我需要遍历给定目录的子目录并搜索文件。如果我得到一个文件,我必须打开它,改变它的内容,用我自己的行替换它。
我试了一下:
import os
rootdir ='C:/Users/sid/Desktop/test'
for subdir, dirs, files in os.walk(rootdir):
for file in files:
f=open(file,'r')
lines=f.readlines()
f.close()
f=open(file,'w')
for line in lines:
newline = "No you are not"
f.write(newline)
f.close()
但是我得到了一个错误。我做错了什么?
另一种返回子目录中所有文件的方法是使用pathlib模块,该模块在Python 3.4中引入,它提供了一种面向对象的方法来处理文件系统路径(pathlib在Python 2.7中也可以通过PyPi上的pathlib2模块使用):
from pathlib import Path
rootdir = Path('C:/Users/sid/Desktop/test')
# Return a list of regular files only, not directories
file_list = [f for f in rootdir.glob('**/*') if f.is_file()]
# For absolute paths instead of relative the current dir
file_list = [f for f in rootdir.resolve().glob('**/*') if f.is_file()]
从Python 3.5开始,glob模块也支持递归文件查找:
import os
from glob import iglob
rootdir_glob = 'C:/Users/sid/Desktop/test/**/*' # Note the added asterisks
# This will return absolute paths
file_list = [f for f in iglob(rootdir_glob, recursive=True) if os.path.isfile(f)]
以上任何一种方法的file_list都可以在不需要嵌套循环的情况下迭代:
for f in file_list:
print(f) # Replace with desired operations
从python >= 3.5开始,你可以使用**,glob。iglob(path/**, recursive=True),这似乎是最python化的解决方案,即:
import glob, os
for filename in glob.iglob('/pardadox-music/**', recursive=True):
if os.path.isfile(filename): # filter dirs
print(filename)
输出:
/pardadox-music/modules/her1.mod
/pardadox-music/modules/her2.mod
...
注:
glob.iglob
glob.iglob(pathname, recursive=False)
Return an iterator which yields the same values as glob() without actually storing them all simultaneously.
If recursive is True, the pattern '**' will match any files and
zero or more directories and subdirectories.
If the directory contains files starting with . they won’t be matched by default. For example, consider a directory containing card.gif and .card.gif:
>>> import glob
>>> glob.glob('*.gif') ['card.gif']
>>> glob.glob('.c*')['.card.gif']
You can also use rglob(pattern),
which is the same as calling glob() with **/ added in front of the given relative pattern.