寻求优雅的Python骰子迭代

Question

是否有一种优雅的方式来迭代最多五个骰子的可能骰子？

我想替换这个hacky Python：

self.rolls[0] = [str(a) for a in range(1,7)]
self.rolls[1] = [''.join([str(a), str(b)])
                 for a in range(1, 7)
                 for b in range(1, 7)
                 if a <= b]
self.rolls[2] = [''.join([str(a), str(b), str(c)])
                 for a in range(1, 7)
                 for b in range(1, 7)
                 for c in range(1, 7)
                 if a <= b <= c]
self.rolls[3] = [''.join([str(a), str(b), str(c), str(d)])
                 for a in range(1, 7)
                 for b in range(1, 7)
                 for c in range(1, 7)
                 for d in range(1, 7)
                 if a <= b <= c <= d]
self.rolls[4] = [''.join([str(a), str(b), str(c), str(d), str(e)])
                 for a in range(1, 7)
                 for b in range(1, 7)
                 for c in range(1, 7)
                 for d in range(1, 7)
                 for e in range(1, 7)
                 if a <= b <= c <= d <= e]

Answer 1

您可以使用itertools'combinations_with_replacement。

例如，使用3个4面骰子（仅因为输出太大）：

>>> from itertools import combinations_with_replacement

>>> dice = 3
>>> faces = 4
>>> list(combinations_with_replacement(range(1, faces+1), dice))
[(1, 1, 1),
 (1, 1, 2),
 (1, 1, 3),
 (1, 1, 4),
 (1, 2, 2),
 (1, 2, 3),
 (1, 2, 4),
 (1, 3, 3),
 (1, 3, 4),
 (1, 4, 4),
 (2, 2, 2),
 (2, 2, 3),
 (2, 2, 4),
 (2, 3, 3),
 (2, 3, 4),
 (2, 4, 4),
 (3, 3, 3),
 (3, 3, 4),
 (3, 4, 4),
 (4, 4, 4)]