improve handling of negative bounds in histogram std dev/var
Signed-off-by: Jeanette Tan <jeanette.tan@grafana.com>
This commit is contained in:
parent
9d32754bc0
commit
22d0f4f114
|
@ -3496,7 +3496,7 @@ func TestNativeHistogram_HistogramStdDevVar(t *testing.T) {
|
|||
},
|
||||
NegativeBuckets: []int64{1, 0},
|
||||
},
|
||||
stdVar: 1544.8582535368798, // actual variance: 1738.4082
|
||||
stdVar: 1844.4651144196398, // actual variance: 1738.4082
|
||||
},
|
||||
{
|
||||
name: "-100000, -10000, -1000, -888, -888, -100, -50, -9, -8, -3",
|
||||
|
@ -3514,7 +3514,7 @@ func TestNativeHistogram_HistogramStdDevVar(t *testing.T) {
|
|||
},
|
||||
NegativeBuckets: []int64{1, 0, 0, 0, 0, 2, -2, 0},
|
||||
},
|
||||
stdVar: 1240930974.5260057, // actual variance: 882690990
|
||||
stdVar: 759352122.1939945, // actual variance: 882690990
|
||||
},
|
||||
{
|
||||
name: "-10 x10",
|
||||
|
@ -3528,7 +3528,7 @@ func TestNativeHistogram_HistogramStdDevVar(t *testing.T) {
|
|||
},
|
||||
NegativeBuckets: []int64{10},
|
||||
},
|
||||
stdVar: 454.2741699796952, // actual variance: 0
|
||||
stdVar: 1.725830020304794, // actual variance: 0
|
||||
},
|
||||
{
|
||||
name: "-50, -8, 0, 3, 8, 9, 100, NaN",
|
||||
|
|
|
@ -1116,6 +1116,9 @@ func funcHistogramStdDev(vals []parser.Value, args parser.Expressions, enh *Eval
|
|||
val = 0
|
||||
} else {
|
||||
val = math.Sqrt(bucket.Upper * bucket.Lower)
|
||||
if bucket.Upper < 0 {
|
||||
val = -val
|
||||
}
|
||||
}
|
||||
delta := val - mean
|
||||
variance, cVariance = kahanSumInc(bucket.Count*delta*delta, variance, cVariance)
|
||||
|
@ -1149,6 +1152,9 @@ func funcHistogramStdVar(vals []parser.Value, args parser.Expressions, enh *Eval
|
|||
val = 0
|
||||
} else {
|
||||
val = math.Sqrt(bucket.Upper * bucket.Lower)
|
||||
if bucket.Upper < 0 {
|
||||
val = -val
|
||||
}
|
||||
}
|
||||
delta := val - mean
|
||||
variance, cVariance = kahanSumInc(bucket.Count*delta*delta, variance, cVariance)
|
||||
|
|
Loading…
Reference in New Issue