You would use a min-max-median heap to find the min, max and median in constant time (and take linear time to build the heap). You can use order-statistics trees to find the kth smallest/largest value. Both of these data structures are described in this paper on min-max heaps [PDF]. Min-max heaps are binary heaps that … Read more
There’s an answer here using min/max and no branches (https://stackoverflow.com/a/14676309/2233603). Actually 4 min/max operations are enough to find the median, there’s no need for xor’s: median = max(min(a,b), min(max(a,b),c)); Though, it won’t give you the median value’s index… Breakdown of all cases: a b c 1 2 3 max(min(1,2), min(max(1,2),3)) = max(1, min(2,3)) = max(1, … Read more
Any random-access container (like std::vector) can be sorted with the standard std::sort algorithm, available in the <algorithm> header. For finding the median, it would be quicker to use std::nth_element; this does enough of a sort to put one chosen element in the correct position, but doesn’t completely sort the container. So you could find the … Read more
I have looked at R’s src/library/stats/src/Trunmed.c a few times as I wanted something similar too in a standalone C++ class / C subroutine. Note that this are actually two implementations in one, see src/library/stats/man/runmed.Rd (the source of the help file) which says \details{ Apart from the end values, the result \code{y = runmed(x, k)} simply … Read more
Looks like other answers are using sorting. That’s not optimal from performance point of view because it takes O(n logn) time. It is possible to calculate median in O(n) time instead. The generalized version of this problem is known as “n-order statistics” which means finding an element K in a set such that we have … Read more
Change your median method to this: function median(values){ if(values.length ===0) throw new Error(“No inputs”); values.sort(function(a,b){ return a-b; }); var half = Math.floor(values.length / 2); if (values.length % 2) return values[half]; return (values[half – 1] + values[half]) / 2.0; } fiddle
There are a number of possibilities. One I like is Hoare’s Select algorithm. The basic idea is similar to a Quicksort, except that when you recurse, you only recurse into the partition that will hold the number(s) you’re looking for. For example, if you want the median of 100 numbers, you’d start by partitioning the … Read more
I think your example input/output does not correspond to typical ways of calculating percentile. If you calculate the percentile as “proportion of data points strictly less than this value”, then the top value should be 0.8 (since 4 of 5 values are less than the largest one). If you calculate it as “percent of data … Read more
If the values are discrete and the number of distinct values isn’t too high, you could just accumulate the number of times each value occurs in a histogram, then find the median from the histogram counts (just add up counts from the top and bottom of the histogram until you reach the middle). Or if … Read more
There is no need to completely sort the vector: std::nth_element can do enough work to put the median in the correct position. See my answer to this question for an example. Of course, that doesn’t help if your teacher forbids using the right tool for the job.