Not my code, but you’re looking for the powerset. Google gave me this solution, which seems t work: public IEnumerable<IEnumerable<T>> GetPowerSet<T>(List<T> list) { return from m in Enumerable.Range(0, 1 << list.Count) select from i in Enumerable.Range(0, list.Count) where (m & (1 << i)) != 0 select list[i]; } Source: http://rosettacode.org/wiki/Power_set#C.23
You can use itertools.combinations() to create the index array, and then use NumPy’s fancy indexing: import numpy as np from itertools import combinations, chain from scipy.special import comb def comb_index(n, k): count = comb(n, k, exact=True) index = np.fromiter(chain.from_iterable(combinations(range(n), k)), int, count=count*k) return index.reshape(-1, k) data = np.array([[1,2,3,4,5],[10,11,12,13,14]]) idx = comb_index(5, 3) print(data[:, idx]) output: … Read more
Generating all the possible permutations and then discarding the repeated ones is highly inefficient. Various algorithms exist to directly generate the permutations of a multiset in lexicographical order or other kind of ordering. Takaoka’s algorithm is a good example, but probably that of Aaron Williams is better http://webhome.csc.uvic.ca/~haron/CoolMulti.pdf moreover, it has been implemented in the … Read more
Your formula is totally wrong, it’s supposed to be fact(n)/fact(r)/fact(n-r), but that is in turn a very inefficient way to compute it. See Fast computation of multi-category number of combinations and especially my comments on that question. (Oh, and please reopen that question also so I can answer it properly) The single-split case is actually … Read more
Ok, thanks @dfan for telling me I was looking in the wrong place. I’ve got it now: from itertools import product def my_product(inp): return (dict(zip(inp.keys(), values)) for values in product(*inp.values()) EDIT: after years more Python experience, I think a better solution is to accept kwargs rather than a dictionary of inputs; the call style is … Read more
Here is code to get all permutations: http://php.net/manual/en/function.shuffle.php#90615 With the code to get the power set, permutations are those of maximal length, the power set should be all combinations. I have no idea what dispositions are, so if you can explain them, that would help.
Yes, there is a “next permutation” algorithm, and it’s quite simple too. The C++ standard template library (STL) even has a function called next_permutation. The algorithm actually finds the next permutation — the lexicographically next one. The idea is this: suppose you are given a sequence, say “32541”. What is the next permutation? If you … Read more
Use itertools.permutations from the standard library: import itertools list(itertools.permutations([1, 2, 3])) Adapted from here is a demonstration of how itertools.permutations might be implemented: def permutations(elements): if len(elements) <= 1: yield elements return for perm in permutations(elements[1:]): for i in range(len(elements)): # nb elements[0:1] works in both string and list contexts yield perm[:i] + elements[0:1] + … Read more
Note you can generate the sequence by recursively generating all combinations with the first element, then all combinations without. In both recursive cases, you drop the first element to get all combinations from n-1 elements. In Python: def combination(l, r): if r == 0: yield [] elif len(l) == r: yield l else: for c … Read more
You are looking for a derangement of your entries. First of all, your algorithm works in the sense that it outputs a random derangement, ie a permutation with no fixed point. However it has a enormous flaw (which you might not mind, but is worth keeping in mind): some derangements cannot be obtained with your … Read more