Question

def common_elements(list1, list2):
    """
    Return a list containing the elements which are in both list1 and list2

    >>> common_elements([1,2,3,4,5,6], [3,5,7,9])
    [3, 5]
    >>> common_elements(["this","this","n","that"],["this","not","that","that"])
    ['this', 'that']
    """

    result = []
    for element in list1:
        if element in list2:
            result.append(element)
    return result

到目前为止我有这个，但它返回重复项，例如：

common_elements(["this","this","n","that"],["this","not","that","that"])

返回为：['this', 'this', 'that']

Answer 1

使用set.intersection()，因为这意味着无需将list2转换为集合

def common_elements(list1, list2):
    return set(list1).intersection(list2)

选择较短的列表转换为集合

更有效

def common_elements(list1, list2):
    short_list, long_list = sorted((list1, list2), key=len)
    return set(short_list).intersection(long_list)

当然要返回一个列表，你会使用

    return list(set(...))

Answer 2

使用sets：

>>> a, b = [1,2,3,4,5,6], [3,5,7,9]
>>> set(a).intersection(b)
set([3, 5])

Answer 3

def common_elements(a, b):
    return list(set(a) & set(b))

在这种情况下，我们采用两个集合的交集，这两个集合又由两个列表构成。每组由每个列表中的唯一项组成。我们最终转换回列表，因为这是所需的返回类型。

Answer 4

>>> a = [1,2,3,4,5,6]
>>> b = [3,5,7,9]
>>> list(set(a).intersection(b))
[3, 5]

编辑：不需要将b转换为集合。谢谢@Johnsyweb

Answer 5

你想使用集合，因为它们有一些非常好的操作：

>>> a = set([1, 2, 3, 4])
>>> b = set([3, 4, 5, 6])
>>> a.intersection(b)
set([3, 4])
>>> a.difference(b)
set([1, 2])
>>> b.intersection(a)
set([5, 6])
>>> a.union(b)
set([1, 2, 3, 4, 5, 6])

如何返回包含没有重复项的公共元素的列表

5 个答案: