Choose a file starting with a given string

Question

In a directory I have a lot of files, named more or less like this:

001_MN_DX_1_M_32
001_MN_SX_1_M_33
012_BC_2_F_23
...
...

In Python, I ha

Answer 1

import os, re
for f in os.listdir('.'):
   if re.match('001_MN_DX', f):
       print f

Answer 2

Try using os.listdir,os.path.join and os.path.isfile.
In long form (with for loops),

import os
path = 'C:/'
files = []
for i in os.listdir(path):
    if os.path.isfile(os.path.join(path,i)) and '001_MN_DX' in i:
        files.append(i)

Code, with list-comprehensions is

import os
path = 'C:/'
files = [i for i in os.listdir(path) if os.path.isfile(os.path.join(path,i)) and \
         '001_MN_DX' in i]

Check here for the long explanation...

Answer 3

You can use the module glob, it follows the Unix shell rules for pattern matching. See more.

from glob import glob

files = glob('*001_MN_DX*')

Answer 4

import os
prefixed = [filename for filename in os.listdir('.') if filename.startswith("prefix")]

Answer 5

You can use the os module to list the files in a directory.

Eg: Find all files in the current directory where name starts with 001_MN_DX

import os
list_of_files = os.listdir(os.getcwd()) #list of files in the current directory
for each_file in list_of_files:
    if each_file.startswith('001_MN_DX'):  #since its all type str you can simply use startswith
        print each_file

Answer 6

import os
 for filename in os.listdir('.'):
    if filename.startswith('criteria here'):
        print filename #print the name of the file to make sure it is what 
                               you really want. If it's not, review your criteria
                #Do stuff with that file