使用python处理文件目录

Question

I have a directory and I want to pick every file within that directory and run a python code on it: So I do the following

for file in os.listdir('/Users/Desktop/Xfiles'):
     os.system('/sw/bin/python2.7 pythonCode.py /Users/Desktop/Xfiles/file')

This does not work, I want to process the "file" from the listdir....how can I do that?

Answer 1

The path passed to the os.system is hard coded. You should pass filename .

dirpath = '/Users/Desktop/Xfiles'
for filename in os.listdir(dirpath):
    os.system('/sw/bin/python2.7 pythonCode.py {}/{}'.format(dirpath, filename))

• Don't use file as variable name. It shadows builtin file function.

Answer 2

You can do string interpolation with

file = "myfilename"
"some text {}".format(file)
# white should result in "some text myfilename"

But for manipulating paths, the best way is to use

os.path.join('/Users/gchella1/Desktop/forGeorge/Xfiles/', file)

Answer 3

did you forget to put "file" out of the quotes ?

for file in os.listdir('/Users/gchella1/Desktop/forGeorge/Xfiles'):
    os.system('/sw/bin/python2.7 pythonCode.py /Users/gchella1/Desktop/forGeorge/Xfiles/'+file)

this works for me.

for file in os.listdir('.'):
    os.system('ls '+file)