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e7e50a3afd
Just query via `HeadAndOOOQuerier`, which will skip series where no in-order chunks are in range. Now we don't need `OOORangeHead`. Signed-off-by: Bryan Boreham <bjboreham@gmail.com>
168 lines
5 KiB
Go
168 lines
5 KiB
Go
// Copyright 2022 The Prometheus Authors
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// Licensed under the Apache License, Version 2.0 (the "License");
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// you may not use this file except in compliance with the License.
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// You may obtain a copy of the License at
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//
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// http://www.apache.org/licenses/LICENSE-2.0
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//
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// Unless required by applicable law or agreed to in writing, software
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// distributed under the License is distributed on an "AS IS" BASIS,
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// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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// See the License for the specific language governing permissions and
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// limitations under the License.
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package tsdb
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import (
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"sort"
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"github.com/prometheus/prometheus/model/histogram"
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"github.com/prometheus/prometheus/tsdb/chunkenc"
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)
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// OOOChunk maintains samples in time-ascending order.
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// Inserts for timestamps already seen, are dropped.
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// Samples are stored uncompressed to allow easy sorting.
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// Perhaps we can be more efficient later.
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type OOOChunk struct {
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samples []sample
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}
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func NewOOOChunk() *OOOChunk {
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return &OOOChunk{samples: make([]sample, 0, 4)}
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}
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// Insert inserts the sample such that order is maintained.
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// Returns false if insert was not possible due to the same timestamp already existing.
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func (o *OOOChunk) Insert(t int64, v float64, h *histogram.Histogram, fh *histogram.FloatHistogram) bool {
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// Although out-of-order samples can be out-of-order amongst themselves, we
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// are opinionated and expect them to be usually in-order meaning we could
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// try to append at the end first if the new timestamp is higher than the
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// last known timestamp.
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if len(o.samples) == 0 || t > o.samples[len(o.samples)-1].t {
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o.samples = append(o.samples, sample{t, v, h, fh})
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return true
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}
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// Find index of sample we should replace.
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i := sort.Search(len(o.samples), func(i int) bool { return o.samples[i].t >= t })
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if i >= len(o.samples) {
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// none found. append it at the end
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o.samples = append(o.samples, sample{t, v, h, fh})
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return true
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}
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// Duplicate sample for timestamp is not allowed.
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if o.samples[i].t == t {
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return false
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}
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// Expand length by 1 to make room. use a zero sample, we will overwrite it anyway.
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o.samples = append(o.samples, sample{})
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copy(o.samples[i+1:], o.samples[i:])
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o.samples[i] = sample{t, v, h, fh}
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return true
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}
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func (o *OOOChunk) NumSamples() int {
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return len(o.samples)
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}
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// ToEncodedChunks returns chunks with the samples in the OOOChunk.
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//
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//nolint:revive // unexported-return.
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func (o *OOOChunk) ToEncodedChunks(mint, maxt int64) (chks []memChunk, err error) {
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if len(o.samples) == 0 {
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return nil, nil
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}
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// The most common case is that there will be a single chunk, with the same type of samples in it - this is always true for float samples.
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chks = make([]memChunk, 0, 1)
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var (
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cmint int64
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cmaxt int64
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chunk chunkenc.Chunk
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app chunkenc.Appender
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)
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prevEncoding := chunkenc.EncNone // Yes we could call the chunk for this, but this is more efficient.
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for _, s := range o.samples {
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if s.t < mint {
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continue
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}
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if s.t > maxt {
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break
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}
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encoding := chunkenc.EncXOR
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if s.h != nil {
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encoding = chunkenc.EncHistogram
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} else if s.fh != nil {
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encoding = chunkenc.EncFloatHistogram
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}
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// prevApp is the appender for the previous sample.
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prevApp := app
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if encoding != prevEncoding { // For the first sample, this will always be true as EncNone != EncXOR | EncHistogram | EncFloatHistogram
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if prevEncoding != chunkenc.EncNone {
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chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
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}
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cmint = s.t
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switch encoding {
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case chunkenc.EncXOR:
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chunk = chunkenc.NewXORChunk()
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case chunkenc.EncHistogram:
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chunk = chunkenc.NewHistogramChunk()
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case chunkenc.EncFloatHistogram:
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chunk = chunkenc.NewFloatHistogramChunk()
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default:
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chunk = chunkenc.NewXORChunk()
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}
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app, err = chunk.Appender()
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if err != nil {
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return
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}
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}
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switch encoding {
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case chunkenc.EncXOR:
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app.Append(s.t, s.f)
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case chunkenc.EncHistogram:
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// Ignoring ok is ok, since we don't want to compare to the wrong previous appender anyway.
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prevHApp, _ := prevApp.(*chunkenc.HistogramAppender)
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var (
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newChunk chunkenc.Chunk
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recoded bool
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)
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newChunk, recoded, app, _ = app.AppendHistogram(prevHApp, s.t, s.h, false)
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if newChunk != nil { // A new chunk was allocated.
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if !recoded {
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chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
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cmint = s.t
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}
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chunk = newChunk
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}
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case chunkenc.EncFloatHistogram:
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// Ignoring ok is ok, since we don't want to compare to the wrong previous appender anyway.
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prevHApp, _ := prevApp.(*chunkenc.FloatHistogramAppender)
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var (
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newChunk chunkenc.Chunk
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recoded bool
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)
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newChunk, recoded, app, _ = app.AppendFloatHistogram(prevHApp, s.t, s.fh, false)
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if newChunk != nil { // A new chunk was allocated.
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if !recoded {
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chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
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cmint = s.t
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}
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chunk = newChunk
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}
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}
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cmaxt = s.t
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prevEncoding = encoding
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}
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if prevEncoding != chunkenc.EncNone {
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chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
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}
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return chks, nil
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}
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