360

How do I count only the files in a directory? This counts the directory itself as a file:

len(glob.glob('*'))
Mateen Ulhaq
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prosseek
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    To leave out directories, you can do '*.fileextension' for whatever file extension you are looking for. –  Mar 24 '18 at 02:28

30 Answers30

407

os.listdir() will be slightly more efficient than using glob.glob. To test if a filename is an ordinary file (and not a directory or other entity), use os.path.isfile():

import os, os.path

# simple version for working with CWD
print len([name for name in os.listdir('.') if os.path.isfile(name)])

# path joining version for other paths
DIR = '/tmp'
print len([name for name in os.listdir(DIR) if os.path.isfile(os.path.join(DIR, name))])
Bruno Bronosky
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Daniel Stutzbach
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    Remember to add the `folder_path` inside `os.path.filename(name)` if you're not on the cwd. http://stackoverflow.com/questions/17893542/why-do-os-path-isfile-return-false – Rafael Oliveira Apr 10 '14 at 14:54
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    This doesn't count the file inside nested folders. – codersofthedark Apr 23 '15 at 09:33
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    For recursively counting files nested inside directories, you might be better off with the os.walk() solution. – Joel B Dec 23 '15 at 21:23
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    What is the benefit of using `os.path.join(DIR, name)` over `DIR + '/' + name`? The latter is shorter and, IMO, more clear than the former. Is there perhaps some OS:es on which the latter would fail? – HelloGoodbye Jun 16 '16 at 13:46
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    @HelloGoodbye That's exactly the reason. – ellockie Nov 29 '16 at 16:28
  • Doing this on a directory with a huge number of files is slow. Is there a way of getting the count in any other way? – Eduardo Pignatelli Apr 13 '18 at 16:17
  • If you need to search for a specific filename, then you can use `'your_filename' in name` like this: `print len([name for name in os.listdir('.') if os.path.isfile(name) and 'your_filename' in name])` – tsveti_iko Aug 10 '18 at 10:01
  • @HelloGoodbye if prettifying bugs you too much - `pathlib`'s (Pure)Path actually has `/` **operator** which does the same OS-aware stuff - `p / 'some' / 'dir'` https://docs.python.org/3/library/pathlib.html#operators – jave.web Nov 13 '20 at 01:28
  • Is this a disk intensive operation to run on unix? I am trying to create a directory serving API and for the UI, I wanna show count of subfolders/files but dont want to wear out the disk too much. – ScipioAfricanus Dec 20 '20 at 18:50
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    For those who uses python3, print(len(os.listdir('DIRECTORY_PATH'))) – Brady Huang Jan 08 '21 at 03:31
  • @DanielStutzbach would you mind updating this answer with Brady Huang 's suggestion ? Edit's queue seems to be full – Timothee W Feb 28 '23 at 14:14
175
import os

_, _, files = next(os.walk("/usr/lib"))
file_count = len(files)
bryant1410
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Luke
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80

For all kind of files, subdirectories included (Python 2):

import os

lst = os.listdir(directory) # your directory path
number_files = len(lst)
print number_files

Only files (avoiding subdirectories):

import os

onlyfiles = next(os.walk(directory))[2] #directory is your directory path as string
print len(onlyfiles)
MattDMo
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Guillermo Pereira
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51

This is where fnmatch comes very handy:

import fnmatch

print len(fnmatch.filter(os.listdir(dirpath), '*.txt'))

More details: http://docs.python.org/2/library/fnmatch.html

ngeek
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    This is much faster (about half the time with my testing on a directory with 10,000 files) if you know the pattern you're looking for, rather then testing each file with `os.path.isfile()` as the accepted answer does. Also significantly faster than `glob.glob()`. – CivFan Apr 27 '16 at 15:54
34

If you want to count all files in the directory - including files in subdirectories, the most pythonic way is:

import os

file_count = sum(len(files) for _, _, files in os.walk(r'C:\Dropbox'))
print(file_count)

We use sum that is faster than explicitly adding the file counts (timings pending)

