itertools and permutations - 合并两个列表以进行所有可能的组合

Question

I have two lists: a and b . a is a list with three or more strings, while b is a list of separators.

I need to generate all the possible combinations of a and than merge the result with all the possible combinations of b (See the example for better understanding).

I want all three items from list a to be in each line of the result, in different order (in order to cover or the possible combinations). Items from list a can not be repeated, while items of list b can be repeated.

I have no idea of how to make it works, so your help would be so appreciated.

Thanks!

a = ["test1", "test2", "test3"]
b = ["_", "-", ".", ""]

# Expected result
test1_test2_test3
test2_test1_test3

. . . .

test1_test2-test3
test1_test2.test3

. . . .

test3test2_test1

. . . .

Edit:

from itertools import permutations, combinations, product, chain

a = ["test1", "test2", "test3"]
b = ["_", "-", ".", ""]

a_permutations = permutations(a)
b_permutations = product(b, repeat=len(a)-1)

for ap in a_permutations:
    for bp in b_permutations:
        result = ''.join([''.join(word) for word in zip(ap, bp)]) + ap[-1]
        print result

For some reason it doesn't work properly..

Answer 1

Since you show no code of your own, I'll just give you some ideas. If you want more details, edit your question to add some more of your own work then ask for details.

As I look at your desired list it seems that you want the 3 items from list a , without repetition, and two items from list b , perhaps with repetition, and to interleave the items from the two lists. I will also assume that the number of items from list b is one less than the length of the given list a . (If those assumptions are not what you want, please state that clearly in your question.)

"Combinations" with different orders without repetition are called permutations. You can get all the permutations of list a with the expression

itertools.permutations(a)

and thus all possible collections of items from list b with repetitions with

itertools.product(b, repeat=len(a)-1)

For each pair of items from each collection, you can interleave those items by using a zip() operation on the two collections. Since the first collection is longer by 1 than the second, that final item will be left out, but you can just add that on to the result.

I could show you a one-liner that does that interleaving but I should leave something for you to do. Use those ideas to try your own code. If you do not completely succeed, show us an attempt and I will be glad to show you some final code.

Note that following my ideas will result in a list in a different order than what you show. The list I would get would start with

test1_test2_test3
test1_test3_test2
test2_test1_test3
test2_test3_test1
test3_test1_test2
test3_test2_test1
test1_test2-test3
test1_test3-test2
test2_test1-test3
test2_test3-test1
test3_test1-test2
test3_test2-test1
test1_test2.test3

and so on. This order fixes the separators then permutes the items from list a . You could do it the other way round, of course, and get a different order.

---

Now that you have shown some work of your own, here is my code. This assumes that lists a and b have no repetitions. This also does a trick to simplify the handling of the problem that there is one more item than separator--I just add the empty separator "" to the end of a tuple of separators. The two lengths (of items and of separators) are then equal and the empty separator at the end of each string does nothing.

import itertools

a = ["test1", "test2", "test3"]
b = ["_", "-", ".", ""]

for separators in itertools.product(b, repeat=len(a) - 1):
    for items in itertools.permutations(a):
        print("".join(v for pair in zip(items, separators + ("",)) for v in pair))

This code gives the printout

test1_test2_test3
test1_test3_test2
test2_test1_test3
test2_test3_test1
test3_test1_test2
test3_test2_test1
test1_test2-test3
test1_test3-test2
test2_test1-test3
test2_test3-test1
test3_test1-test2
test3_test2-test1
test1_test2.test3
test1_test3.test2
test2_test1.test3
test2_test3.test1
test3_test1.test2
test3_test2.test1
test1_test2test3
test1_test3test2
test2_test1test3
test2_test3test1
test3_test1test2
test3_test2test1
test1-test2_test3
test1-test3_test2
test2-test1_test3
test2-test3_test1
test3-test1_test2
test3-test2_test1
test1-test2-test3
test1-test3-test2
test2-test1-test3
test2-test3-test1
test3-test1-test2
test3-test2-test1
test1-test2.test3
test1-test3.test2
test2-test1.test3
test2-test3.test1
test3-test1.test2
test3-test2.test1
test1-test2test3
test1-test3test2
test2-test1test3
test2-test3test1
test3-test1test2
test3-test2test1
test1.test2_test3
test1.test3_test2
test2.test1_test3
test2.test3_test1
test3.test1_test2
test3.test2_test1
test1.test2-test3
test1.test3-test2
test2.test1-test3
test2.test3-test1
test3.test1-test2
test3.test2-test1
test1.test2.test3
test1.test3.test2
test2.test1.test3
test2.test3.test1
test3.test1.test2
test3.test2.test1
test1.test2test3
test1.test3test2
test2.test1test3
test2.test3test1
test3.test1test2
test3.test2test1
test1test2_test3
test1test3_test2
test2test1_test3
test2test3_test1
test3test1_test2
test3test2_test1
test1test2-test3
test1test3-test2
test2test1-test3
test2test3-test1
test3test1-test2
test3test2-test1
test1test2.test3
test1test3.test2
test2test1.test3
test2test3.test1
test3test1.test2
test3test2.test1
test1test2test3
test1test3test2
test2test1test3
test2test3test1
test3test1test2
test3test2test1

Answer 2

This can be achived in the pythonic way using permutations , product and combinations module from itertools below is the working snippet:

from itertools import permutations, combinations, product

a = ["test1", "test2", "test3"]
b = ["_", "-", ".", ""]

for com in combinations(b, len(a) - 1):
    for per in product(com, repeat=len(a) - 1):
        for ear_per in permutations(a):
            out = ''.join(map(''.join, zip(list(ear_per[:-1]), per)))
            print(out + list(ear_per)[-1])