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

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/"
os.chdir(search_dir)
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))

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

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.

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.

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.

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 

os.chdir(directory)
sorted(filter(os.path.isfile, os.listdir('.')), key=os.path.getmtime)

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:
    os.chdir(dirpath)
    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

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

dataset_path='/mydir/'   
files = glob.glob(dataset_path+"/morepath/*.extension")   
files.sort(key=os.path.getmtime)

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

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(datetime.now().timetuple())

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

In python 3.5+

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

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 re.search(r'.*.({})$'.format(regex_ext), f)]
    files_to_evaluate.sort(key=os.path.getmtime, reverse=sort_reverse)
    return files_to_evaluate

# *** the shortest and best way ***
# getmtime --> sort by modified time
# getctime --> sort by created time

import glob,os

lst_files = glob.glob("*.txt")
lst_files.sort(key=os.path.getmtime)
print("n".join(lst_files))

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

import os
sorted(os.scandir('/tmp/test'), key=lambda d: d.stat().st_mtime)

from pathlib import Path
import os

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

or

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

or

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

where m time is modified time.

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

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

import os
last_modified=os.listdir()[::-1]

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)  
os.chdir(path)    
saved_file=glob.glob('*.xlsx')
saved_file.sort(key=os.path.getmtime)

print(saved_file)