如何返回不包含重复项的公共元素列表

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

我现在有这个代码，但它返回的结果有重复的，比如：

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

返回的结果是：['this', 'this', 'that']

Answer 1

使用集合：

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

Answer 2

使用 set.intersection() 是因为这样就不需要把 list2 转换成集合了

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

选择较短的列表来转换成集合会更有效率

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

当然，如果你想返回一个列表，可以使用

    return list(set(...))

Answer 3

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

编辑: 不需要把 b 转换成集合。谢谢 @Johnsyweb