The p95 of a batch of latencies
Dashboards quote p95 rather than the average, so one pathological request does not hide behind a thousand fast ones.
- Use the nearest-rank method: sort ascending, then take the value at position ceil(rank percent of the count), counting from 1.
- The rank is a percentage from 1 to 100; anything outside that gives 0.
- No readings gives 0.
percentileValue(values: list<int>, rank: int) → int
Go needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
func percentileValue(values []int, rank int) int {
}Worked examples
| Call | Result |
|---|---|
percentileValue([]int{1, 2, 3, 4, 5}, 50) | 3 |
percentileValue([]int{1, 2, 3, 4, 5}, 100) | 5 |
percentileValue([]int{1, 2, 3, 4, 5}, 1) | 1 |
percentileValue([]int{10, 20, 30, 40, 50, 60, 70, 80, 90, 100}, 95) | 100 |
Hint
In integers, ceil(rank * n / 100) is (rank * n + 99) / 100. Then subtract one for a zero-based index.
Reference solution in Go
func percentileValue(values []int, rank int) int {
if len(values) == 0 || rank < 1 || rank > 100 {
return 0
}
s := append([]int{}, values...)
sort.Ints(s)
pos := (rank*len(s) + 99) / 100
return s[pos-1]
}