Generating permutations with repetitions

问题

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

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

回答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:

import itertools
x = [1, 2, 3, 4, 5, 6]
[p for p in itertools.product(x, repeat=2)]
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), 
 (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 
 (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), 
 (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

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

import random
random.choice([p for p in itertools.product(x, repeat=2)])
(6, 3)

回答2:

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

from itertools import product
for roll in product([1, 2, 3, 4, 5, 6], repeat = 2):
    print(roll)

回答3:

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

>>> list(itertools.combinations_with_replacement([1, 2, 3, 4, 5, 6], 2))
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), 
 (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),
 (5, 5), (5, 6), (6, 6)]

回答4:

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

Given

import itertools as it


iter_ = range(1, 7)
r = 2

Code

list(it.product(iter_, repeat=r))

---

Details

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

• with replacement: one can produce all permutations n**r via product
• without replacement: one can filter from the latter

Permutations with replacement, n**r

[x for x in it.product(iter_, repeat=r)]

Permutations without replacement, n!

[x for x in it.product(iter_, repeat=r) if len(set(x)) == r]

# Equivalent
list(it.permutations(iter_, r))  

回答5:

First, you'll want to turn the generator returned by itertools.permutations(list) into a list first. Then secondly, you can use set() to remove duplicates Something like below:

def permutate(a_list):
    import itertools
    return set(list(itertools.permutations(a_list)))

来源：https://stackoverflow.com/questions/3099987/generating-permutations-with-repetitions