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I have a numpy array [0, 1, 1, 2, 2, 0, 1, ...] which only contains the numbers 0-k. I would like to create a new array that contains the n possible arrays of permutations of 0-k. A small example with k=2 and n=6:
a = [0, 1, 0, 2]
permute(a)
result = [[0, 1, 0, 2]
[0, 2, 0, 1]
[1, 0, 1, 2]
[2, 1, 2, 0]
[1, 2, 1, 0]
[2, 0, 2, 1]]
Does anyone have any ideas/solutions as to how one could achieve this?
932
To calculate the number of permutations, take the number of possibilities for each event and then multiply that number by itself X times, where X equals the number of events in the sequence. For example, with four-digit PINs, each digit can range from 0 to 9, giving us 10 possibilities for each digit.
Overview. The Random module in NumPy helps make permutations of elements of an array. Permutation refers to the arrangement of elements in an array. For example, [3, 1, 2] and [1, 3, 2] are permutations of the array elements in [1, 2, 3] and vice versa.
To find all possible permutations of a given string, you can use the itertools module which has a useful method called permutations(iterable[, r]). This method return successive r length permutations of elements in the iterable as tuples.
Your a is what combinatorists call a multiset. The sympy library has various routines for working with them.
>>> from sympy.utilities.iterables import multiset_permutations
>>> import numpy as np
>>> a = np.array([0, 1, 0, 2])
>>> for p in multiset_permutations(a):
... p
...
[0, 0, 1, 2]
[0, 0, 2, 1]
[0, 1, 0, 2]
[0, 1, 2, 0]
[0, 2, 0, 1]
[0, 2, 1, 0]
[1, 0, 0, 2]
[1, 0, 2, 0]
[1, 2, 0, 0]
[2, 0, 0, 1]
[2, 0, 1, 0]
[2, 1, 0, 0]
if your permutations fit in the memory, you could store them in a set and thus only get the distinguishable permutations.
from itertools import permutations
a = [0, 1, 0, 2]
perms = set(permutations(a))
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