getting file list using glob in python

Question

I have a list of csv files in mydir.I want to get the list of file names. however using glob as below is returning an empty list.

import glob

mydir = "C:\Data"

file_list = glob(mydir + "*.csv")
print('file_list {}'.format(file_list))

Answer 1

Looks like you just need to include your slash to search in the correct directory.

import glob

mydir = "C:\Data"

file_list = glob.glob(mydir + "/*.csv") # Include slash or it will search in the wrong directory!!
print('file_list {}'.format(file_list))

Answer 2

Try fnmatch :

import os
from fnmatch import fnmatch

mydir = "C:/Data"
file_list = [file for file in os.listdir(mydir) if fnmatch(file, '*.csv')]

print('file_list {}'.format(file_list))

Also, use RegEx:

import os
import re

mydir = "C:/Data"
file_list = [file for file in os.listdir(mydir) if re.search('.*\.png', file)]  

print('file_list {}'.format(file_list))

---

By the way, glob is a module, you should use glob.glob() like this:

from glob import glob

mydir = "C:/Data"

file_list = glob(mydir + "/*.csv")
print('file_list {}'.format(file_list))

Answer 3

You are missing a backslash between the filename and the directory. BUT you can't end a string with a single backslash, because it will think you are trying to escape the endquote. You can put it in the call to glob instead. (Note, that you'll want to make both strings raw, it's good practice in case your directory starts with an "n" or a "t", which in this case is interpreted as spacing characters.):

import glob
mydir = r"C:\Data"
file_list = glob.glob(mydir + r"\*.csv")
print('file_list {}'.format(file_list))

Also you might want to try using pylint to check for common errors or warnings. It would have thrown a warning for the unescaped directory name.

Update:

In fact I would just simplify it to this:

import glob
file_list = glob.glob(r"C:\Data\*.csv")
print('file_list {}'.format(file_list))