Python比较两个字符串的差异 [英] Python comparing two strings to differences

问题描述

我目前有两个列表:

str1 = ['test1.jpg', 'test2.jpg', 'test3.jpg']
str2 = ['test1.jpg', 'test2.jpg']

您可能会注意到，str2中缺少test3.jpg.有什么办法可以将str2与str1进行比较并找到缺失的片段?

As you may well notice, test3.jpg is missing from str2. Is there a way I can compare str2 with str1 and find the missing pieces?

推荐答案

您可以使用内置类型 set

str1 = ['test1.jpg', 'test2.jpg', 'test3.jpg']
str2 = ['test1.jpg', 'test2.jpg']
s1 = set(str1)
s2 = set(str2)
print(s1-s2)

您可以从那里轻松创建函数:

From there you can easily create function:

def difference(list1,list2):
    return set(list1)-set(list2)

如果您要在两者之间建立差异，请执行以下操作:对于数据:

If you want to create difference between both of them you can do the following: For data :

str1 = ['test1.jpg', 'test2.jpg', 'test3.jpg']
str2 = ['test1.jpg', 'test2.jpg', 'test4.jpg']

功能将是:

def difference(list1,list2):
    return (set(list1)-set(list2)) | (set(list2)-set(list1))

或:

def difference(list1,list2):
    return set(list1).symmetric_difference(set(list2))

输出:

{'test3.jpg'，'test4.jpg'}

{'test3.jpg', 'test4.jpg'}

此处记录了所有内容: Python3

All is documented here: Python3 Python2

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