promql: remove histogramMean from groupedAggregation
Re-use histogramValue since we don't need them separately. Tidy up initialization. Signed-off-by: Bryan Boreham <bjboreham@gmail.com>
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@ -2733,7 +2733,6 @@ type groupedAggregation struct {
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floatValue float64
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histogramValue *histogram.FloatHistogram
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floatMean float64
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histogramMean *histogram.FloatHistogram
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groupCount int
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heap vectorByValueHeap
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reverseHeap vectorByReverseValueHeap
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@ -2762,20 +2761,14 @@ func (ev *evaluator) aggregation(e *parser.AggregateExpr, q float64, inputMatrix
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floatMean: f,
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groupCount: 1,
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}
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switch {
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case h == nil:
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group.hasFloat = true
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case op == parser.SUM:
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group.histogramValue = h.Copy()
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group.hasHistogram = true
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case op == parser.AVG:
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group.histogramMean = h.Copy()
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group.hasHistogram = true
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case op == parser.STDVAR || op == parser.STDDEV:
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group.groupCount = 0
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}
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switch op {
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case parser.SUM, parser.AVG:
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if h == nil {
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group.hasFloat = true
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} else {
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group.histogramValue = h.Copy()
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group.hasHistogram = true
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}
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case parser.STDVAR, parser.STDDEV:
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group.floatValue = 0
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case parser.QUANTILE:
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@ -2807,11 +2800,11 @@ func (ev *evaluator) aggregation(e *parser.AggregateExpr, q float64, inputMatrix
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group.groupCount++
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if h != nil {
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group.hasHistogram = true
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if group.histogramMean != nil {
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if group.histogramValue != nil {
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left := h.Copy().Div(float64(group.groupCount))
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right := group.histogramMean.Copy().Div(float64(group.groupCount))
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right := group.histogramValue.Copy().Div(float64(group.groupCount))
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toAdd := left.Sub(right)
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group.histogramMean.Add(toAdd)
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group.histogramValue.Add(toAdd)
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}
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// Otherwise the aggregation contained floats
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// previously and will be invalid anyway. No
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@ -2886,7 +2879,7 @@ func (ev *evaluator) aggregation(e *parser.AggregateExpr, q float64, inputMatrix
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continue
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}
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if aggr.hasHistogram {
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aggr.histogramValue = aggr.histogramMean.Compact(0)
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aggr.histogramValue = aggr.histogramValue.Compact(0)
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} else {
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aggr.floatValue = aggr.floatMean
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}
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