Python3的字符串列表算术

Question

我该怎么做？

>>> a
['eins', 'zwei', 'drei', 'vier', 'fünf']
>>> b
['eins', 'zwei', 'drei']
>>> c = a - b

这也行不通。

c = a[:].remove(b)

Answer 1

如果您不需要重复元素，可以使用set s作为jadelord说

如果列表a包含重复的元素并且您需要它们，则应使用

c = list(filter(lambda x: x not in b, a))

或者

c = [x for x in a if x not in b]

Answer 2

您正在寻找的是与python中的set数据类型相关联的功能。不是lists！

这可以解决您的问题：

c = list(set(a) - set(b))

Answer 3

Set和ImmutableSet的实例都提供以下操作：

Operation        Equivalent     Result
len(s)                          cardinality of set s
x in s                          test x for membership in s
x not in s                      test x for non-membership in s
s.issubset(t)      s <= t       test whether every element in s is in t
s.issuperset(t)    s >= t       test whether every element in t is in s
s.union(t)          s | t       new set with elements from both s and t
s.intersection(t)   s & t       new set with elements common to s and t
s.difference(t)     s - t       new set with elements in s but not in t
s.symmetric_difference(t) s ^ t new set with elements in either s or t but not both
s.copy()                        new set with a shallow copy of s