552 lines
13 KiB
Markdown
552 lines
13 KiB
Markdown
# 如何用 Python 列出一个目录下的文件？
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> 原文：<https://www.askpython.com/python/examples/list-files-in-a-directory-using-python>
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在本教程中，我们将介绍如何使用 Python 列出目录中的文件。
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Python 是一种通用语言，用于各种领域，如数据科学、机器学习，甚至是 Web 开发。Python 语言的应用似乎没有限制。
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因此，似乎很简单的 Python 可以用来列出任何系统中的文件和目录。这篇文章的目的是向读者阐明使用 Python 在系统中列出文件的方法。
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## 使用 Python 列出目录中的所有文件
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为了使用 Python 与系统中的目录进行交互，使用了`os`库。
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### 1.使用“操作系统”库
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我们要练习的方法是 listdir()。顾名思义，它用于[列出目录](https://www.askpython.com/python/examples/python-directory-listing)中的项目。
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```py
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# Importing the os library
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import os
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# The path for listing items
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path = '.'
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# The list of items
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files = os.listdir(path)
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# Loop to print each filename separately
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for filename in files:
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print(filename)
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```
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**输出:**
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```py
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game_file.py
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hi-lo_pygame.py
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Journaldev
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list_files1.py
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hi_lo_pygame.mp4
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test.py
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list_files.py
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my_program.cpp
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a.out
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cut.cpp
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```
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Linux 用户可以在终端上使用标准的`ls`命令轻松匹配上面的输出。
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List Items using ‘ls’ command
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正如我们所看到的，每个方法的输出都是匹配的。
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* * *
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### 2.使用“glob”库
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`glob`主要是一个文件名模式匹配库，但是它可以通过以下方式列出当前目录中的项目:
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```py
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# Importing the glob library
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import glob
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# Path to the directory
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path = ''
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# or
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# path = './'
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# Extract the list of filenames
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files = glob.glob(path + '*', recursive=False)
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# Loop to print the filenames
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for filename in files:
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print(filename)
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```
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**输出:**
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```py
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game_file.py
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hi-lo_pygame.py
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Journaldev
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list_files1.py
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hi_lo_pygame.mp4
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test.py
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list_files.py
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my_program.cpp
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a.out
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cut.cpp
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```
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通配符`'*'`用于匹配当前目录中的所有项目。因为我们希望显示当前目录的条目，所以我们需要关闭`glob()`函数的递归特性。
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* * *
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### 3.仅列出当前目录中的文件
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在上面的方法中，python 代码返回当前目录中的所有项目，而不管它们的性质如何。我们可以使用`os`库中的`path.isfile()`函数只提取文件。
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```py
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# Importing the os library
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import os
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# The path for listing items
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path = '.'
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# List of only files
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files = [f for f in os.listdir(path) if os.path.isfile(f)]
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# Loop to print each filename separately
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for filename in files:
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print(filename)
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```
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**输出:**
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```py
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game_file.py
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hi-lo_pygame.py
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list_files1.py
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hi_lo_pygame.mp4
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test.py
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list_files.py
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my_program.cpp
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a.out
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cut.cpp
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```
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在上面的代码片段中， [List Comprehension](https://www.askpython.com/python/list/python-list-comprehension) 用于过滤掉那些实际上是文件的项目。
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这里需要注意的另一个关键点是，上面的代码不适用于其他目录，因为变量`'f'`不是绝对路径，而是当前目录的相对路径。
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* * *
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## 递归列出目录中的所有文件
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为了打印目录及其子目录中的文件，我们需要递归地遍历它们。
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### 1.使用“操作系统”库
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在`walk()`方法的帮助下，我们可以逐个遍历目录中的每个子目录。
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```py
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# Importing the os library
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import os
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# The path for listing items
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path = './Documents/'
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# List of files in complete directory
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file_list = []
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"""
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Loop to extract files inside a directory
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path --> Name of each directory
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folders --> List of subdirectories inside current 'path'
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files --> List of files inside current 'path'
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"""
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for path, folders, files in os.walk(path):
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for file in files:
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file_list.append(os.path.join(path, file))
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# Loop to print each filename separately
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for filename in file_list:
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print(filename)
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```
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**输出:**
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```py
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./Documents/game_file.py
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./Documents/hi-lo_pygame.py
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./Documents/list_files1.py
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./Documents/hi_lo_pygame.mp4
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./Documents/test.py
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./Documents/list_files.py
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./Documents/my_program.cpp
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./Documents/a.out
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./Documents/cut.cpp
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./Documents/Journaldev/mastermind.py
