Python iterate over array of arrays special way

Question

I have this array of arrays:

[[[1111, 2222, 3333], [4444, 5555, 6666]], [[7777]], [[8888, 9999]], [[0000]], [[1122, 2233]]]

I need to get all the possible combinations IN ORDER of the elements of the list.

Expected output for this example:

• [1111,2222,3333,7777,8888,9999,0000,1122,2233]
• [4444, 5555, 6666,7777,8888,9999,0000,1122,2233]

As you may see in this example the first array of the array is the only one who has two options.

I've tried with something like:

for i in array_input:
        for j in i:           
            print ', '.join([str(x) for x in j]),

but that is not getting me the expected output because is firstly iterating through the two array[0] options instead of picking one of them and filling the rest and then picking the other one and filling it.

So I want some loop that gets array[0][0],array[1][0],array[2][0] ... instead of: array[0][0],array[0][1],array[1][0] ...

thank you

Answer 1

Your elements are integers, so 0000 will be shown as 0 :

import itertools as it

lst = [[[1111, 2222, 3333], [4444, 5555, 6666]], [[7777]], [[8888, 9999]], [[0000]], [[1122, 2233]]]
out = [list(it.chain.from_iterable(x)) for x in it.product(*lst)]

First itertools.product generates a product of all arguments. Then you need itertools.chain.from_iterable to flatten the list of lists.

Answer 2

You could use itertools.combinations, lets simplify your example to the list `l = ['a', 'b', 'c', 'd'] then you can generate all combinations of length r with:(here ie r = 3)

import itertools as it

l = ['a', 'b', 'c', 'd']
for e in it.combinations(l, 3):
    print e
=>   ('a', 'b', 'c')
     ('a', 'b', 'd')
     ('a', 'c', 'd')
     ('b', 'c', 'd')

if you want to get all combinations just loop your r over the length of l

to overcome the problem with the two arrays you could maybe flatten your list ie with something like

new_list = [item for sublist in l for item in sublist]

full example:

import itertools as it

l = [[[1111, 2222, 3333], [4444, 5555, 6666]], [[7777]], [[8888, 9999]], [[0000]], [[1122, 2233]]]
l = [item for sublist in l for item in sublist]

for n in range(1, len(l)):
    for e in it.combinations(l, n):
        print e

Answer 3

You can just do the following ...

import operator
reduce(operator.__add__, zip(*array_input)[0])

The zip(*array_input) basically unzips the array to ...

[([1111, 2222, 3333], [7777], [8888, 9999], [0], [1122, 2233])]

zip(*array_input)[0] simply remove one level of nesting ...

Finally the reduce operator adds the lists to a single list. You could also have done

reduce(lambda m, n: m+n, zip(*array_input)[0]) but I don't like using lambda functions here ...

Answer 4

list1 = [[[1111, 2222, 3333], [4444, 5555, 6666]], [[7777]], [[8888, 9999]], [[0000]], [[1122, 2233]]]
a = []
for i in list1:
 for j in i:
  for k in j:
   a.append(k)
print(a)