生成所有三个数的集合，总和为 12 python

Question

I would like to generate all the tuples of three numbers which add up to 12. For example: ((12, 0, 0), (11, 1, 0)) , and so on. How would you generate this entire list in python? I tried:

x = []
for a in range(0, 13):
    for b in range(0, 13):
        for c in range(0, 13):
            x.append((a, b, c))

I also tried to only append to the list if the sum was 12, but I feel like this is a very inefficient way to complete the task, because it loops over way more iterations than necessary.

Answer 1

You can use itertools.product for this to avoid nested loops (based on @User 12692182's answer )

from itertools import product

[(a, b, 12-a-b) for a, b in product(range(13), repeat=2) if 12-a-b >=0]

Results:

[(0, 0, 12), (0, 1, 11), (0, 2, 10), (0, 3, 9), (0, 4, 8), (0, 5, 7), (0, 6, 6), (0, 7, 5), (0, 8, 4), (0, 9, 3), (0, 10, 2), (0, 11, 1), (0, 12, 0), (1, 0, 11), (1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7), (1, 5, 6), (1, 6, 5), (1, 7, 4), (1, 8, 3), (1, 9, 2), (1, 10, 1), (1, 11, 0), (2, 0, 10), (2, 1, 9), (2, 2, 8), (2, 3, 7), (2, 4, 6), (2, 5, 5), (2, 6, 4), (2, 7, 3), (2, 8, 2), (2, 9, 1), (2, 10, 0), (3, 0, 9), (3, 1, 8), (3, 2, 7), (3, 3, 6), (3, 4, 5), (3, 5, 4), (3, 6, 3), (3, 7, 2), (3, 8, 1), (3, 9, 0), (4, 0, 8), (4, 1, 7), (4, 2, 6), (4, 3, 5), (4, 4, 4), (4, 5, 3), (4, 6, 2), (4, 7, 1), (4, 8, 0), (5, 0, 7), (5, 1, 6), (5, 2, 5), (5, 3, 4), (5, 4, 3), (5, 5, 2), (5, 6, 1), (5, 7, 0), (6, 0, 6), (6, 1, 5), (6, 2, 4), (6, 3, 3), (6, 4, 2), (6, 5, 1), (6, 6, 0), (7, 0, 5), (7, 1, 4), (7, 2, 3), (7, 3, 2), (7, 4, 1), (7, 5, 0), (8, 0, 4), (8, 1, 3), (8, 2, 2), (8, 3, 1), (8, 4, 0), (9, 0, 3), (9, 1, 2), (9, 2, 1), (9, 3, 0), (10, 0, 2), (10, 1, 1), (10, 2, 0), (11, 0, 1), (11, 1, 0), (12, 0, 0)]

Answer 2

You would have to use subtraction, to avoid not going above 12, or below it. First, you would use a range of 0 to 12, independent from everything else. Then, you would use another range, however with this one, to avoid going over, you need to limit the second part of the iteration to 12-(the first number). The last number will be completely dependant on the others, giving you 12-(the first number)-(the second number), its purpose only to fill the gap between the first number, the second number, and 12. In python, this would look like:

x = []
for a in range(0, 13):
    for b in range(0, (13-a)):
        c = 12-a-b
        x.append((a, b, c))