![]() No need to create the list in memory if you just want to push it to a file. 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. Note New code should use the permutation method of a Generator instance instead please see the Quick Start. I am trying to append all possible permutations of the list, candidates, to the list called res. ![]() If x is a multi-dimensional array, it is only shuffled along its first index. Just use itertools.permutations and write the objects to the file on-the-fly. random.permutation(x) Randomly permute a sequence, or return a permuted range. Second, if you just want to "save" the items to some kind of massive file, you do NOT need to realize the list in memory. Given that I'm not too sure what you would do with a saved list of 6B things, but I think you have 2 strategies if you want to go forward.įirst, you could reduce the size of the things in the list by substituting something like a numpy unit8 for each item, which would reduce the size of the resulting list by a LOT, but you would not have the format you want. How can I make this code more memory efficient so that it doesn't crash for a large list? If I need to, I can run the code separately for each item separator (commas, periods, directly joined). Names_t = list(set(names_t)) # remove all duplicates Perms = filter(None, perms) # remove empty lists, to prevent two symbols in a row or a symbol before/after for i in range(inarr.rank) for i in range(len(inarr)): perm list(permutations(index)) permnumber 0 for. Perms = # remove empty nested elements, to prevent two symbols in a row or a symbol before/after Lets discuss certain ways in which one can perform the task of getting all the permutations of N lists. def permutations (nums): if len (nums) 0: return if len (nums) 1: return nums result for i in range (len (nums)): m nums i remlist nums :i + nums i+1: for p in permutations (remlist): result.append ( m + p) return result nums 1,2,3,4 print ('Original list:') print (nums) print ('Permutations of the m. ![]() Even when using a machine with a large amount of RAM. But then, this fails with a MemoryError when removing empty elements after converting to a list. ![]() I've used itertools.permutations to create an itertools.chain ( perms). # Also creating groups of length 1, 2, and 3 # Avoiding a symbol at the beginning or end of a resulting string 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). for all the possible groups of 4, order is important As the comments say, you have mixed up the order of the for loops. # ideal output would be a list that contains strings like the following:Īpple-banana-bread (no dashes before or after!)Īpple-banana (by combining with the first item in the list, I also get shorter groups but need to delete empty items before joining) This should work: import itertools def great (l, n): function that counts the permutations of list l with n great indices def countofgreat (k): function that counts the great indices of list k c1 because first element is always great for i in range (1,len (k)): if k i>max (k :i): if the value is greater than. The goal is to pair items in groups of four or less, separated by 1) periods, 2) dashes, and 3) without any separation. Permutations differ from combinations, which are selections of some members of a set regardless of order.I'm trying to run a very large permutation using Python. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. from itertools import permutations for i in permutations (yourlist, 3): print i. Mathematical version of an order change Each of the six rows is a different permutation of three distinct balls
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |