Robert Floyd invented a sampling algorithm for just such situations. It’s generally superior to shuffling then grabbing the first x elements since it doesn’t require O(y) storage. As originally written it assumes values from 1..N, but it’s trivial to produce 0..N and/or use non-contiguous values by simply treating the values it produces as subscripts into … Read more
Sort within the rows first, then use duplicated, see below: # example data dat = matrix(scan(‘data.txt’), ncol = 3, byrow = TRUE) # Read 90 items dat[ !duplicated(apply(dat, 1, sort), MARGIN = 2), ] # [,1] [,2] [,3] # [1,] 1 2 3 # [2,] 1 2 4 # [3,] 1 2 5 # [4,] … Read more
Here is recursive solution: function combinations($arrays, $i = 0) { if (!isset($arrays[$i])) { return array(); } if ($i == count($arrays) – 1) { return $arrays[$i]; } // get combinations from subsequent arrays $tmp = combinations($arrays, $i + 1); $result = array(); // concat each array from tmp with each element from $arrays[$i] foreach ($arrays[$i] as … Read more
Here is a quite simple and short one using a recursive helper function: function cartesian(…args) { var r = [], max = args.length-1; function helper(arr, i) { for (var j=0, l=args[i].length; j<l; j++) { var a = arr.slice(0); // clone arr a.push(args[i][j]); if (i==max) r.push(a); else helper(a, i+1); } } helper([], 0); return r; } … Read more
you need itertools.product: >>> import itertools >>> a = [[1,2,3],[4,5,6],[7,8,9,10]] >>> list(itertools.product(*a)) [(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 4, 10), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 5, 10), (1, 6, 7), (1, 6, 8), (1, 6, 9), (1, 6, 10), (2, 4, 7), (2, 4, 8), (2, … Read more
The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired … Read more
Finding all possible combinations of numbers to reach a given sum
Algorithm to return all combinations of k elements from n
This answer missed one aspect: the OP asked for ALL combinations… not just combinations of length “r”. So you’d either have to loop through all lengths “L”: import itertools stuff = [1, 2, 3] for L in range(0, len(stuff)+1): for subset in itertools.combinations(stuff, L): print(subset) Or — if you want to get snazzy (or bend … Read more
$result = $data1 . $data2; This is called string concatenation. Your example lacks a space though, so for that specifically, you would need: $result = $data1 . ‘ ‘ . $data2;