如何从Java或Linux命令行中的一组数字计算90％的分配值？

Question

I ran my integration test and got a list of execution time back. I need to find the 90% distribution value from that set of result not the average. Because it represents what the user will encounter 90% of the time. Is there a linux command line that does that? If not is there a java library that does that?

Thanks,

Sean

Answer 1

Assuming you have a file with each measurement on a separate line, with the actual measurement in the first column, count the lines, call that number L; sort reversed, print the first field from the L/10th line. How you round a non-integer L/10 is up to you; you could also interpolate a value between n(floor(L/10)) and n(ceil(L/10)).

#!/bin/sh
L=$(wc -l <data.txt)
sort -r -n data.txt |
awk "NR >= $L/10"'{print $1; exit}'

This assumes fields are whitespace-separated.

Answer 2

You are looking for a 90th percentile . Simply sort the execution times from shortest to longest and pick the one lying 10% from the end. The quoted article explains other methods.

You can easily implement this using Java or Unix commands: sort , wc , head and tail or sed .

Answer 3

If you store your times in a List or an array you can do this.

If not is there a java library that does that?

Its just a couple of lines of Java code. I would just add it to your Java program so you have less output to deal with.

List<Long> times = new ArrayList<>();
// add times
Collections.sort(times);
System.out.printf("The typical, 90%% and 99%%tile times were %,d / %,d / %,d %n",
   times.get(times.size()/2), times.get(times.size()*9/10), times.get(times.size()*99/100));

or

long[] times = new long[SAMPLES];
// add times and
Arrays.sort(times);
System.out.printf("The typical, 90%% and 99%%tile times were %,d / %,d / %,d %n",
   times[SAMPLES/2], times[SAMPLES*9/10], times[SAMPLES*99/100]);

Because it represents what the user will encounter 90% of the time.

Actually the 90th percentile is the time its will less than 90% of the time. The user will only experience this delay (or more) 10% of the time.