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./Documents/Journaldev/blackjack_terminal.py
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./Documents/Journaldev/lcm.cpp
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./Documents/Journaldev/super.cpp
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./Documents/Journaldev/blackjack_pygame.py
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./Documents/Journaldev/test.java
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```
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`os.walk()`方法简单地跟随每个子目录，默认情况下以自顶向下的方式提取文件。有三个迭代器用于遍历`os.walk()`函数的输出:
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* `**path**`–该变量包含函数在某次迭代中观察的当前目录
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* `**folders**`–该变量是`'path'`目录中的目录列表。
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* `**files**`–目录`'path'`中的文件列表。
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`join()`方法用于将文件名与其父目录连接起来，为我们提供文件的相对路径。
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* * *
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### 2.使用“glob”库
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类似于上面的过程，`glob`可以递归地访问每个目录，提取所有项目并返回。
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```py
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# Importing the glob library
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import glob
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# Importing the os library
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import os
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# Path to the directory
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path = './Documents/'
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# Extract all the list of items recursively
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files = glob.glob(path + '**/*', recursive=True)
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# Filter only files
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files = [f for f in files if os.path.isfile(f)]
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# Loop to print the filenames
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for filename in files:
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print(filename)
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```
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**输出:**
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```py
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./Documents/game_file.py
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./Documents/hi-lo_pygame.py
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./Documents/list_files1.py
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./Documents/hi_lo_pygame.mp4
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./Documents/test.py
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./Documents/list_files.py
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./Documents/my_program.cpp
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./Documents/a.out
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./Documents/cut.cpp
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./Documents/Journaldev/mastermind.py
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./Documents/Journaldev/blackjack_terminal.py
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./Documents/Journaldev/lcm.cpp
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./Documents/Journaldev/super.cpp
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./Documents/Journaldev/blackjack_pygame.py
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./Documents/Journaldev/test.java
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```
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与 path 变量一起使用的符号`'**'`告诉`glob()`函数匹配任何子目录中的文件。`'*'`告诉函数匹配一个目录中的所有项目。
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因为我们希望只提取完整目录中的文件，所以我们使用之前使用的`isfile()`函数过滤掉这些文件。
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* * *
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## 列出一个目录中的所有子目录
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我们可以列出特定目录中的所有子目录，而不是列出文件。
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```py
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# Importing the os library
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import os
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# The path for listing items
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path = './Documents/'
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# List of folders in complete directory
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folder_list = []
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"""
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Loop to extract folders inside a directory
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path --> Name of each directory
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folders --> List of subdirectories inside current 'path'
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files --> List of files inside current 'path'
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"""
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for path, folders, files in os.walk(path):
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for folder in folders:
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folder_list.append(os.path.join(path, folder))
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# Loop to print each foldername separately
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for foldername in folder_list:
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print(foldername)
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```
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**输出:**
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```py
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./Documents/Journaldev
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```
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列出文件和目录的微小区别在于`os.walk()`函数过程中迭代器的选择。对于文件，我们迭代 files 变量。这里，我们循环遍历 folders 变量。
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* * *
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## 用绝对路径列出目录中的文件
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一旦我们知道如何列出目录中的文件，那么显示绝对路径就是小菜一碟。`abspath()`方法为我们提供了文件的绝对路径。
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```py
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# Importing the os library
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import os
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# The path for listing items
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path = './Documents/'
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# List of files in complete directory
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file_list = []
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"""
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Loop to extract files inside a directory
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path --> Name of each directory
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folders --> List of subdirectories inside current 'path'
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files --> List of files inside current 'path'
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"""
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for path, folders, files in os.walk(path):
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for file in files:
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file_list.append(os.path.abspath(os.path.join(path, file)))
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# Loop to print each filename separately
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for filename in file_list:
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print(filename)
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```
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**输出:**
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```py
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/home/aprataksh/Documents/game_file.py
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/home/aprataksh/Documents/hi-lo_pygame.py
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/home/aprataksh/Documents/list_files1.py
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/home/aprataksh/Documents/hi_lo_pygame.mp4
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/home/aprataksh/Documents/test.py
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/home/aprataksh/Documents/list_files.py
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/home/aprataksh/Documents/my_program.cpp
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/home/aprataksh/Documents/a.out
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/home/aprataksh/Documents/cut.cpp
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/home/aprataksh/Documents/Journaldev/mastermind.py
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/home/aprataksh/Documents/Journaldev/blackjack_terminal.py
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/home/aprataksh/Documents/Journaldev/lcm.cpp
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/home/aprataksh/Documents/Journaldev/super.cpp
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/home/aprataksh/Documents/Journaldev/blackjack_pygame.py
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/home/aprataksh/Documents/Journaldev/test.java
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```
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这里需要注意的一点是，`abspath()`必须提供文件的相对路径，这也是`join()`函数的目的。
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* * *
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## 通过匹配模式列出目录中的文件