Mr_and_Mrs_D
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    Hi, I was trying to understand this code (the code works perfect), I know we can use `_` in a `for` loop. `os.walk` also I know. But not sure what's going on with underscores inside the `sum` function, could you please elaborate. Thanks! – Ejaz Jan 02 '18 at 13:22
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    Unsderscore is just a variable name @Ejaz, by convention used when we ignore the variable - that's what we do here - we call walk and only count the number of files in each directory, ignoring the root and dirs walk return values – Mr_and_Mrs_D Jan 03 '18 at 18:58
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    This is completely recursive and probably the best answer here. – SomJura Sep 22 '20 at 21:25
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    This should be the most appropriate answer, to also count the files in any subfolders.. – NoobCat Oct 13 '20 at 13:54
24

An answer with pathlib and without loading the whole list to memory:

from pathlib import Path

path = Path('.')

print(sum(1 for _ in path.glob('*')))  # Files and folders, not recursive
print(sum(1 for _ in path.rglob('*')))  # Files and folders, recursive

print(sum(1 for x in path.glob('*') if x.is_file()))  # Only files, not recursive
print(sum(1 for x in path.rglob('*') if x.is_file()))  # Only files, recursive
Paul
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22

Short and simple

import os
directory_path = '/home/xyz/'
No_of_files = len(os.listdir(directory_path))
19

I am surprised that nobody mentioned os.scandir:

def count_files(dir):
    return len([1 for x in list(os.scandir(dir)) if x.is_file()])
qed
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12
def directory(path,extension):
  list_dir = []
  list_dir = os.listdir(path)
  count = 0
  for file in list_dir:
    if file.endswith(extension): # eg: '.txt'
      count += 1
  return count
Marcus Riemer
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ninjrok
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12
import os
print len(os.listdir(os.getcwd()))
rash
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8

This uses os.listdir and works for any directory:

import os
directory = 'mydirpath'

number_of_files = len([item for item in os.listdir(directory) if os.path.isfile(os.path.join(directory, item))])

this can be simplified with a generator and made a little bit faster with:

import os
isfile = os.path.isfile
join = os.path.join

directory = 'mydirpath'
number_of_files = sum(1 for item in os.listdir(directory) if isfile(join(directory, item)))
joaquin
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8

While I agree with the answer provided by @DanielStutzbach: os.listdir() will be slightly more efficient than using glob.glob.

However, an extra precision, if you do want to count the number of specific files in folder, you want to use len(glob.glob()). For instance if you were to count all the pdfs in a folder you want to use:

pdfCounter = len(glob.glob1(myPath,"*.pdf"))
LBes
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6

This is an easy solution that counts the number of files in a directory containing sub-folders. It may come in handy:

import os
from pathlib import Path

def count_files(rootdir):
    '''counts the number of files in each subfolder in a directory'''
    for path in pathlib.Path(rootdir).iterdir():
        if path.is_dir():
            print("There are " + str(len([name for name in os.listdir(path) \
            if os.path.isfile(os.path.join(path, name))])) + " files in " + \
            str(path.name))
            
 
count_files(data_dir) # data_dir is the directory you want files counted.

You should get an output similar to this (with the placeholders changed, of course):

There are {number of files} files in {name of sub-folder1}
There are {number of files} files in {name of sub-folder2}
MLDev
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5
def count_em(valid_path):
   x = 0
   for root, dirs, files in os.walk(valid_path):
       for f in files:
            x = x+1
print "There are", x, "files in this directory."
return x

Taked from this post

Kristian Damian
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5

one liner and recursive:

def count_files(path):
    return sum([len(files) for _, _, files in os.walk(path)])

count_files('path/to/dir')
juan Isaza
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4

Luke's code reformat.

import os

print len(os.walk('/usr/lib').next()[2])
okobaka
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4
import os

def count_files(in_directory):
    joiner= (in_directory + os.path.sep).__add__
    return sum(
        os.path.isfile(filename)
        for filename
        in map(joiner, os.listdir(in_directory))
    )

>>> count_files("/usr/lib")
1797
>>> len(os.listdir("/usr/lib"))
2049
tzot
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4

