Python looping combinations of 8 objects into 3 groups, 3-3-2

Question

Let's say I have a list of 8 objects, numbered that 1-8.

The objects are put into three boxes, 3 in one box, 3 in another box, 2 in the last box. By mathematics, there are 8C3*5C3=560 ways to do this. I want to loop through there 560 items. Is there any way in Python to do so?

The result should look like this:

list=['12','345',678'], ['12','346','578'], ..., etc.

Note that ['12','345','678'] and ['12','354',876'] are considered the same for this purpose.

I want to make a for-loop this list. Is there any way in Python to do so?

Here is the solution I get, but it seems ugly.

import itertools
for c1,c2 in itertools.combinations(range(8),2):
            l2=list(range(8))
            l2.pop(c2)
            l2.pop(c1)
            for c3,c4,c5 in itertools.combinations(l2,3):
                l3=l2[:]
                l3.remove(c5)
                l3.remove(c4)
                l3.remove(c3)
                c6,c7,c8=l3
                print(c1,c2,c3,c4,c5,c6,c7,c8)

You could itertools http://docs.python.org/2/library/itertools.html and add some of your own logic. — Pramod, May 02 '13 at 05:15

jamylak · Accepted Answer · 2013-05-02T06:21:08.137

def F(seq, parts, indexes=None, res=[], cur=0):
    if indexes is None: # indexes to use for combinations
        indexes = range(len(seq))

    if cur >= len(parts): # base case
        yield [[seq[i] for i in g] for g in res]
        return    

    for x in combinations(indexes, r=parts[cur]):
        set_x = set(x)
        new_indexes = [i for i in indexes if i not in set_x]
        for comb in F(seq, parts, new_indexes, res=res + [x], cur=cur + 1):
            yield comb

it = F('12345678', parts=(2,3,3))
for i in range(10):
    print [''.join(g) for g in next(it)]

['12', '345', '678']
['12', '346', '578']
['12', '347', '568']
['12', '348', '567']
['12', '356', '478']
['12', '357', '468']
['12', '358', '467']
['12', '367', '458']
['12', '368', '457']
['12', '378', '456']

Another example:

for c in F('1234', parts=(2,2)):
    print [''.join(g) for g in c]

['12', '34']
['13', '24']
['14', '23']
['23', '14']
['24', '13']
['34', '12']

I'm getting IndexError: list index out of range when running the code. — Pramod, May 02 '13 at 07:34

score 0 · Answer 2 · edited May 23 '17 at 12:33

You could just permute all your 8 values (like shown on previous answers). for that use this previous answer (also on the following code).

Then assign each combination as a tuple, so they can be hashed and unique, for that you'll have to order them, so they can also be compare uniquely.

def all_perms(elements):
    if len(elements) <=1:
        yield elements
    else:
        for perm in all_perms(elements[1:]):
            for i in range(len(elements)):
                #nb elements[0:1] works in both string and list contexts
                yield perm[:i] + elements[0:1] + perm[i:]


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

a = {}
for i in all_perms(v):
    k = (tuple(sorted([i[0],i[1]])) , tuple(sorted([i[2],i[3],i[4]])) , tuple(sorted([i[5],i[6],i[7]])))

    if k not in a:
        a[k] = [str(i[0])+str(i[1]), str(i[2])+str(i[3])+str(i[4]), str(i[5])+str(i[6]) + str(i[7])]

x = 0
for i in a.values():
    print x, i
    x+=1

For your example on 8 values, this gives 560 combinations.

he removed it for some reason, it's back in the question now — jamylak, May 02 '13 at 06:18

HennyH · Answer 3 · 2013-05-02T05:30:07.143

-1

l would be a list of eight objects, in this example strings:

l = ["O1","02","03","04","04","06","07","08"]
for group in [l[:3],l[3:6],l[6:]]: #get 3 slices of the list into 3's and a 2
    print(group)

Produces:

>>> 
['O1', '02', '03']
['04', '04', '06']
['07','08']

edited May 02 '13 at 05:30

answered May 02 '13 at 05:24

HennyH

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Python looping combinations of 8 objects into 3 groups, 3-3-2

3 Answers3