在多个列表中进行迭代时，跳过特定列表中的迭代

Question

这个问题与这个问题有些相似，但不同之处在于它包含多个列表：

我有三个清单：

a = [1, 2, 3, 4, 5, 6, 7]
b = ['ball', 'cat', 'dog', 'elephant', 'baboon', 'crocodile']
c = [6, 3, 5, 4, 3, 2, 1]

我正在按如下方式迭代列表：

for (x, y, z) in itertools.product(a, b, c):
print("The combination is: " + str(x) + ", " + y  + ", " str(z))

我想添加一个用以下伪代码表示的条件：

for (x, y, z) in itertools.product(a, b, c):
if x != z:
print("The combination is: " + str(x) + ", " + y  + ", " str(z))
if x == z:
skip to the next element in list "a" without skipping elements of "b" & "c" # this is pseudo code

我怎样才能做到这一点？itertools中有没有一个函数可以用来实现这一点？

Answer 1

假设我正确地解释了你的问题(你想跳到下一个a元素，而不重置product(a, b, c)的b/c周期中的偏移量(，这应该会奏效。它并不像你喜欢的那样专注于itertools，但它确实有效，而且速度不应该太慢。据我所知，itertools没有任何东西可以满足您的要求，至少不是开箱即用的。大多数函数(islice是一个主要的例子(只是跳过元素，这仍然会消耗时间。

from itertools import product
def mix(a, b, c):
a_iter = iter(a)
for x in a_iter:
for y, z in product(b, c):
while x == z:
x = next(a_iter)
yield x, y, z

输出：

>>> a = [1, 2, 3, 4, 5, 6, 7]
>>> b = ['ball', 'cat', 'dog', 'elephant', 'baboon', 'crocodile']
>>> c = [6, 3, 5, 4, 3, 2, 1]
>>> pprint(list(mix(a, b, c)))
[(1, 'ball', 6),
(1, 'ball', 3),
(1, 'ball', 5),
(1, 'ball', 4),
(1, 'ball', 3),
(1, 'ball', 2),
(2, 'ball', 1),
(2, 'cat', 6),
(2, 'cat', 3),
(2, 'cat', 5),
(2, 'cat', 4),
(2, 'cat', 3),
(3, 'cat', 2),
(3, 'cat', 1),
(3, 'dog', 6),
(4, 'dog', 3),
(4, 'dog', 5),
(5, 'dog', 4),
(5, 'dog', 3),
(5, 'dog', 2),
(5, 'dog', 1),
(5, 'elephant', 6),
(5, 'elephant', 3),
(6, 'elephant', 5),
(6, 'elephant', 4),
(6, 'elephant', 3),
(6, 'elephant', 2),
(6, 'elephant', 1),
(7, 'baboon', 6),
(7, 'baboon', 3),
(7, 'baboon', 5),
(7, 'baboon', 4),
(7, 'baboon', 3),
(7, 'baboon', 2),
(7, 'baboon', 1),
(7, 'crocodile', 6),
(7, 'crocodile', 3),
(7, 'crocodile', 5),
(7, 'crocodile', 4),
(7, 'crocodile', 3),
(7, 'crocodile', 2),
(7, 'crocodile', 1)]