Python 3 删除文件字符串的一部分

Question

我有如下所示的声音文件列表

Mix = ['/home/pi/Music/The Very Best Of The Stone Roses/02 She Bangs The Drums.mp3',
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-20 Sheila Take A Bow.mp3',
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-21 Girlfriend In A Coma.mp3',
                   "/home/pi/Music/The Smiths/The Sound Of The Smiths/1-22 I Started Something I Couldn't.mp3",
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-23 Last Night I Dreamt That Somebo.mp3']

我想做的是删除歌曲名称之前的所有内容，并在我的 PiFace Controll & Display 2 上仅显示歌曲名称，代码为 `cad.lcd.write(Song_Name).

我想我对如何创建一个单独的字符串和我想要删除的内容有点了解，如下所示

remove = '/home/pi/Music/The Very Best Of The Stone Roses/01'

然而，问题是每首歌都有一个单独的编号，例如 1-20 和 1-21，然后我将添加的不同专辑将有不同的专辑位置，所以我不确定如何实现这个想法。

如有任何帮助，我们将不胜感激。

如果我让任何人感到困惑，我有一个音乐文件例如；

'/home/pi/Music/The Smiths/The Sound Of The Smiths/1-20 Sheila Take A Bow.mp3',

我想删除

'/home/pi/Music/The Smiths/The Sound Of The Smiths/1-20

所以我只剩下

Sheila Take A Bow.mp3',

Answer 1

标准模块 os.path 拥有操作文件路径所需的一切。 os.path.basename(filename) 满足您的需求。

Answer 2

对于您要解决的问题，有两种常见的解决方案。

1.正则表达式

在更一般的情况下，如果您正在寻找的内容或您正在处理的字符串显示出规律性，您可以使用 python 的正则表达式库，re。这是使用正则表达式提取最后一个正斜杠后所有内容的示例。

import re
x = "this/stuff/should/go/leave this.ext"
re.search('([^/]+$)',x).group(0)

要了解有关 python 中正则表达式的更多信息，请尝试 documentation for re in Python 3。

2。 os.path

在这种特殊情况下，最简单的选择是使用 os.path，因为您要处理的是文件名。

import os.path
x = "this/stuff/should/go/leave this.ext"
os.path.basename(x)

Answer 3

你可以试试这个方法：

import re

pattern=r'\/[0-9-]+\s(\w.+?mp3)'
Mix = ['/home/pi/Music/The Very Best Of The Stone Roses/02 She Bangs The Drums.mp3',
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-20 Sheila Take A Bow.mp3',
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-21 Girlfriend In A Coma.mp3',
                   "/home/pi/Music/The Smiths/The Sound Of The Smiths/1-22 I Started Something I Couldn't.mp3",
                   '/home/pi/Music/The Smiths/The Sound Of The Smiths/1-23 Last Night I Dreamt That Somebo.mp3']

for i in Mix:
    print(re.findall(pattern,i)[0])

输出：

She Bangs The Drums.mp3
Sheila Take A Bow.mp3
Girlfriend In A Coma.mp3
I Started Something I Couldn't.mp3
Last Night I Dreamt That Somebo.mp3

正则表达式信息：

\/ matches the character / literally (case sensitive)
Match a single character present in the list below [0-9-]+
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
0-9 a single character in the range between 0 (index 48) and 9 (index 57) (case sensitive)
- matches the character - literally (case sensitive)
\s matches any whitespace character (equal to [\r\n\t\f\v ])
1st Capturing Group (\w.+?mp3)
\w matches any word character (equal to [a-zA-Z0-9_])
.+? matches any character (except for line terminators)
+? Quantifier — Matches between one and unlimited times, as few times as possible, expanding as needed (lazy)
mp3 matches the characters mp3 literally (case sensitive)

Answer 4

import os
fileName = ' '.join("/path/to/file".split(os.path.sep)[-1].split(" ")[1:])

Python包os提供函数[split], to split the path (or we can say string) and [join]，加入路径

首先你需要拆分路径，用“/”分隔符连接并存储返回输出的最后一个元素。

filePath = '/home/pi/Music/The Very Best Of The Stone Roses/02 She Bangs The Drums.mp3'
headName = filePath.split(os.path.sep)[-1]

在这里，headName 将包含 02 She Bangs The Drums.mp3

现在你需要以同样的方式拆分 headName，由 space ( ) 连接。存储除第 0th 个之外的所有元素。

splittedHeadName = headName.split(" ")[1:]

在这里，splittedHeadName 将包含 ["She", "Bangs", "The", "Drums.mp3"]。

现在您需要做的就是用 space 分隔符加入 splittedHeadName。

fileName = ' '.join (splittedHeadName)

就是这样。现在 fileName 包含 She Bangs The Drums.mp3，这是您想要的输出所必需的。

现在，只需迭代 filePath 并获得受人尊敬的 fileName。您可以使用我在顶部提到的代码段。

Python 3 删除文件字符串的一部分

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