How to count the number of files in a directory using Python

Question

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

len(glob.glob('*'))

Answer 1

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))])

Answer 2

import os

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

Answer 3

For all kind of files, subdirectories included:

import os

list = os.listdir(dir) # dir is your directory path
number_files = len(list)
print number_files

Only files (avoiding subdirectories):

import os

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

Answer 4

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

Answer 5

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)

Answer 6

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()])

Answer 7

Short and simple

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

Answer 8

import os
print len(os.listdir(os.getcwd()))

Answer 9

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

Answer 10

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.glob('**/*')))  # 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.glob('**/*') if x.is_file()))  # Only files, recursive

Answer 11

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)))

Answer 12

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"))

Answer 13

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}

Answer 14

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

Answer 15

Luke's code reformat.

import os

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

Answer 16

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

Answer 17

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

print int(os.popen("ls | wc -l").read())

Answer 18

one liner and recursive:

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

count_files('path/to/dir')

Answer 19

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"))

Answer 20

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.

Answer 21

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)

Answer 22

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))

Answer 23

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) 

Answer 24

Simpler one:

import os
number_of_files = len(os.listdir(directory))
print(number_of_files)

Answer 25

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))

Answer 26

I solved this problem while calculating the number of files in a google drive directory through Google Colab by directing myself into the directory folder by

import os                                                                                                
%cd /content/drive/My Drive/  
print(len([x for x in os.listdir('folder_name/']))  

Normal user can try

 import os                                                                                                     
 cd Desktop/Maheep/                                                     
 print(len([x for x in os.listdir('folder_name/']))  

Answer 27

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.

Answer 28

Convert to list after that you can Len

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