Here is my version, which works purely in integers (the division by k always produces an integer quotient) and is fast at O(k): function choose(n, k) if k == 0 return 1 return (n * choose(n – 1, k – 1)) / k I wrote it recursively because it’s so simple and pretty, but you … Read more
Or if you’d like to create combinations with an arbitrary number of arrays of arbitrary sizes…(I’m sure you can do this recursively, but since this isn’t a job interview, I’m instead using an iterative “odometer” for this…it increments a “number” with each digit a “base-n” digit based on the length of each array)…for example… combineArrays([ … Read more
Use Path.GetFullPath string path = Path.Combine(@”c:\alpha\beta”, @”..\gamma”); Console.WriteLine(Path.GetFullPath(path)); or the DirectoryInfo class: string path = Path.Combine(@”c:\alpha\beta”, @”..\gamma”); DirectoryInfo info = new DirectoryInfo(path); Console.WriteLine(info.FullName); Both will output: c:\alpha\gamma
>>> def pascal(n): … line = [1] … for k in range(n): … line.append(line[k] * (n-k) / (k+1)) … return line … >>> pascal(9) [1, 9, 36, 84, 126, 126, 84, 36, 9, 1] This uses the following identity: C(n,k+1) = C(n,k) * (n-k) / (k+1) So you can start with C(n,0) = 1 and … Read more
Try [p,q] = meshgrid(vec1, vec2); pairs = [p(:) q(:)]; See the MESHGRID documentation. Although this is not exactly what that function is for, but if you squint at it funny, what you are asking for is exactly what it does.
You need itertools.combinations: >>> from itertools import combinations >>> L = [1, 2, 3, 4] >>> [“,”.join(map(str, comb)) for comb in combinations(L, 3)] [‘1,2,3’, ‘1,2,4’, ‘1,3,4’, ‘2,3,4’]
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
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
Simply use itertools.combinations. For example: import itertools lst = [1, 2, 3] combs = [] for i in xrange(1, len(lst)+1): combs.append(i) els = [list(x) for x in itertools.combinations(lst, i)] combs.append(els) Now combs holds this value: [1, [[1], [2], [3]], 2, [[1, 2], [1, 3], [2, 3]], 3, [[1, 2, 3]]] Yes, it’s slightly different from … Read more