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()])
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)
import os
path, dirs, files = next(os.walk("/usr/lib"))
file_count = len(files)
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)
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)
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))
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))
Short and simple
import os
directory_path = '/home/xyz/'
No_of_files = len(os.listdir(directory_path))
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/']))
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
one liner and recursive:
def count_files(path):
return sum([len(files) for _, _, files in os.walk(path)])
count_files('path/to/dir')
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
import os
print len(os.listdir(os.getcwd()))
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
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.
Luke's code reformat.
import os
print len(os.walk('/usr/lib').next()[2])
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
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
Here is a simple one-line command that I found useful:
print int(os.popen("ls | wc -l").read())
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)))
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"))
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)
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"))
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.
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}
Source: Stackoverflow.com