如何grep文件名并将其存储在python输出中

Question

I have a directory that has multiple files and I want to grep only test.uwsgi.log files and remove the .uwsgi.log part. After greping the names I store it in a output file.

Here is what I got:

import subprocess
testlist = “cd /root/test/ && ls | grep ‘uwsgi.log' |sed 's/\.uwsgi.log\>//g' > /root/test/testlist.txt"
output = subprocess.check_output(['bash','-c', testlist])

and the output is :

test1
test2
test3

Can someone educate me on how to achieve this without the subprocess module and use python only?

Answer 1

Using Python alone:

import os

for _, dirs, files in os.walk('/root/test/'):
    with open('output.txt', 'w') as out_file:
        out_file.writelines("{}\n".format(f) for f in files if f.endswith('uwsgi.log'))

output.txt is as expected:

test1
test2
test3

Notice I added \\n to each element in the list comprehension . as writelines does not do it automatically.