How to count the number of files in a directory us

问题:

I need to count the number of files in a directory using Python. I guess the easiest way is len(glob.glob('*')), but that also counts the directory as a file.

Is there any way to count only the files in a directory?

回答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))])

回答2:

import os

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

回答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)

回答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

回答5:

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

回答6:

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

回答7:

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)))

回答8:

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()])

回答9:

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)

回答10:

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

回答11:

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

回答12:

Luke's code reformat.

import os

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

回答13:

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

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

回答14:

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)

回答15:

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))

回答16:

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) 

回答17:

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"))

回答18:

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))

回答19:

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.

回答20:

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"))