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有多种方法可以过滤出符合特定模式的文件名。让我们一个一个地看一遍。
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### 1.使用“fnmatch”库
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顾名思义，`fnmatch`是一个文件名模式匹配库。将`fnmatch`与我们的标准文件名提取库一起使用，我们可以过滤出那些匹配特定模式的文件。
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```py
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# Importing the os and fnmatch library
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import os, fnmatch
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# The path for listing items
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path = './Documents/'
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# List of files in complete directory
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file_list = []
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"""
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Loop to extract files containing word "file" inside a directory
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path --> Name of each directory
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folders --> List of subdirectories inside current 'path'
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files --> List of files inside current 'path'
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"""
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print("List of files containing \"file\" in them")
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for path, folders, files in os.walk(path):
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for file in files:
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if fnmatch.fnmatch(file, '*file*'):
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file_list.append(os.path.join(path, file))
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# Loop to print each filename separately
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for filename in file_list:
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print(filename)
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```
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**输出:**
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```py
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List of files containing "file" in them
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./Documents/game_file.py
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./Documents/list_files1.py
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./Documents/list_files.py
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```
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`fnmatch()`函数接受两个参数，文件名后面跟着要匹配的模式。在上面的代码中，我们查看了所有包含单词`file`的文件。
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* * *
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### 2.使用“glob”库
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正如我们之前提到的，`glob's`的主要目的是文件名模式匹配。
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```py
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# Importing the glob library
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import glob
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# Importing the os library
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import os
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# Path to the directory
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path = './Documents/'
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# Extract items containing numbers in name
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files = glob.glob(path + '**/*[0-9]*.*', recursive=True)
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# Filter only files
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files = [f for f in files if os.path.isfile(f)]
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# Loop to print the filenames
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for filename in files:
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print(filename)
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```
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**输出:**
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```py
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./Documents/list_files1.py
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```
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上面的模式匹配[正则表达式](https://www.askpython.com/python/regular-expression-in-python) `'**/*[0-9]*.*'`可以解释为:
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* `**'**'**`–遍历路径内的所有子目录
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* `**'/*'**`–文件名可以以任何字符开头
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* `**'[0-9]'**`–文件名中包含一个数字
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* `**'*.*'**`–文件名可以以任何字符结尾，可以有任何扩展名
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* * *
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### 3.使用“pathlib”库
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遵循面向对象的方式与文件系统交互。库中的`rglob()`函数可用于通过某个路径对象递归提取文件列表。
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这些文件列表可以使用`rglob()`函数中的模式进行过滤。
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```py
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# Importing the pathlib library
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import pathlib
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# Creating a Path object
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path = pathlib.Path('./Documents/')
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# Extracting a list of files starting with 'm'
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files = path.rglob('m*')
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# Loop to print the files separately
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for file in files:
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print(file)
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```
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**输出:**
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```py
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Documents/my_program.cpp
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Documents/Journaldev/mastermind.py
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```
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上面的代码片段用于列出所有以字母`'m'`开头的文件。
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* * *
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## 列出目录中具有特定扩展名的文件
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在 Python 中列出带有特定扩展名的文件有点类似于模式匹配。为此，我们需要创建一个关于文件扩展名的模式。
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```py
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# Importing the os and fnmatch library
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import os, fnmatch
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# The path for listing items
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path = './Documents/'
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# List to store filenames
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file_list = []
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"""
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Loop to extract python files
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path --> Name of each directory
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folders --> List of subdirectories inside current 'path'
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files --> List of files inside current 'path'
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"""
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print("List of python files in the directory:")
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for path, folders, files in os.walk(path):
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for file in files:
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if fnmatch.fnmatch(file, '*.py'):
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file_list.append(os.path.join(path, file))
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# Loop to print each filename separately
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for filename in file_list:
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print(filename)
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```
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**输出:**
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```py
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List of python files in the directory:
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./Documents/game_file.py
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./Documents/hi-lo_pygame.py
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./Documents/list_files1.py
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./Documents/test.py
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./Documents/list_files.py
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./Documents/Journaldev/mastermind.py
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./Documents/Journaldev/blackjack_terminal.py
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./Documents/Journaldev/blackjack_pygame.py
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```
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`fnmatch()`函数过滤掉那些以`'.py'`结尾的文件，也就是 python 文件。如果我们想要提取不同扩展名的文件，那么我们必须修改这部分代码。例如，为了只获取 C++文件，必须使用`'.cpp'`。
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这总结了使用 Python 获取目录中文件列表的方法。
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* * *
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## 结论
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可以有多种方法来解决手头的任何问题，最方便的方法并不总是答案。关于这篇文章，Python 程序员必须知道我们可以在目录中列出文件的每一种方式。
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我们希望这篇文章容易理解。如有任何疑问或建议，欢迎在下方留言。 |