Python: get pathname of files using regex pattern recursively

Question

If one want to get the (absolute) file paths (file listings) of all files with names of certain pattern how can this be done in Python (v 3.5 on unix). Something similar to the bash command find -regex 'pattern'. I have been looking as the os, glob, os.path and this SO but cannot get it together.

Say you want absolute paths to the files that matches /.*[pat.txt]$/ and you have the below diretories:

/home/me/dir1/dir1a/filepat.txt #1
/home/me/dir1/dir1a/file.txt
/home/me/dir1/dir1a/filepat.png
/home/me/dir2/filepat.txt #2
/home/me/dir3/dir3a/dir3ab/filepat
/home/me/dir3/dir3a/dir3ac/filepat.txt #3 
/home/me/dir3/dir3a/dir3ac/filepat.png

Then you would get want the three indicated paths:

/home/me/dir1/dir1a/filepat.txt
/home/me/dir2/filepat.txt
/home/me/dir3/dir3a/dir3ac/filepat.txt

One try was:

import fnmatch
import os 
start_path = "/home/me/" 
for root, dirs, files in os.walk(start_path):
    for filename in fnmatch.filter(files, ".*pat.txt"):
        print(os.path.join(start_path, filename))

Python is misspelled in the post title :-)

user1094081
– user1094081

2017-04-20 06:26:54 +00:00
Commented Apr 20, 2017 at 6:26 — user1094081
– user1094081, Commented Apr 20, 2017 at 6:26
Embarrassing - but good catch

user3375672
– user3375672

2017-04-20 06:33:35 +00:00
Commented Apr 20, 2017 at 6:33 — user3375672
– user3375672, Commented Apr 20, 2017 at 6:33
don't worry, I was in "debug mode" ;-)

user1094081
– user1094081

2017-04-20 07:10:39 +00:00
Commented Apr 20, 2017 at 7:10 — user1094081
– user1094081, Commented Apr 20, 2017 at 7:10

redacted · Accepted Answer · 2017-04-20 06:30:29Z

3

Here is one using regexes, but for the simple case I would go with akash`s answer using in operator

import re
pattern = re.compile(r'.*pat\.txt$')

import fnmatch
import os 
start_path = "/home/me/" 
for root, dirs, files in os.walk(start_path):
    for filename in files:
        if pattern.find(filename):
            print(os.path.join(start_path, filename))

answered Apr 20, 2017 at 6:30

redacted

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1 Comment

user3375672 Over a year ago

I was close I see. I think yours is very general though (+1)

akash karothiya · Accepted Answer · 2017-04-20 06:27:19Z

1

You can use basename and in operator

x = given list
>>> [i for i in x if 'pat.txt' in os.path.basename(i)]
['/home/me/dir1/dir1a/filepat.txt',
 '/home/me/dir2/filepat.txt',
 '/home/me/dir3/dir3a/dir3ac/filepat.txt']

answered Apr 20, 2017 at 6:27

akash karothiya

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5 Comments

user3375672 Over a year ago

Oh I completely missed basename

Wiktor Stribiżew Over a year ago

However, this solution will also fetch files like filepat.txt.bak

redacted Over a year ago

To fix that, just use i.endswith('pat.txt')

akash karothiya Over a year ago

@WiktorStribiżew , we can still achieve this using endswith('pat.txt') if required :)

Wiktor Stribiżew Over a year ago

- Yes, but then it would fail to fetch FILEPAT.TXT. - We may use lower().

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