有没有更好的方法来使用glob。Glob在python中获取多个文件类型的列表,如.txt, .mdown和.markdown?现在我有这样的东西:

projectFiles1 = glob.glob( os.path.join(projectDir, '*.txt') )
projectFiles2 = glob.glob( os.path.join(projectDir, '*.mdown') )
projectFiles3 = glob.glob( os.path.join(projectDir, '*.markdown') )

当前回答

要glob多种文件类型,需要在循环中多次调用glob()函数。因为这个函数返回一个列表,所以需要连接这些列表。

例如,这个函数是这样的:

import glob
import os


def glob_filetypes(root_dir, *patterns):
    return [path
            for pattern in patterns
            for path in glob.glob(os.path.join(root_dir, pattern))]

简单的用法:

project_dir = "path/to/project/dir"
for path in sorted(glob_filetypes(project_dir, '*.txt', '*.mdown', '*.markdown')):
    print(path)

你也可以使用glob.iglob()来拥有一个迭代器:

返回一个迭代器,该迭代器产生与glob()相同的值,但实际上不会同时存储它们。

def iglob_filetypes(root_dir, *patterns):
    return (path
            for pattern in patterns
            for path in glob.iglob(os.path.join(root_dir, pattern)))

其他回答

from glob import glob

files = glob('*.gif')
files.extend(glob('*.png'))
files.extend(glob('*.jpg'))

print(files)

如果你需要指定一个路径,循环匹配模式,并保持连接在循环中简单:

from os.path import join
from glob import glob

files = []
for ext in ('*.gif', '*.png', '*.jpg'):
   files.extend(glob(join("path/to/dir", ext)))

print(files)
import glob
import pandas as pd

df1 = pd.DataFrame(columns=['A'])
for i in glob.glob('C:\dir\path\*.txt'):
    df1 = df1.append({'A': i}, ignore_index=True)
for i in glob.glob('C:\dir\path\*.mdown'):
    df1 = df1.append({'A': i}, ignore_index=True)
for i in glob.glob('C:\dir\path\*.markdown):
    df1 = df1.append({'A': i}, ignore_index=True)

来这里寻求帮助后,我有了自己的解决方案,想和大家分享。它基于user2363986的答案,但我认为这更具可伸缩性。这意味着,即使您有1000个扩展,代码仍然看起来很优雅。

from glob import glob

directoryPath  = "C:\\temp\\*." 
fileExtensions = [ "jpg", "jpeg", "png", "bmp", "gif" ]
listOfFiles    = []

for extension in fileExtensions:
    listOfFiles.extend( glob( directoryPath + extension ))

for file in listOfFiles:
    print(file)   # Or do other stuff
import os
import glob

projectFiles = [i for i in glob.glob(os.path.join(projectDir,"*")) if os.path.splitext(i)[-1].lower() in ['.txt','.markdown','.mdown']]

Os.path.splitext将返回filename & .extension

filename, .extension = os.path.splitext('filename.extension')

.lower()将字符串转换为小写

你可以用这个:

project_files = []
file_extensions = ['txt','mdown','markdown']
for file_extension in file_extensions:
    project_files.extend(glob.glob(projectDir  + '*.' + file_extension))