到目前为止我所拥有的几乎什么都没做
def dress_me(shirt, tie, suit): # if type(shirt) != list or type(tie) != list or type(suit) != list: # return None combinations = dress_me(shirt, tie, suit) for combo in combinations: print(combo)
到目前为止我所拥有的几乎什么都没做
def dress_me(shirt, tie, suit): # if type(shirt) != list or type(tie) != list or type(suit) != list: # return None combinations = dress_me(shirt, tie, suit) for combo in combinations: print(combo)
def dress_me(shirt, tie, suit):
if type(shirt) != list or type(tie) != list or type(suit) != list:
return None
return list(itertools.product(shirt, tie, suit))
演示:
>>> dress_me([1,2,3],[4,5,6],[7,8,9])
[(1, 4, 7), (1, 4, 8), (1, 4, 9), (1, 5, 7), (1, 5, 8), (1, 5, 9), (1, 6, 7), (1, 6, 8), (1, 6, 9), (2, 4, 7), (2, 4, 8), (2, 4, 9), (2, 5, 7), (2, 5, 8), (2, 5, 9), (2, 6, 7), (2, 6, 8), (2, 6, 9), (3, 4, 7), (3, 4, 8), (3, 4, 9), (3, 5, 7), (3, 5, 8), (3, 5, 9), (3, 6, 7), (3, 6, 8), (3, 6, 9)]
或者,为了完整起见,在没有额外功能的生成器样式中:
import itertools
for combination in itertools.product(shirts, ties, suits):
whatever_you_want_to_do(combination)
def dress_me(shirt, tie, suit):
all_combinations = []
for s in shirt:
for t in tie:
for su in suit:
all_combinations.append((s,t,su))
return all_combination
也许有一种更pythonic的方式来做到这一点:)
由于看起来您正在尝试提出递归解决方案,因此这是它的一般形式:
def all_perms(thing):
if len(thing) <=1:
yield thing
else:
for perm in all_perms(thing[1:]):
for i in range(len(perm)+1):
yield perm[:i] + thing[0:1] + perm[i:]
这适用于大多数类型的迭代。演示:
In [5]: list(all_perms(('shirt','tie','suit')))
Out[5]:
[('shirt', 'tie', 'suit'),
('tie', 'shirt', 'suit'),
('tie', 'suit', 'shirt'),
('shirt', 'suit', 'tie'),
('suit', 'shirt', 'tie'),
('suit', 'tie', 'shirt')]
递归一开始很难理解,但一般形式是:
if simplest_case:
return simplest_case
else:
#recurse
在这种情况下,return
被替换为yield
是为了制作一个对内存更友好的生成器。您仍然不应该期望这是性能最好的解决方案,但我将其包括在内是为了完整性,因为“使用 ITERTOOLS”除了 itertools 很酷之外,最终并没有教给您很多东西。