Python 比较列表。哪些元素不在列表中？

我有以下两个列表，我想获得列表2中不在列表1中的元素：

list1 = ["0100","0300","0500"]
list2 = ["0100","0200","0300","0400","0500"]

我的输出应该是：

list3 = ["0200","0400"]

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我正在寻找一种从另一个中减去一个的方法，但到目前为止，我无法得到我想要的列表3

这个解决方案对您有效吗

list3 = []
for i in range(len(list2)):
    if list2[i] not in list1:
        list3.append(list2[i])

或者，如果不关心顺序，可以将列表转换为集合：

set(list2) - set(list1)

然后，您还可以将其转换回列表：

list3 = list(set(list2) - set(list1))

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我相信这里已经回答了这个问题：

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set

函数将帮助您用几行代码解决问题

set1=set(["0100","0300","0500"])
set2=set(["0100","0200","0300","0400","0500"])
set3=set2-set1
print(list(set3))

set

提供了比列表更快的Python实现

list1 = ["0100","0300","0500"]
list2 = ["0100","0200","0300","0400","0500"]

list3 = list(filter(lambda e: e not in list1,list2))
print(list3)

import numpy as np

list1 = ["0100","0300","0500"]
list2 = ["0100","0200","0300","0400","0500"]

list3 = np.setdiff1d(list2,list1)

print(list3)

set1=set(["0100","0300","0500"])
set2=set(["0100","0200","0300","0400","0500"])
set3=set2-set1
print(list(set3))