使用python在列表列表中查找匹配的值

问题描述

我试图遍历python 2.7.5中的列表列表，并返回在第二个列表中找到第一个值的列表，如下所示:

I'm trying to iterate over a list of lists in python 2.7.5 and return those where the first value is found in a second list, something like this:

#python 2.7.5
list1 = ['aa', 'ab', 'bb', 'bc', 'cc']
list2 = [['aa', 1, 3, 7],['de', 2, 2, 1],['bc', 3, 4, 4]]

list3 = []
for x in list1:
    for y in list2:
        if x == y:
            list3.append(y)

所以我希望list3包含 [[''aa'，1,3,7]，['bc'，3，4，4]] ，但是我只是得到了整个清单2.

So I would want list3 to contain [['aa',1,3,7],['bc', 3, 4, 4]] but instead I just get the whole of list2.

推荐答案

尝试更接近您想要的更简单的方法:

Try a more simple approach that is closer to what you want:

for e in list2:
    if e[0] in list1:
        list3.append(e)

您需要 e [0] ，因为 list2 是列表列表.您也可以使用 filter()函数

You need e[0] since list2 is a list of lists. You can also write this in a single line using the filter() function:

list3 = filter(lambda e: e[0] in list1, list2)

或使用列表理解:

list3 = [e for e in list2 if e[0] in list1]

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