列表列表中的排列

Question

So, say I have a list of lists like

l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

How do I get all possible permutations with the restriction that I can only pick 1 item per list? Meaning that 147 or 269 would be possible permutations, whereas 145 would be wrong since 4 and 5 are in the same list. Also, how does this work for a list containing any number of lists?

Answer 1

This works for me in Python 2.7 and 3.5

import itertools
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(list(itertools.product(*l)))

it returns

[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9)]

Answer 2

this worked in python 3, see the comment on the last line for python 2

l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
row1, row2, row3 = l

# get all the permutations as lists [1,4,7], etc.
permutations = ([x, y, z] for x in row1 for y in row2 for z in row3)

# get strings to have a more easily readable output
permutation_strings = (''.join(map(str, permutation))
                       for permutation in permutations)

print(*permutation_strings)
# in python2 you can use: print list(permutation_strings)

Answer 3

I wouldn't call what you are looking for permutations, but the following recursive algorithm should return what I assume you would like to see

def get_all_possibilities(S, P=[]):
    if S == []:
        return P

    s = S[0]
    if P == []:
        for x in s:
            P.append(str(x))
        return get_all_possibilities(S[1:], P)
    else:
        new_P = []
        for x in s:
            for p in P:
                new_P.append(p + str(x))

        return get_all_possibilities(S[1:], new_P)

print get_all_possibilities([[1, 2, 3], [4, 5, 6], [7, 8, 9]])

My output was the following 27 items which could later be converted back to integers if you like;

['147', '247', '347', '157', '257', '357', '167', '267', '367', '148', '248', '348', '158', '258', '358', '168', '268', '368', '149', '249', '349', '159', '259', '359', '169', '269', '369']

Answer 4

You could use recursion.

l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

def permutate(w, l):
    for x in l[0]:
        if len(l) > 1:
            permutate(w + str(x), l[1:])
        else:
            print w + str(x)

permutate("", l)

Answer 5

You can use itertools for this!

from itertools import product
l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print(list(product(*l)))

A few things to note:

• I'm passing *l instead of simply l as product expects iterables as arguments, not a list of iterables; another way to write this would have been:

   product([1, 2, 3], [4, 5, 6], [7, 8, 9]) 

  ie, passing every list as a single argument. *l unpacks the elements of l into arguments for you.

• product does not return a list, but a generator. You can pass that to anything that expects an iterable. "Printing" the generator would not be helpful (you would not see the content of the resulting list, but <itertools.product object...> which is only mildly interesting); this is why I'm forcing a conversion to a list using list()

• using print() with parentheses allows this code to be compatible with Python 2 & 3.