我正在迭代其中的目录和文件,同时我修改了每个文件。我希望能够立即读取新修改的文件。 这是我的代码,带有描述性注释:
# go through each directory based on their ids
for id in id_list:
id_dir = os.path.join(ouput_dir, id)
os.chdir(id_dir)
# go through all files (with a specific extension)
for filename in glob('*' + ext):
# modify the file by replacing all new-line characters with an empty space
with fileinput.FileInput(filename, inplace=True) as f:
for line in f:
print(line.replace('\n', ' '), end='')
# here I would like to read the NEW modified file
with open(filename) as newf:
content = newf.read()
按照目前的情况,newf
不新修改后的版本,而是原始f
。我想我明白为什么会这样,但是我发现很难克服这个问题。
我总是可以进行2次单独的迭代(根据他们的ID查看每个目录,浏览所有文件(具有特定扩展名)并修改文件,然后重复迭代以阅读其中的每一个)但我希望如果有更有效的方法来解决它。也许如果在修改发生之后重新启动第二个for
循环,然后发生read
,那么可能是为了避免至少重复外部{{ {1}}循环)。
以干净有效的方式实现上述目标的任何想法/设计?
答案 0 :(得分:1)
我并不是说你这样做的方式是不正确的,但我觉得你过于复杂了。这是我的超级简单解决方案。
import glob, fileinput
for filename in glob('*' + ext):
f_in = (x.rstrip() for x in open(filename, 'rb').readlines()) #instead of trying to modify in place we instead read in data and replace raw_values.
with open(filename, 'wb') as f_out: # we then write the data stream back out
#extra modification to the data can go here, i just remove the /r and /n and write back out
for i in f_in:
f_out.write(i)
#now there is no need to read the data back in because we already have a static referance to it.
答案 1 :(得分:1)
对我来说,它适用于此代码:
#!/usr/bin/env python3
import os
from glob import glob
import fileinput
id_list=['1']
ouput_dir='.'
ext = '.txt'
# go through each directory based on their ids
for id in id_list:
id_dir = os.path.join(ouput_dir, id)
os.chdir(id_dir)
# go through all files (with a specific extension)
for filename in glob('*' + ext):
# modify the file by replacing all new-line characters with an empty space
for line in fileinput.FileInput(filename, inplace=True):
print(line.replace('\n', ' ') , end="")
# here I would like to read the NEW modified file
with open(filename) as newf:
content = newf.read()
print(content)
注意我如何迭代这些线!