Here is a simple one-line command that I found useful:

print int(os.popen("ls | wc -l").read())
  • Parsing the output of `ls` is generally frowned upon (it can frequently cause issues), though this is not a bad "quick-and-dirty" method at the shell. You should use `ls -1`, though, so it guarantees one line per file. – Bloodgain Apr 03 '20 at 22:45
2

I used glob.iglob for a directory structure similar to

data
└───train
│   └───subfolder1
│   |   │   file111.png
│   |   │   file112.png
│   |   │   ...
│   |
│   └───subfolder2
│       │   file121.png
│       │   file122.png
│       │   ...
└───test
    │   file221.png
    │   file222.png

Both of the following options return 4 (as expected, i.e. does not count the subfolders themselves)

  • len(list(glob.iglob("data/train/*/*.png", recursive=True)))
  • sum(1 for i in glob.iglob("data/train/*/*.png"))
user799188
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2

It is simple:

print(len([iq for iq in os.scandir('PATH')]))

it simply counts number of files in directory , i have used list comprehension technique to iterate through specific directory returning all files in return . "len(returned list)" returns number of files.

Agha Saad
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    Welcome to Stack Overflow. The quality of this answer can be improved by adding an explanation: [How to Answer](https://stackoverflow.com/help/how-to-answer) – Elletlar Jul 29 '18 at 10:44
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    Thankyou Elletlar , i have edited my answer , i will make sure to respond in more comprehensive manner :D – Agha Saad Jul 31 '18 at 15:05
1
import os

total_con=os.listdir('<directory path>')

files=[]

for f_n in total_con:
   if os.path.isfile(f_n):
     files.append(f_n)


print len(files)
Mohit Dabas
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1

If you'll be using the standard shell of the operating system, you can get the result much faster rather than using pure pythonic way.

Example for Windows:

import os
import subprocess

def get_num_files(path):
    cmd = 'DIR \"%s\" /A-D /B /S | FIND /C /V ""' % path
    return int(subprocess.check_output(cmd, shell=True))
styler
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1

I found another answer which may be correct as accepted answer.

for root, dirs, files in os.walk(input_path):    
for name in files:
    if os.path.splitext(name)[1] == '.TXT' or os.path.splitext(name)[1] == '.txt':
        datafiles.append(os.path.join(root,name)) 


print len(files) 
Ismail
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1

A simple utility function I wrote that makes use of os.scandir() instead of os.listdir().

import os 

def count_files_in_dir(path: str) -> int:
    file_entries = [entry for entry in os.scandir(path) if entry.is_file()]

    return len(file_entries)

The main benefit is that, the need for os.path.is_file() is eliminated and replaced with os.DirEntry instance's is_file() which also removes the need for os.path.join(DIR, file_name) as shown in other answers.

Kinyugo
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0

i did this and this returned the number of files in the folder(Attack_Data)...this works fine.

import os
def fcount(path):
    #Counts the number of files in a directory
    count = 0
    for f in os.listdir(path):
        if os.path.isfile(os.path.join(path, f)):
            count += 1

    return count
path = r"C:\Users\EE EKORO\Desktop\Attack_Data" #Read files in folder
print (fcount(path))
jkalden
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0

Simpler one:

import os
number_of_files = len(os.listdir(directory))
print(number_of_files)
Mayur Gupta
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0

I find that sometimes I don't know if I will receive filenames or the path to the file. So I printed the os walk solution output:

def count_number_of_raw_data_point_files(path: Union[str, Path], with_file_prefix: str) -> int:
    import os
    path: Path = force_expanduser(path)

    _, _, files = next(os.walk(path))
    # file_count = len(files)
    filename: str
    count: int = 0
    for filename in files:
        print(f'-->{filename=}')  # e.g. print -->filename='data_point_99.json'
        if with_file_prefix in filename:
            count += 1
    return count

out:

