I am making product bundling using a list. Here is the list:

list_id_category = ['c1', 'c2', 'c3', 'c4', 'c5', 'c6', 'c7', 'c8', 'c9', 'c10']

Here are the condition that I need:

1. Each of bundling contain 3 products (ex: c1-c2-c3)
2. Can't only have 1 type of category in a bundling (ex: c1-c1-c1 => it doesn't allowed)
3. Each of bundling may have 2 same categories (ex: c1-c1-c2)
4. The sequence of each bundling doesn't make a difference (ex: c1-c2-c3 = c1-c3-c2 and soon)

Here is my code:

bundling_list = []
for i in list_id_category:
catefory1 = i
for j in list_id_category:
    if j != i:
        category2 = j
        bundle = i   '-'   j
    for k in list_id_category:
        if i != k and j != k:
            category3 = k
            bundle = i '-' j '-' k
            bundling_list.append(bundle)

But my code just fill the 1,2,3 condition. It doesn't fill the 4th condition. Anybody know how to fix it? Thank you.

CodePudding user response：

The following generator function will work without overgenerating and filtering:

def combos(lst):
    n = len(lst)
    for i in range(n-1):
        for j in range(i, n):
            for k in range(max(i 1, j), n):
                yield lst[i], lst[j], lst[k]

list(combos([1,2,3])]
# [(1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3), (1, 3, 3), (2, 2, 3), (2, 3, 3)]

list(map("-".join, combos(['c1', 'c2', 'c3', 'c4', 'c5', 'c6', 'c7', 'c8', 'c9', 'c10'])))
# ['c1-c1-c2', 'c1-c1-c3', 'c1-c1-c4', 'c1-c1-c5', 'c1-c1-c6', 
#  'c1-c1-c7', 'c1-c1-c8', 'c1-c1-c9', 'c1-c1-c10', 'c1-c2-c2', 
#  'c1-c2-c3', ... , 'c9-c10-c10']

CodePudding user response：

You can use itertools.combinations_with_replacement(iterable, r) to build the list of all combinations with replacement, then filter to remove the combinations in which a same element is repeated 3 times:

from itertools import combinations_with_replacement

list_id_category = ['c1', 'c2', 'c3', 'c4', 'c5', 'c6', 'c7', 'c8', 'c9', 'c10']

allbundlings  = combinations_with_replacement(list_id_category, 3)

bundlings = [b for b in allbundlings if len(set(b)) > 1]

I prefered if len(set(b)) > 1 to if not (b[0] == b[1] and b[1] == b[2]) to generalize