values to fall in the range 1 < x <= 2.
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eval instant at 5m histogram_fraction(1, 2, incr_histogram)
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{} 0.6
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eval instant at 5m histogram_quantile(0.5, incr_histogram)
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{} 1.5
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eval instant at 50m incr_histogram
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{__name__="incr_histogram"} {{count:14 sum:24 buckets:[1 12 1]}}
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eval instant at 50m histogram_count(incr_histogram)
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{} 14
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eval instant at 50m histogram_sum(incr_histogram)
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{} 24
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eval instant at 50m histogram_avg(incr_histogram)
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{} 1.7142857142857142
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# We expect 12/14ths of the values to fall in the range 1 < x <= 2.
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eval instant at 50m histogram_fraction(1, 2, incr_histogram)
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{} 0.8571428571428571
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eval instant at 50m histogram_quantile(0.5, incr_histogram)
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{} 1.5
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# Per-second average rate of increase should be 1/(5*60) for count and buckets, then 2/(5*60) for sum.
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eval instant at 50m rate(incr_histogram[5m])
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{} {{count:0.0033333333333333335 sum:0.006666666666666667 offset:1 buckets:[0.0033333333333333335]}}
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# Calculate the 50th percentile of observations over the last 10m.
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eval instant at 50m histogram_quantile(0.5, rate(incr_histogram[10m]))
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{} 1.5
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# Schema represents the histogram resolution, different schema have compatible bucket boundaries, e.g.:
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# 0: 1 2 4 8 16 32 64 (higher resolution)
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# -1: 1 4 16 64 (lower resolution)
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#
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# Histograms can be merged as long as the histogram to the right is same resolution or higher.
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load 5m
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low_res_histogram {{schema:-1 sum:4 count:1 buckets:[1] offset:1}}+{{schema:0 sum:4 count:4 buckets:[2 2] offset:1}}x1
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eval instant at 5m low_res_histogram
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{__name__="low_res_histogram"} {{schema:-1 count:5 sum:8 offset:1 buckets:[5]}}
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eval instant at 5m histogram_count(low_res_histogram)
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{} 5
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eval instant at 5m histogram_sum(low_res_histogram)
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{} 8
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eval instant at 5m histogram_avg(low_res_histogram)
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{} 1.6
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# We expect all values to fall into the lower-resolution bucket with the range 1 < x <= 4.
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eval instant at 5m histogram_fraction(1, 4, low_res_histogram)
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{} 1
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# z_bucket:1 means there is one observation in the zero bucket and z_bucket_w:0.5 means the zero bucket has the range
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# 0 < x <= 0.5. Sum and count are expected to represent all observations in the histogram, including those in the zero bucket.
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load 5m
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single_zero_histogram {{schema:0 z_bucket:1 z_bucket_w:0.5 sum:0.25 count:1}}
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eval instant at 5m histogram_count(single_zero_histogram)
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{} 1
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eval instant at 5m histogram_sum(single_zero_histogram)
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{} 0.25
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eval instant at 5m histogram_avg(single_zero_histogram)
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{} 0.25
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# When only the zero bucket is populated, or there are negative buckets, the distribution is assumed to be equally
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# distributed around zero; i.e. that there are an equal number of positive and negative observations. Therefore the
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# entire distribution must lie within the full range of the zero bucket, in this case: -0.5 < x <= +0.5.
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eval instant at 5m histogram_fraction(-0.5, 0.5, single_zero_histogram)
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{} 1
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# Half of the observations are estimated to be zero, as this is the midpoint between -0.5 and +0.5.
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eval instant at 5m histogram_quantile(0.5, single_zero_histogram)
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{} 0
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# Let's turn single_histogram upside-down.
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load 5m
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negative_histogram {{schema:0 sum:-5 count:4 n_buckets:[1 2 1]}}
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eval instant at 5m histogram_count(negative_histogram)
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{} 4
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eval instant at 5m histogram_sum(negative_histogram)
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{} -5
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eval instant at 5m histogram_avg(negative_histogram)
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{} -1.25
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# We expect half of the values to fall in the range -2 < x <= -1.
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eval instant at 5m histogram_fraction(-2, -1, negative_histogram)
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{} 0.5
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eval instant at 5m histogram_quantile(0.5, negative_histogram)
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{} -1.5
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# Two histogram samples.
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load 5m
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two_samples_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}} {{schema:0 sum:-4 count:4 n_buckets:[1 2 1]}}
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# We expect to see the newest sample.
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eval instant at 10m histogram_count(two_samples_histogram)
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{} 4
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eval instant at 10m histogram_sum(two_samples_histogram)
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{} -4
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eval instant at 10m histogram_avg(two_samples_histogram)
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{} -1
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eval instant at 10m histogram_fraction(-2, -1, two_samples_histogram)
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{} 0.5
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eval instant at 10m histogram_quantile(0.5, two_samples_histogram)
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{} -1.5
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# Add two histograms with negated data.
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load 5m
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balanced_histogram {{schema:0 sum:4 count:4 buckets:[1 2 1]}}+{{schema:0 sum:-4 count:4 n_buckets:[1 2 1]}}x1
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eval instant at 5m histogram_count(balanced_histogram)
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{} 8
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eval instant at 5m histogram_sum(balanced_histogram)
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{} 0
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eval instant at 5m histogram_avg(balanced_histogram)
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{} 0
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eval instant at 5m histogram_fraction(0, 4, balanced_histogram)
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{} 0.5
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# If the quantile happens to be located in a span of empty buckets, the actually returned value is the lower bound of
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# the first populated bucket after the span of empty buckets.
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eval instant at 5m histogram_quantile(0.5, balanced_histogram)
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{} 0.5
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# Add histogram to test sum(last_over_time) regression
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load 5m
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incr_sum_histogram{number="1"} {{schema:0 sum:0 count:0 buckets:[1]}}+{{schema:0 sum:1 count:1 buckets:[1]}}x10
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incr_sum_histogram{number="2"} {{schema:0 sum:0 count:0 buckets:[1]}}+{{schema:0 sum:2 count:1 buckets:[1]}}x10
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eval instant at 50m histogram_sum(sum(incr_sum_histogram))
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{} 30
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eval instant at 50m histogram_sum(sum(last_over_time(incr_sum_histogram[5m])))
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{} 30
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