10/6/2023 0 Comments Python permutations of a listIf you want to restore the original order, use sorted (set (perm)), since permutations returns in lexicographical order (if your original string was in sorted order). Now len() is 2, hence the else branch of the if is taken. from itertools import permutations perm permutations AABB, 4) for i in set (perm): print (i) This may result in a different order to the printout. import itertools alist 1, 2, 3 perm list (x) for i in range (1, len (alist)+1) for x in itertools. The first for is the outer loop, the second for is the inner loop (read left to right in a list comprehension). 4 Answers Sorted by: 119 You don't need to know n in advance to use itertools.product > import itertools > s 'a', 'b', 'c', 'd', 'e', 'f' > list (itertools. Either the list is empty, in which case the only permutation is the list itself otherwise. This is trivial, since len(') is 1 then the first branch of the if condition is taken and the return value is itself (this complies with rule 2.1). As the comments say, you have mixed up the order of the for loops. Thats a straightforward recursion over the structure of the list. Now we try to 'mental execute' it in order to shed some light for the absolute beginner. I am trying to append all possible permutations of the list, candidates, to the list called res. class Solution (object): def permute (self, nums): res self.permuteHelper ( nums, 0, res) return res def permuteHelper (self, nums, start, results): helper method if start > len (nums): results.append (nums :) else: for i in range (start, len (nums)): nums i, nums start nums start, nums i self.permuteHelper (num. ![]() The program is a straightforward (possibly inefficient, see the remarks at the bottom of the page) implementation of the recursive algorithm. Trying to Understand the Code by 'Mental Execution' Such code is not mine, it is the original snippet the OP was asking for explanations (as a matter of fact, I am more inclined towards Lua than Python and I had to lookup the enumerate function in Python's documentation in order to fully understand the code). ![]() So you have to convert the given string into list using list() function. ![]() You can use it as follows Example import itertools perms list(itertools. Finally, the permutations variable is returned. You can use the itertools package's permutations method to find all permutations of a list in Python. This is done by invoking the insert_char() function inside a for loop. You can simply use itertools. Then we put back the first character (that was taken out) back in every possible position in every string in smaller_permutations. That will give us a list of permutations, which is stored in variable “smaller_permutations”. We strip out the first character and call this function recursively with the shortened string (s)). If “s” has two or more characters, that is when the bulk of the work lies. ![]() These two base cases are covered in the first two clauses. It turns out that when it pops the element of curr, the sub-list of res will also be popped. If ‘s” is either the empty string or a string containing only one character, then we simply return because there is either no permutation possible or only one permutation possible. 1 I am trying to append all possible permutations of the list, candidates, to the list called res. The list “permutations” keeps a running tally of all permutations created and this is the returned value from this function. Suppose I have the following list: x1,2,3,4,5,6 I want to generate permutations of the list in a specialized way: divide the list into blocks containing 2 consecutive elements: (1,2) (3,4) (5,6) and then permute each block independently. In the above function, permute, we pass the string to be permuted as an argument in variable “s”.
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