如何使用glob搜索子文件夹。水珠在Python模块?

I want to open a series of subfolders in a folder and find some text files and print some lines of the text files. I am using this:

我想在文件夹中打开一系列子文件夹，找到一些文本文件并打印一些文本文件。我用这个:

configfiles = glob.glob('C:/Users/sam/Desktop/file1/*.txt')

But this cannot access the subfolders as well. Does anyone know how I can use the same command to access subfolders as well?

但是这也不能访问子文件夹。有人知道我如何使用相同的命令来访问子文件夹吗?

9 个解决方案

#1

In Python 3.5 and newer use the new recursive **/ functionality:

在Python 3.5和更新版本中，使用新的递归**/功能:

configfiles = glob.glob('C:/Users/sam/Desktop/file1/**/*.txt', recursive=True)

When recursive is set, ** followed by a path separator matches 0 or more subdirectories.

设置递归时，**后跟路径分隔符匹配0或更多子目录。

In earlier Python versions, glob.glob() cannot list files in subdirectories recursively.

在早期的Python版本中，glob.glob()不能递归地列出子目录中的文件。

In that case I'd use os.walk() combined with fnmatch.filter() instead:

在这种情况下，我将使用os.walk()和fnmatch.filter()相结合:

import os
import fnmatch

path = 'C:/Users/sam/Desktop/file1'

configfiles = [os.path.join(dirpath, f)
    for dirpath, dirnames, files in os.walk(path)
    for f in fnmatch.filter(files, '*.txt')]

This'll walk your directories recursively and return all absolute pathnames to matching .txt files. In this specific case the fnmatch.filter() may be overkill, you could also use a .endswith() test:

这将递归地遍历目录，并将所有绝对路径名返回到匹配的.txt文件中。在这种特定的情况下，fnmatch.filter()可能会过度，您还可以使用.endswith()测试:

import os

path = 'C:/Users/sam/Desktop/file1'

configfiles = [os.path.join(dirpath, f)
    for dirpath, dirnames, files in os.walk(path)
    for f in files if f.endswith('.txt')]

#2

The glob2 package supports wild cards and is reasonably fast

glob2包支持通配符，并且速度相当快

code = '''
import glob2
glob2.glob("files/*/**")
'''
timeit.timeit(code, number=1)

On my laptop it takes approximately 2 seconds to match >60,000 file paths.

在我的笔记本电脑上，大约需要2秒的时间来匹配>万条文件路径。

#3

To find files in immediate subdirectories:

在直接子目录中查找文件:

configfiles = glob.glob(r'C:\Users\sam\Desktop\*\*.txt')

For a recursive version that traverse all subdirectories, you could use ** and pass recursive=True since Python 3.5:

对于遍历所有子目录的递归版本，可以使用**并传递递归=True，因为Python 3.5:

configfiles = glob.glob(r'C:\Users\sam\Desktop\**\*.txt', recursive=True)

Both function calls return lists. You could use glob.iglob() to return paths one by one. Or use pathlib:

两个函数都调用返回列表。可以使用glob.iglob()逐个返回路径。或者使用pathlib:

from pathlib import Path

path = Path(r'C:\Users\sam\Desktop')
txt_files_only_subdirs = path.glob('*/*.txt')
txt_files_all_recursively = path.rglob('*.txt') # including the current dir

Both methods return iterators (you can get paths one by one).

这两个方法都返回迭代器(您可以逐个获得路径)。

#4

You can use Formic with Python 2.6

您可以在Python 2.6中使用Formic

import formic
fileset = formic.FileSet(include="**/*.txt", directory="C:/Users/sam/Desktop/")

Disclosure - I am the author of this package.

披露-我是这个包裹的作者。

#5

Here is a adapted version that enables glob.glob like functionality without using glob2.

这里有一个适合的版本，它支持glob。glob喜欢不使用glob2的功能。

def find_files(directory, pattern='*'):
    if not os.path.exists(directory):
        raise ValueError("Directory not found {}".format(directory))

    matches = []
    for root, dirnames, filenames in os.walk(directory):
        for filename in filenames:
            full_path = os.path.join(root, filename)
            if fnmatch.filter([full_path], pattern):
                matches.append(os.path.join(root, filename))
    return matches

So if you have the following dir structure

如果你有下面的dir结构

tests/files
├── a0
│   ├── a0.txt
│   ├── a0.yaml
│   └── b0
│       ├── b0.yaml
│       └── b00.yaml
└── a1

You can do something like this

你可以这样做

files = utils.find_files('tests/files','**/b0/b*.yaml')
> ['tests/files/a0/b0/b0.yaml', 'tests/files/a0/b0/b00.yaml']

Pretty much fnmatch pattern match on the whole filename itself, rather than the filename only.

几乎是整个文件名本身的fnmatch模式匹配，而不仅仅是文件名。

#6

configfiles = glob.glob('C:/Users/sam/Desktop/**/*.txt")

configfile = glob.glob(“C:/用户/ sam /桌面/ * * / * . txt”)

Doesn't works for all cases, instead use glob2

并不适用于所有情况，而是使用glob2

configfiles = glob2.glob('C:/Users/sam/Desktop/**/*.txt")

#7

If you can install glob2 package...

如果您可以安装glob2软件包…

import glob2
filenames = glob2.glob("C:\\top_directory\\**\\*.ext")  # Where ext is a specific file extension
folders = glob2.glob("C:\\top_directory\\**\\")

All filenames and folders:

所有的文件名和文件夹:

all_ff = glob2.glob("C:\\top_directory\\**\\**")

#8

If you're running Python 3.4+, you can use the pathlib module. The Path.glob() method supports the ** pattern, which means “this directory and all subdirectories, recursively”. It returns a generator yielding Path objects for all matching files.

如果您正在运行Python 3.4+，您可以使用pathlib模块。Path.glob()方法支持**模式，它的意思是“递归地包含这个目录和所有子目录”。它返回一个生成器，生成所有匹配文件的路径对象。

from pathlib import Path
configfiles = Path("C:/Users/sam/Desktop/file1/").glob("**/*.txt")

#9

As pointed out by Martijn, glob can only do this through the **operator introduced in Python 3.5. Since the OP explicitly asked for the glob module, the following will return a lazy evaluation iterator that behaves similarly

正如Martijn所指出的，glob只能通过Python 3.5中引入的**操作符来实现这一点。由于OP显式地请求glob模块，下面将返回一个行为类似的延迟计算迭代器

import os, glob, itertools

configfiles = itertools.chain.from_iterable(glob.iglob(os.path.join(root,'*.txt'))
                         for root, dirs, files in os.walk('C:/Users/sam/Desktop/file1/'))

Note that you can only iterate once over configfiles in this approach though. If you require a real list of configfiles that can be used in multiple operations you would have to create this explicitly by using list(configfiles).

注意，在这种方法中，只能对configfiles进行一次迭代。如果您需要一个可以在多个操作中使用的真正的configfile列表，那么必须使用list(configfiles)显式地创建这个列表。

#1