How do you get a directory listing sorted by creation date in python?


What is the best way to get a list of all files in a directory, sorted by date [created | modified], using python, on a windows machine?

Asked By: Liza



I’ve done this in the past for a Python script to determine the last updated files in a directory:

import glob
import os

search_dir = "/mydir/"
# remove anything from the list that is not a file (directories, symlinks)
# thanks to J.F. Sebastion for pointing out that the requirement was a list 
# of files (presumably not including directories)  
files = list(filter(os.path.isfile, glob.glob(search_dir + "*")))
files.sort(key=lambda x: os.path.getmtime(x))

That should do what you’re looking for based on file mtime.

EDIT: Note that you can also use os.listdir() in place of glob.glob() if desired – the reason I used glob in my original code was that I was wanting to use glob to only search for files with a particular set of file extensions, which glob() was better suited to. To use listdir here’s what it would look like:

import os

search_dir = "/mydir/"
files = filter(os.path.isfile, os.listdir(search_dir))
files = [os.path.join(search_dir, f) for f in files] # add path to each file
files.sort(key=lambda x: os.path.getmtime(x))
Answered By: Jay

Maybe you should use shell commands. In Unix/Linux, find piped with sort will probably be able to do what you want.

Answered By: stephanea

Here’s a one-liner:

import os
import time
from pprint import pprint

pprint([(x[0], time.ctime(x[1].st_ctime)) for x in sorted([(fn, os.stat(fn)) for fn in os.listdir(".")], key = lambda x: x[1].st_ctime)])

This calls os.listdir() to get a list of the filenames, then calls os.stat() for each one to get the creation time, then sorts against the creation time.

Note that this method only calls os.stat() once for each file, which will be more efficient than calling it for each comparison in a sort.

Answered By: Greg Hewgill

Here’s my version:

def getfiles(dirpath):
    a = [s for s in os.listdir(dirpath)
         if os.path.isfile(os.path.join(dirpath, s))]
    a.sort(key=lambda s: os.path.getmtime(os.path.join(dirpath, s)))
    return a

First, we build a list of the file names. isfile() is used to skip directories; it can be omitted if directories should be included. Then, we sort the list in-place, using the modify date as the key.

Answered By: efotinis
sorted(filter(os.path.isfile, os.listdir('.')), 
    key=lambda p: os.stat(p).st_mtime)

You could use os.walk('.').next()[-1] instead of filtering with os.path.isfile, but that leaves dead symlinks in the list, and os.stat will fail on them.

Answered By: Alex Coventry

Update: to sort dirpath‘s entries by modification date in Python 3:

import os
from pathlib import Path

paths = sorted(Path(dirpath).iterdir(), key=os.path.getmtime)

(put @Pygirl’s answer here for greater visibility)

If you already have a list of filenames files, then to sort it inplace by creation time on Windows (make sure that list contains absolute path):


The list of files you could get, for example, using glob as shown in @Jay’s answer.

old answer
Here’s a more verbose version of @Greg Hewgill‘s answer. It is the most conforming to the question requirements. It makes a distinction between creation and modification dates (at least on Windows).

#!/usr/bin/env python
from stat import S_ISREG, ST_CTIME, ST_MODE
import os, sys, time

# path to the directory (relative or absolute)
dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'

# get all entries in the directory w/ stats
entries = (os.path.join(dirpath, fn) for fn in os.listdir(dirpath))
entries = ((os.stat(path), path) for path in entries)

# leave only regular files, insert creation date
entries = ((stat[ST_CTIME], path)
           for stat, path in entries if S_ISREG(stat[ST_MODE]))
#NOTE: on Windows `ST_CTIME` is a creation date 
#  but on Unix it could be something else
#NOTE: use `ST_MTIME` to sort by a modification date
for cdate, path in sorted(entries):
    print time.ctime(cdate), os.path.basename(path)


$ python
Thu Feb 11 13:31:07 2009
Answered By: jfs

There is an os.path.getmtime function that gives the number of seconds since the epoch
and should be faster than os.stat.

import os 

sorted(filter(os.path.isfile, os.listdir('.')), key=os.path.getmtime)
Answered By: gypaetus

this is a basic step for learn:

import os, stat, sys
import time

dirpath = sys.argv[1] if len(sys.argv) == 2 else r'.'

listdir = os.listdir(dirpath)

for i in listdir:
    data_001 = os.path.realpath(i)
    listdir_stat1 = os.stat(data_001)
    listdir_stat2 = ((os.stat(data_001), data_001))
    print time.ctime(listdir_stat1.st_ctime), data_001
Answered By: cumulus13

Here’s my answer using glob without filter if you want to read files with a certain extension in date order (Python 3).

files = glob.glob(dataset_path+"/morepath/*.extension")   
Answered By: dinos66

Without changing directory:

import os    

path = '/path/to/files/'
name_list = os.listdir(path)
full_list = [os.path.join(path,i) for i in name_list]
time_sorted_list = sorted(full_list, key=os.path.getmtime)

print time_sorted_list

# if you want just the filenames sorted, simply remove the dir from each
sorted_filename_list = [ os.path.basename(i) for i in time_sorted_list]
print sorted_filename_list
Answered By: Nic

Alex Coventry’s answer will produce an exception if the file is a symlink to an unexistent file, the following code corrects that answer:

import time
import datetime
sorted(filter(os.path.isfile, os.listdir('.')), 
    key=lambda p: os.path.exists(p) and os.stat(p).st_mtime or time.mktime(

When the file doesn’t exist, now() is used, and the symlink will go at the very end of the list.

Answered By: Paolo Benvenuto

In python 3.5+

from pathlib import Path
sorted(Path('.').iterdir(), key=lambda f: f.stat().st_mtime)
Answered By: ignorant

Here is a simple couple lines that looks for extention as well as provides a sort option

def get_sorted_files(src_dir, regex_ext='*', sort_reverse=False): 
    files_to_evaluate = [os.path.join(src_dir, f) for f in os.listdir(src_dir) if'.*.({})$'.format(regex_ext), f)]
    files_to_evaluate.sort(key=os.path.getmtime, reverse=sort_reverse)
    return files_to_evaluate
Answered By: TXN_747
# *** the shortest and best way ***
# getmtime --> sort by modified time
# getctime --> sort by created time

import glob,os

lst_files = glob.glob("*.txt")
Answered By: Arash

For completeness with os.scandir (2x faster over pathlib):

import os
sorted(os.scandir('/tmp/test'), key=lambda d: d.stat().st_mtime)
Answered By: n1nj4
from pathlib import Path
import os

sorted(Path('./').iterdir(), key=lambda t: t.stat().st_mtime)


sorted(Path('./').iterdir(), key=os.path.getmtime)


sorted(os.scandir('./'), key=lambda t: t.stat().st_mtime)

where m time is modified time.

Answered By: Pygirl

This was my version:

import os

folder_path = r'D:Moviesextranewdramas' # your path
os.chdir(folder_path) # make the path active
x = sorted(os.listdir(), key=os.path.getctime)  # sorted using creation time

folder = 0

for folder in range(len(x)):
    print(x[folder]) # print all the foldername inside the folder_path
    folder = +1
Answered By: haqrafiul

Turns out os.listdir sorts by last modified but in reverse so you can do:

import os
Answered By: Mayank

Add the file directory/folder in path, if you want to have specific file type add the file extension, and then get file name in chronological order.
This works for me.

import glob, os
from pathlib import Path
path = os.path.expanduser(file_location+"/"+date_file)  

Answered By: Aps
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