How to get the latest file in a folder?

Question

I need to get the latest file of a folder using python. While using the code:

max(files, key = os.path.getctime)

I am getting the below error:

FileNotFoundError: [WinError 2] The system cannot find the file specified: 'a'

Answer 1

Whatever is assigned to the files variable is incorrect. Use the following code.

import glob
import os

list_of_files = glob.glob('/path/to/folder/*') # * means all if need specific format then *.csv
latest_file = max(list_of_files, key=os.path.getctime)
print(latest_file)

Answer 2

max(files, key = os.path.getctime)

is quite incomplete code. What is files? It probably is a list of file names, coming out of os.listdir().

But this list lists only the filename parts (a. k. a. "basenames"), because their path is common. In order to use it correctly, you have to combine it with the path leading to it (and used to obtain it).

Such as (untested):

def newest(path):
    files = os.listdir(path)
    paths = [os.path.join(path, basename) for basename in files]
    return max(paths, key=os.path.getctime)

Answer 3

I lack the reputation to comment but ctime from Marlon Abeykoons response did not give the correct result for me. Using mtime does the trick though. (key=os.path.getmtime))

import glob
import os

list_of_files = glob.glob('/path/to/folder/*') # * means all if need specific format then *.csv
latest_file = max(list_of_files, key=os.path.getmtime)
print(latest_file)

I found two answers for that problem:

python os.path.getctime max does not return latest Difference between python - getmtime() and getctime() in unix system

Answer 4

I've been using this in Python 3, including pattern matching on the filename.

from pathlib import Path

def latest_file(path: Path, pattern: str = "*"):
    files = path.glob(pattern)
    return max(files, key=lambda x: x.stat().st_ctime)

Answer 5

I would suggest using glob.iglob() instead of the glob.glob(), as it is more efficient.

glob.iglob() Return an iterator which yields the same values as glob() without actually storing them all simultaneously.

Which means glob.iglob() will be more efficient.

I mostly use below code to find the latest file matching to my pattern:

LatestFile = max(glob.iglob(fileNamePattern),key=os.path.getctime)

---

NOTE: There are variants of max function, In case of finding the latest file we will be using below variant: max(iterable, *[, key, default])

which needs iterable so your first parameter should be iterable. In case of finding max of nums we can use beow variant : max (num1, num2, num3, *args[, key])

Answer 6

Try to sort items by creation time. Example below sorts files in a folder and gets first element which is latest.

import glob
import os

files_path = os.path.join(folder, '*')
files = sorted(
    glob.iglob(files_path), key=os.path.getctime, reverse=True) 
print files[0]

Answer 7

Most of the answers are correct but if there is a requirement like getting the latest two or three latest then it could fail or need to modify the code.

I found the below sample is more useful and relevant as we can use the same code to get the latest 2,3 and n files too.

import glob
import os

folder_path = "/Users/sachin/Desktop/Files/"
files_path = os.path.join(folder_path, '*')
files = sorted(glob.iglob(files_path), key=os.path.getctime, reverse=True) 
print (files[0]) #latest file 
print (files[0],files[1]) #latest two files

Answer 8

A much faster method on windows (0.05s), call a bat script that does this:

get_latest.bat

@echo off
for /f %%i in ('dir \\directory\in\question /b/a-d/od/t:c') do set LAST=%%i
%LAST%

where \\directory\in\question is the directory you want to investigate.

get_latest.py

from subprocess import Popen, PIPE
p = Popen("get_latest.bat", shell=True, stdout=PIPE,)
stdout, stderr = p.communicate()
print(stdout, stderr)

if it finds a file stdout is the path and stderr is None.

Use stdout.decode("utf-8").rstrip() to get the usable string representation of the file name.

Answer 9

(Edited to improve answer)

First define a function get_latest_file

def get_latest_file(path, *paths):
    fullpath = os.path.join(path, paths)
    ...
get_latest_file('example', 'files','randomtext011.*.txt')

You may also use a docstring !

def get_latest_file(path, *paths):
    """Returns the name of the latest (most recent) file 
    of the joined path(s)"""
    fullpath = os.path.join(path, *paths)

If you use Python 3, you can use iglob instead.

Complete code to return the name of latest file:

def get_latest_file(path, *paths):
    """Returns the name of the latest (most recent) file 
    of the joined path(s)"""
    fullpath = os.path.join(path, *paths)
    files = glob.glob(fullpath)  # You may use iglob in Python3
    if not files:                # I prefer using the negation
        return None                      # because it behaves like a shortcut
    latest_file = max(files, key=os.path.getctime)
    _, filename = os.path.split(latest_file)
    return filename

Answer 10

I have tried to use the above suggestions and my program crashed, than I figured out the file I'm trying to identify was used and when trying to use 'os.path.getctime' it crashed. what finally worked for me was:

    files_before = glob.glob(os.path.join(my_path,'*'))
    **code where new file is created**
    new_file = set(files_before).symmetric_difference(set(glob.glob(os.path.join(my_path,'*'))))

this codes gets the uncommon object between the two sets of file lists its not the most elegant, and if multiple files are created at the same time it would probably won't be stable