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
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
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
Here is a solution that works on both even and odd length array and won’t alter the array: def median(array) return nil if array.empty? sorted = array.sort len = sorted.length (sorted[(len – 1) / 2] + sorted[len / 2]) / 2.0 end
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
You can use the Median of Medians algorithm to find median of an unsorted array in linear time.
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
Ongoing work SPARK-30569 – Add DSL functions invoking percentile_approx Spark 2.0+: You can use approxQuantile method which implements Greenwald-Khanna algorithm: Python: df.approxQuantile(“x”, [0.5], 0.25) Scala: df.stat.approxQuantile(“x”, Array(0.5), 0.25) where the last parameter is a relative error. The lower the number the more accurate results and more expensive computation. Since Spark 2.2 (SPARK-14352) it supports estimation … Read more
I use these incremental/recursive mean and median estimators, which both use constant storage: mean += eta * (sample – mean) median += eta * sgn(sample – median) where eta is a small learning rate parameter (e.g. 0.001), and sgn() is the signum function which returns one of {-1, 0, 1}. (Use a constant eta if … Read more