如何返回包含没有重复项的公共元素的列表

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()因为它意味着没有必要转换list2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,你会用

    return list(set(...))

Answer 2

>>> 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