Question

我需要构建一个包含2个元组的函数，并将它们配对为所有可能的对。

例如，我需要学习元组：

first_tuple = (1, 2)
second_tuple = (4, 5)

结果必须是：

((1, 4), (4, 1), (1, 5), (5, 1), (2, 4), (4, 2), (2, 5), (5, 2))

Answer 1

您可以使用itertools.product和itertools.chain，其想法是接受所有可能的产品订购，并且由于元组的大小为2，因此只需要翻转它们即可：

>>> from itertools import product, chain
>>> first_tuple = (1, 2)
>>> second_tuple = (4, 5)
>>> half = list(product(first_tuple, second_tuple))
>>> half
[(1, 4), (1, 5), (2, 4), (2, 5)]
>>> list(chain(half, map(lambda x: (x[1], x[0]), half)))
[(1, 4), (1, 5), (2, 4), (2, 5), (4, 1), (5, 1), (4, 2), (5, 2)]

对于任意的元组大小，您可以使用（@ Aran-Fei想法）：

[perm for tup in half for perm in itertools.permutations(tup)]

Answer 2

您首先要使用itertools.product创建初始配对，然后使用该配对创建另一个配对，其中元组元素将被交换

from itertools import product
first_tuple = (1, 2)
second_tuple = (4, 5)

#First pairing
prod_1 = list(product(first_tuple, second_tuple))

#Pairing with swapped positions of tuple
prod_2 = [(t[1], t[0]) for t in prod_1]

#Final output
res = prod_1 + prod_2
print(res)

输出将为

[(1, 4), (1, 5), (2, 4), (2, 5), (4, 1), (5, 1), (4, 2), (5, 2)]

Answer 3

from itertools import chain,product

first_tuple = (1, 2)
second_tuple = (4, 5)

combined = list(chain(*[[(f,s),(s,f)] for f in first_tuple for s in second_tuple]))
print (combined)

输出：

[(1, 4), (4, 1), (1, 5), (5, 1), (2, 4), (4, 2), (2, 5), (5, 2)]

。

itertools.product（）

此工具计算输入可迭代项的笛卡尔积。它是   等效于嵌套的for循环。例如，product（A，B）返回   与（A中x等于B中y等于（（x，y））。

要获得所需列表的第一部分是：

list(product(first_tuple, second_tuple)) # (1, 4), (4, 1), (1, 5), (5, 1)

第二部分足以反转：

list(product(second_tuple, first_tuple)) # (2, 4), (4, 2), (2, 5), (5, 2)

。

combined1 = list(product(first_tuple, second_tuple)) + list(product(second_tuple, first_tuple))
print (combined1)

输出：

[(1, 4), (1, 5), (2, 4), (2, 5), (4, 1), (4, 2), (5, 1), (5, 2)]

如何在python中配对两个元组？

3 个答案: