![]() ![]() 3, 2, 1 is a permutation of 1, 2, 3 and vice-versa. Is there a way to do this? My code is below: temp = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", A permutation refers to an arrangement of elements. ![]() I tried to use map, but am not sure if this works because my p圜harm doesn't result anything. I am not sure how to access each individual permutation at a time without going through another list, since I am not able to store it in a variable, because itertools.permutations returns a permutations object. I don't necessarily want to go through another list of permuted strings and try and access all of them. A common example (and a good story on how common usage. Basically what I want to do is after I get a permutation, I want to hash it using hashlib and check it against a certain value, which I am getting from a password.txt file before moving on to the next permutation. OP wants to permute bits in msg using iptable index permutation e.g., 56-th (zero-based) bit from msg should go to 0-th bit in the result (enc), 48-th bit from msg should go to 1-th bit in enc, 40-> 2, 32-> 3, etc. A permutation is a set of objects selected from a base set of elements, however here the order matters. I want to find all the possible permutations of them, and I am doing so with itertools.permutations. But since Pythons for loops are pretty slow, you actually get better performance for all but very tall matrices. See also which should be used for new code. Returns: outndarray Permuted sequence or array range. If x is an array, make a copy and shuffle the elements randomly. If you wish to implement permutation function without using. Parameters: xint or arraylike If x is an integer, randomly permute np.arange (x). ![]() Note that this has sub-optimal asymptotic time complexity, since the sort takes time O(n m log m) for an array of size m x n. The length of the list of the permutations gives n n factorial where n is the number of elements. I am using Python, and I have a list of strings (around 30). This produces a random permutation of each columns indices. ![]()
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