I've used a different example array in this case - your version will yield an identical output after performing the row/column swaps which makes it difficult to understand what's going on. You can use the same indexing approach to swap columns. In this particular case you could avoid the copy by using slice indexing, which returns a view rather than a copy: b = b # invert the row order Note that array indexing always returns a copy rather than a view - there's no way to swap arbitrary rows/columns of an array without generating a copy. Python indexing starts at 0 rather than 1) You can perform the swap in a one-liner using integer array indexing: a = np.array(, Furthermore, I need to do an arbitrary number of permutations (more than one). That doesn't work for me because the matrices are adjacency matrices (representing graphs), and I need to do the permutations which will give me a graph which is isomorphic with the original graph. ![]() numpy.shuffle and numpy.permutation seem to permute only the rows of the matrix (not the columns at the same time). Now, an incredibly naive (and memory costly) way of doing so might be: a2 = deepcopy(a1)īut, I would like to know if there is something more efficient that does this. Permute 2 2 2 7 Setup Permute 2 2 2 7 Permute 2 2 2 7 X 6 5 Last Updated on Octoby. That doesn't work for me because the matrices are adjacency matrices (representing graphs), and I need to do the permutations which will give me a. Set j = last-2 and find first j such that a =.Assuming that I have the following matrix/array: array(,Īnd I want to apply the following permutation: 1 -> 5 But, I would like to know if there is something more efficient that does this. So 6 is next larger and 2345(least using numbers other than 6) Find lexicogrpahically least way to extend the new aĬonsider example array state of for sorted Īfter 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5.Increase a by smallest feasible amount.Find largest j such that a can be increased.There are n! permutations at most and hasNextPermutation(.) runs in O(n) time complexity This is the asymptotically optimal way O(n*n!) of generating permutations after initial sorting. It was enough when I needed it, but it's no itertools.permutations by a long shot. This method is non-recursive, but it is slightly slower on my computer and xrange raises an error when n! is too large to be converted to a C long integer (n=13 for me). NewPermutation.append(available.pop(index)) This way the numbers 0 through n!-1 correspond to all possible permutations in lexicographic order. You have n choices for the first item, n-1 for the second, and only one for the last, so you can use the digits of a number in the factorial number system as the indices. I used an algorithm based on the factorial number system- For a list of length n, you can assemble each permutation item by item, selecting from the items left at each stage. ![]() Indices, indices = indices, indicesĪnd another, based on itertools.product: def permutations(iterable, r=None):įor indices in product(range(n), repeat=r): If len(elements) AB AC AD BA BC BD CA CB CD DA DB DC Use itertools.permutations from the standard library: import itertoolsĪdapted from here is a demonstration of how itertools.permutations might be implemented: def permutations(elements):
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