获得常见的元组元素

Question

我该怎么做呢：

(("and", "dog"), ("a", "dog"))

进入这个：

("and", "dog", "a")

这意味着只需要使用一次公共元素"dog"。

Answer 1

>>> tuple(set(("and", "dog")) | set(("a", "dog")))
('and', 'a', 'dog')

或者，通常：

import operator
tuple(reduce(operator.__or__, map(set, mytuples)))

Answer 2

您可以将其转换为set。

diff = (set(("a","dog")) - set(("and","dog")))
result = list(("and","dog")) + list((diff))
print(result)  #['and', 'dog', 'a']

Answer 3

如果订单无关紧要：

>>> tuple(set.union(*map(set, tuples)))
('and', 'a', 'dog')

如果订单确实重要：

>>> seen = set()
>>> tuple(elem for tupl in tuples for elem in tupl
          if elem not in seen and not seen.add(elem))
('and', 'dog', 'a')