在许多文件上运行python脚本

Question

I have a set of files aaa_cntrl.txt , bbb_cntrl.txt ,.... zzz_cntrl.txt . I want to run a python script script.py on each of these files and produce output aaa_out.txt , bbb_out.txt ,.... zzz_out.txt .

my python script is

import sys
file_in = sys.argv[0]
file_out = sys.argv[1]
print "This is the input file", file_in
print "This is the output file", file_out

Command line is python script.py aaa_cntrl.txt aaa_out.txt

But I want to automatically specifiy the input as *_cntrl.txt and get the output as *_out.txt . How do I do this?

Answer 1

You need to loop over all input files and determine the output name inside the script then as your shell (at least on linux / unix) will expand the wildcard for you.

import sys

for file_in in sys.argv[1:]:
    # probably some more reliable way is required here in production
    file_out = file_in.split('_', 1)[0] + '_out.txt'

    print "This is the input file", file_in
    print "This is the output file", file_out

Answer 2

I've just done that a couple of day ago using argparse .

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('files', nargs='+')
args = parser.parse_args()

for f in args.files:
    process(f)

Then, just have to call your script with ./myscript.py *_cntrl.txt , and your shell will perform the expansion, as pointed out by Willem Van Onsem in the comments.
You can also take a look at the argparse.FileType to improve this code.

If your files names are well formated, I suggest you to automatically create the name of your output file depending on the input file. I mean, if xxx_cntrl.txt is always transformed into xxx_out.txt , you can simply do file_out = file_in.replace("cntrl", "out") .

Otherwise, you can do something like

parser.add_argument('files', nargs='+')
parser.add_argument('-o', nargs='+')
args = parser.parse_args()

for in_file, out_file in zip(args.file, args.o):
    process(in_file, out_file)

Anyway, I really recommend you the argparse module instead of a manual parsing of sys.argv .