Permute 2.2 For Macos

Permute 2.2 For Macos Operating System
Permute 2.2 For Macos 10.13
Permute 2.2 For Macos High Sierra

Question or problem about Python programming:

I know about itertools, but it seems it can only generate permutations without repetitions.

For example, I’d like to generate all possible dice rolls for 2 dice. So I need all permutations of size 2 of [1, 2, 3, 4, 5, 6] including repetitions: (1, 1), (1, 2), (2, 1)… etc

Note that there are n! Permutations and it requires O(n) time to print a a permutation. Note: The above solution prints duplicate permutations if there are repeating characters in input string. Please see below link for a solution that prints only distinct permutations even if there are duplicates in input. Storyist 4 is a powerful writing environment for novelists and screenwriters. With Storyist, you can: Produce submission-ready manuscripts and screenplays. Storyist provides a rich text editor with. Things is the award-winning personal task manager that helps you achieve your goals. This all-new version has been rethought from the ground up: it’s got an all-new design, delightful new interactions, and powerful new features. OS X 10.9 or later (Intel only). Convert video, audio, and image files to different formats, modify them in various ways,.

If possible I don’t want to implement this from scratch

How to solve the problem:

Solution 1:

You are looking for the Cartesian Product.

In mathematics, a Cartesian product (or product set) is the direct product of two sets.

In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}.
itertools can help you there:

To get a random dice roll (in a totally inefficient way):

Solution 2:

You’re not looking for permutations – you want the Cartesian Product. For this use product from itertools:

Solution 3:

In python 2.7 and 3.1 there is a itertools.combinations_with_replacement function:

Solution 4:

In this case, a list comprehension is not particularly needed.

Given

Code

Details

Unobviously, Cartesian product can generate subsets of permutations. However, it follows that:

with replacement: produce all permutations n^r via product
without replacement: filter from the latter

Permutations with replacement, n^r

Permutations without replacement, n!

Consequently, all combinatoric functions could be implemented from product:

combinations_with_replacement implemented from product
combinations implemented from permutations, which can be implemented with product (see above)

Solution 5:

I think I found a solution using only lambdas, map and reduce.

Essentially I’m mapping a first lambda function that given a row, iterates the columnns

then this is used as the output of a new lambda function

which is mapped across all the possible rows

and then we reduce all the resulting lists into one.