python - 所有正整数元组

Question

如何创建一个生成器来返回所有正整数组合的元组，例如生成三元组。

(1, 1, 1)
(2, 1, 1)
(1, 2, 1)
(1, 1, 2)
(2, 2, 1)
(2, 1, 2)
(2, 2, 2)
(3, 2, 2)
(2, 3, 2)
# and so on...

Answer 1

此代码使用与 Paul Hankin 类似的方法，但更通用，因为它会生成任何所需宽度的元组，而不仅仅是 3。

from itertools import combinations, count

def compositions(num, width):
    m = num - 1
    first, last = (-1,), (m,)
    for t in combinations(range(m), width - 1):
        yield tuple(v - u for u, v in zip(first + t, t + last))

def ordered_compositions(width):
    for n in count(width):
        yield from compositions(n, width)

# test

width = 3
for i, t in enumerate(ordered_compositions(width), 1):
    print(i, t)
    if i > 30:
        break

输出

1 (1, 1, 1)
2 (1, 1, 2)
3 (1, 2, 1)
4 (2, 1, 1)
5 (1, 1, 3)
6 (1, 2, 2)
7 (1, 3, 1)
8 (2, 1, 2)
9 (2, 2, 1)
10 (3, 1, 1)
11 (1, 1, 4)
12 (1, 2, 3)
13 (1, 3, 2)
14 (1, 4, 1)
15 (2, 1, 3)
16 (2, 2, 2)
17 (2, 3, 1)
18 (3, 1, 2)
19 (3, 2, 1)
20 (4, 1, 1)
21 (1, 1, 5)
22 (1, 2, 4)
23 (1, 3, 3)
24 (1, 4, 2)
25 (1, 5, 1)
26 (2, 1, 4)
27 (2, 2, 3)
28 (2, 3, 2)
29 (2, 4, 1)
30 (3, 1, 3)
31 (3, 2, 2)

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compositions 的算法源自维基百科文章 compositions 中用于计算作品数量的技术。 .这本质上是众所周知的 Stars and Bars 的变体。技术。