Fast algorithm for repeated calculation of percentile?

Question

In an algorithm I have to calculate the 75th percentile of a data set whenever I add a value. Right now I am doing this:

1. Get value x
2. Inse

Answer 1

If you have a known set of values, following will be very fast:

Create a large array of integers (even bytes will work) with number of elements equal to maximum value of your data. For example, if the maximum value of t is 100,000 create an array

int[] index = new int[100000]; // 400kb

Now iterate over the entire set of values, as

for each (int t : set_of_values) {
  index[t]++;
}

// You can do a try catch on ArrayOutOfBounds just in case :)

Now calculate percentile as

int sum = 0, i = 0;
while (sum < 0.9*set_of_values.length) {
  sum += index[i++];
}

return i;

You can also consider using a TreeMap instead of array, if the values don't confirm to these restrictions.

Answer 2

You can use binary search to do find the correct position in O(log n). However, shifting the array up is still O(n).

Answer 3

A simple Order Statistics Tree is enough for this.

A balanced version of this tree supports O(logn) time insert/delete and access by Rank. So you not only get the 75% percentile, but also the 66% or 50% or whatever you need without having to change your code.

If you access the 75% percentile frequently, but only insert less frequently, you can always cache the 75% percentile element during an insert/delete operation.

Most standard implementations (like Java's TreeMap) are order statistic trees.

Answer 4

You can do it with two heaps. Not sure if there's a less 'contrived' solution, but this one provides O(logn) time complexity and heaps are also included in standard libraries of most programming languages.

First heap (heap A) contains smallest 75% elements, another heap (heap B) - the rest (biggest 25%). First one has biggest element on the top, second one - smallest.

1. Adding element.

See if new element x is <= max(A). If it is, add it to heap A, otherwise - to heap B.
Now, if we added x to heap A and it became too big (holds more than 75% of elements), we need to remove biggest element from A (O(logn)) and add it to heap B (also O(logn)).
Similar if heap B became too big.

2. Finding "0.75 median"

Just take the largest element from A (or smallest from B). Requires O(logn) or O(1) time, depending on heap implementation.

edit
As Dolphin noted, we need to specify precisely how big each heap should be for every n (if we want precise answer). For example, if size(A) = floor(n * 0.75) and size(B) is the rest, then, for every n > 0, array[array.size * 3/4] = min(B).

Answer 5

Here is a javaScript solution . Copy-paste it in browser console and it works . $scores contains the List of scores and , $percentilegives the n-th percentile of the list . So 75th percentile is 76.8 and 99 percentile is 87.9.

function get_percentile($percentile, $array) {
    $array = $array.sort();
    $index = ($percentile/100) * $array.length;
    if (Math.floor($index) === $index) {
         $result = ($array[$index-1] + $array[$index])/2;
    }
    else {
        $result = $array[Math.floor($index)];
    }
    return $result;
}

$scores = [22.3, 32.4, 12.1, 54.6, 76.8, 87.3, 54.6, 45.5, 87.9];

get_percentile(75, $scores);
get_percentile(90, $scores);

Answer 6

If you can do with an approximate answer, you can use a histogram instead of keeping entire values in memory.

For each new value, add it to the appropriate bin. Calculate percentile 75th by traversing bins and summing counts until 75% of the population size is reached. Percentile value is between bin's (which you stopped at) low bound to high bound.

This will provide O(B) complexity where B is the count of bins, which is range_size/bin_size. (use bin_size appropriate to your user case).

I have implemented this logic in a JVM library: https://github.com/IBM/HBPE which you can use as a reference.