替换列表元素的一部分

Question

我有一个列表变量，它的元素如下：['Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131121.csv', 'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131150.csv', 'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131160.csv', 'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131169.csv', 'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131189.csv']

这是包含文件路径的文件名。想要建立一个只有文件名的列表，我该怎么做？输出应为：

去掉字符串'Cordial/contactactivity/export/bounce/0-ContactActivity-‘的['bounce-20211109-131121.csv', 'bounce-20211109-131150.csv', 'bounce-20211109-131160.csv', 'bounce-20211109-131169.csv', 'bounce-20211109-131189.csv'] ie

Answer 1

您可以使用os.path.basename (docs here)并获取最后一个段。然后你只需要修剪掉你不想要的部分。

import os
data = [
    'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131121.csv', 
    'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131150.csv',
    'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131160.csv',
    'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131169.csv',
    'Cordial/contactactivity/export/bounce/0-ContactActivity-bounce-20211109-131189.csv'
]
data = [os.path.basename(i).replace('0-ContactActivity-', '') for i in data]

结果是：

>>> data
['bounce-20211109-131121.csv', 'bounce-20211109-131150.csv', 'bounce-20211109-131160.csv', 'bounce-20211109-131169.csv', 'bounce-20211109-131189.csv']