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

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

其他回答

一个glob,许多扩展…但不完美的解决方案(可能匹配其他文件)。

filetypes = ['tif', 'jpg']

filetypes = zip(*[list(ft) for ft in filetypes])
filetypes = ["".join(ch) for ch in filetypes]
filetypes = ["[%s]" % ch for ch in filetypes]
filetypes = "".join(filetypes) + "*"
print(filetypes)
# => [tj][ip][fg]*

glob.glob("/path/to/*.%s" % filetypes)
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)
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 os    
import glob
import operator
from functools import reduce

types = ('*.jpg', '*.png', '*.jpeg')
lazy_paths = (glob.glob(os.path.join('my_path', t)) for t in types)
paths = reduce(operator.add, lazy_paths, [])

https://docs.python.org/3.5/library/functools.html#functools.reduce https://docs.python.org/3.5/library/operator.html#operator.add

这招对我很管用:

import glob
images = glob.glob('*.JPG' or '*.jpg' or '*.png')