2个列表的公共元素比较 [英] Common elements comparison between 2 lists

问题描述

def common_elements(list1, list2):"""返回包含 list1 和 list2 中的元素的列表>>>common_elements([1,2,3,4,5,6], [3,5,7,9])[3, 5]>>>common_elements(['this','this','n','that'],['this','not','that','that'])['这个那个']"""对于 list1 中的元素:如果列表 2 中的元素:返回列表(元素)

到此为止，但似乎无法让它工作！

有什么想法吗?

解决方案

使用 Python 的 设置交集:

<预><代码>>>>list1 = [1,2,3,4,5,6]>>>列表 2 = [3, 5, 7, 9]>>>列表(设置(列表1).交叉点(列表2))[3, 5]

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']
    """
    for element in list1:
        if element in list2:
            return list(element)

Got that so far, but can't seem to get it to work!

Any ideas?

解决方案

Use Python's set intersection:

>>> list1 = [1,2,3,4,5,6]
>>> list2 = [3, 5, 7, 9]
>>> list(set(list1).intersection(list2))
[3, 5]

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