How can I solve this problem using itertools?

Question

Hello I have tried using this tool:

list(combinations('01', 3))

but I get this result:

[]

I would like to have these results:

Separately, I would also like to have different cases.

for instance, given 111 I expect these results:

Is it possible to do these two things using itertools?

The keyword you're missing for your searches is "partitioning". You haven't sufficiently described your problem, but it looks as if you want to partition the outputs into groups and sum the groups. — Prune, Oct 07 '19 at 17:04
Seems like something similar to this? https://stackoverflow.com/questions/39192777/how-to-split-a-list-into-n-groups-in-all-possible-combinations-of-group-length-a — DenverCoder1, Oct 07 '19 at 17:21

score 0 · Answer 1 · answered Oct 07 '19 at 18:09

Here's a solution which is partially using the solution in this answer.

The partition function from alexis' answer (see link) returns the list partitioned into groups (i.e. [[[1,1,1]],[[1,1],[1]],[[1],[1,1]],[[1],[1],[1]]]
To get the numbers in your question, I added a part that stores only the sum of each inner list
list(dict.fromkeys(...)) is used to remove duplicates

Code

def partition(collection):
    if len(collection) == 1:
        yield [ collection ]
        return

    first = collection[0]
    for smaller in partition(collection[1:]):
        # insert `first` in each of the subpartition's subsets
        for n, subset in enumerate(smaller):
            yield smaller[:n] + [[ first ] + subset]  + smaller[n+1:]
        # put `first` in its own subset 
        yield [ [ first ] ] + smaller


elements = list([1,1,1])

lst = list(dict.fromkeys([ tuple([sum(p[1][i]) for i in range(len(p[1]))]) for p in enumerate(partition(elements), 1) ]))

print(sorted(lst))

Output:

[(1, 1, 1), (1, 2), (2, 1), (3,)]

If you want it to print as `111 12 21 3`, you can do `print(' '.join([''.join([str(i) for i in j]) for j in lst]))` — DenverCoder1, Oct 07 '19 at 18:20

pylang · Answer 2 · 2019-10-09T23:02:20.043

You seem to be asking two separate questions. Based on your results:

000 001 011 111 101 100 010

Use a permutations_with_replacement combinatoric via itertools.product.

111 12 21 3

Use a partitions algorithm via more_itertools.partitions.

Given

import itertools as it

import more_itertools as mit

Code

# 1 - Permutation w Replacement
>>> list(it.product("01", repeat=3))
[('0', '0', '0'),
 ('0', '0', '1'),
 ('0', '1', '0'),
 ('0', '1', '1'),
 ('1', '0', '0'),
 ('1', '0', '1'),
 ('1', '1', '0'),
 ('1', '1', '1')]

# 2 - Partitions
>>> list(mit.partitions("111"))
[[['1', '1', '1']],
 [['1'], ['1', '1']],
 [['1', '1'], ['1']],
 [['1'], ['1'], ['1']]]

Details

To achieve your specific results, use list comprehensions:

# 1
>>> ["".join(x) for x in it.product("01", repeat=3)]

['000', '001', '010', '011', '100', '101', '110', '111']

# 2
>>> [[sum(map(int, x)) for x in sub] for sub in mit.partitions("111")]

[[3], [1, 2], [2, 1], [1, 1, 1]]

more_itertools is a third-party package. Install with > pip install more-itertools.

How can I solve this problem using itertools?

2 Answers2

Code