6/29/2023 0 Comments Permute a matrix python![]() Right: movie plot (“Predestination”, 2014). Three images from an Internet search for “too complicated”. Let me emphasize that the “clever” approach would never be used in practice - it’s just way too complicated. I used the built-in NumPy argsort() function to get a permutation of (3, 2, 0, 1, 5, 6, 4) which when applied to the array will give an array that’s sorted in alphabetical order which makes it easy to tell if the demo code is working correctly or not. I coded a demo where I set an array = (“cow”, “dog”, “bat”, “ant”, “gnu”, “elk”, “fox”). There is a slightly different version of this algorithm that reduces the number iterations in the while-loop at the expense of having to track changes by modifying the perm parameter. The algorithm is quite subtle and the only way to really understand it (for me at least) is to walk through the code line by line and examine the values of i, j, and the source array. ![]() An old interview question asks how to apply a permutation to an array in place, without using a result array. Notice the code above creates a result array and then copies into that array. Set roundup to True to get a binary matrix, i.e. Start your free 7-days trial now NumPy Random Generators permutation() method return a new array with the values shuffled. The best way to implement this as a function is very simple. ![]() In words, “the value in array cell 4 goes to result cell 0, the value in array cell 2 goes to result cell 1, the value in array cell 0 goes to result cell 2, the value in array cell 1 goes to result cell 3, the value in array cell 3 goes to result cell 4”. scalebool, optional The selector to turn on and off the scaling. permutebool, optional The selector to define whether permutation of A is also performed prior to scaling. Suppose you have an array of (“C”, “E”, “A”, “B”, “D”) and the problem is to apply the permutation to the array. Parameters: A(n, n) arraylike Square data matrix for the balancing. For example, one permutation of order 5 is (4, 2, 0, 1, 3). Set j = last-2 and find first j such that a =.A permutation of order n is an arrangement of the numbers 0 through n-1. 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. For an n-D array, if axes are given, their order indicates how the axes are permuted (see Examples). Indices, indices = indices, indicesĪnd another, based on itertools.product: def permutations(iterable, r=None):įor indices in product(range(n), repeat=r): For a 2-D array, this is the standard matrix transpose. 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): The permutation matrices can be constructed: > Pr cscmatrix( (np.ones(4), (lu.permr, np.arange(4)))) > Pc cscmatrix( (np.ones(4), (np.arange(4), lu.permc))) We can reassemble the original matrix: > (Pr.T (lu.L lu.U) Pc.T).A array ( 1., 2., 0., 4., 1., 0., 0., 1., 1., 0., 2., 1., 2., 2., 1., 0. ![]()
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