-->filename='data_point_780.json'
-->filename='data_point_781.json'
-->filename='data_point_782.json'
-->filename='data_point_783.json'
-->filename='data_point_784.json'
-->filename='data_point_785.json'
-->filename='data_point_786.json'
-->filename='data_point_787.json'
-->filename='data_point_788.json'
-->filename='data_point_789.json'
-->filename='data_point_79.json'
-->filename='data_point_790.json'
-->filename='data_point_791.json'
-->filename='data_point_792.json'
-->filename='data_point_793.json'
-->filename='data_point_794.json'
-->filename='data_point_795.json'
-->filename='data_point_796.json'
-->filename='data_point_797.json'
-->filename='data_point_798.json'
-->filename='data_point_799.json'
-->filename='data_point_8.json'
-->filename='data_point_80.json'
-->filename='data_point_800.json'
-->filename='data_point_801.json'
-->filename='data_point_802.json'
-->filename='data_point_803.json'
-->filename='data_point_804.json'
-->filename='data_point_805.json'
-->filename='data_point_806.json'
-->filename='data_point_807.json'
-->filename='data_point_808.json'
-->filename='data_point_809.json'
-->filename='data_point_81.json'
-->filename='data_point_810.json'
-->filename='data_point_811.json'
-->filename='data_point_812.json'
-->filename='data_point_813.json'
-->filename='data_point_814.json'
-->filename='data_point_815.json'
-->filename='data_point_816.json'
-->filename='data_point_817.json'
-->filename='data_point_818.json'
-->filename='data_point_819.json'
-->filename='data_point_82.json'
-->filename='data_point_820.json'
-->filename='data_point_821.json'
-->filename='data_point_822.json'
-->filename='data_point_823.json'
-->filename='data_point_824.json'
-->filename='data_point_825.json'
-->filename='data_point_826.json'
-->filename='data_point_827.json'
-->filename='data_point_828.json'
-->filename='data_point_829.json'
-->filename='data_point_83.json'
-->filename='data_point_830.json'
-->filename='data_point_831.json'
-->filename='data_point_832.json'
-->filename='data_point_833.json'
-->filename='data_point_834.json'
-->filename='data_point_835.json'
-->filename='data_point_836.json'
-->filename='data_point_837.json'
-->filename='data_point_838.json'
-->filename='data_point_839.json'
-->filename='data_point_84.json'
-->filename='data_point_840.json'
-->filename='data_point_841.json'
-->filename='data_point_842.json'
-->filename='data_point_843.json'
-->filename='data_point_844.json'
-->filename='data_point_845.json'
-->filename='data_point_846.json'
-->filename='data_point_847.json'
-->filename='data_point_848.json'
-->filename='data_point_849.json'
-->filename='data_point_85.json'
-->filename='data_point_850.json'
-->filename='data_point_851.json'
-->filename='data_point_852.json'
-->filename='data_point_853.json'
-->filename='data_point_86.json'
-->filename='data_point_87.json'
-->filename='data_point_88.json'
-->filename='data_point_89.json'
-->filename='data_point_9.json'
-->filename='data_point_90.json'
-->filename='data_point_91.json'
-->filename='data_point_92.json'
-->filename='data_point_93.json'
-->filename='data_point_94.json'
-->filename='data_point_95.json'
-->filename='data_point_96.json'
-->filename='data_point_97.json'
-->filename='data_point_98.json'
-->filename='data_point_99.json'
854

note you might have to sort.

Charlie Parker
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I would like to extend the reply from @Mr_and_Mrs_D:

import os
folder = 'C:/Dropbox'
file_count = sum(len(files) for _, _, files in os.walk(folder))
print(file_count)

This counts all the files in the folder and its subfolders. However, if you want to do some filtering - like only counting the files ending in .svg, you can do:

import os
file_count = sum(len([f for f in files if f.endswith('.svg')]) for _, _, files in os.walk(folder))
print(file_count)

You basically replace:

  • len(files)

with:

  • len([f for f in files if f.endswith('.svg')])
K.Mulier
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I suppose, using just a shell command is a short and simple solution (based on):

import os

no_files = os.popen(f"ls -F {directory_name} | grep -v / | wc -l")
print(no_files)
  • ls -F - shows the contents of a directory and adds an indicator character to entry names: *, /, =, >, @ or |. In our case, we're most interested in / to mark directories.
  • grep -v / - keeps in the output all strings not containing /.
  • wc -l - counts output's lines.

directory_name is optional.

maciejwww
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-1

Convert it to a list, after that you can make use of the len() function:

len(list(glob.glob('*')))
Skully
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Eslamspot
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