Calculating a directory's size using Python?

Question

This walks all sub-directories; summing file sizes:

import os

def get_size(start_path = '.'):
    total_size = 0
    for dirpath, dirnames, filenames in os.walk(start_path):
        for f in filenames:
            fp = os.path.join(dirpath, f)
            # skip if it is symbolic link
            if not os.path.islink(fp):
                total_size += os.path.getsize(fp)

    return total_size

print(get_size(), 'bytes')

And a oneliner for fun using os.listdir (Does not include sub-directories):

import os
sum(os.path.getsize(f) for f in os.listdir('.') if os.path.isfile(f))

Reference:

• os.path.getsize - Gives the size in bytes
• os.walk
• os.path.islink

Updated To use os.path.getsize, this is clearer than using the os.stat().st_size method.

Thanks to ghostdog74 for pointing this out!

os.stat - st_size Gives the size in bytes. Can also be used to get file size and other file related information.

import os

nbytes = sum(d.stat().st_size for d in os.scandir('.') if d.is_file())

Update 2018

If you use Python 3.4 or previous then you may consider using the more efficient walk method provided by the third-party scandir package. In Python 3.5 and later, this package has been incorporated into the standard library and os.walk has received the corresponding increase in performance.

Update 2019

Recently I've been using pathlib more and more, here's a pathlib solution:

from pathlib import Path

root_directory = Path('.')
sum(f.stat().st_size for f in root_directory.glob('**/*') if f.is_file())

Some of the approaches suggested so far implement a recursion, others employ a shell or will not produce neatly formatted results. When your code is one-off for Linux platforms, you can get formatting as usual, recursion included, as a one-liner. Except for the print in the last line, it will work for current versions of python2 and python3:

du.py
-----
#!/usr/bin/python3
import subprocess

def du(path):
    """disk usage in human readable format (e.g. '2,1GB')"""
    return subprocess.check_output(['du','-sh', path]).split()[0].decode('utf-8')

if __name__ == "__main__":
    print(du('.'))

is simple, efficient and will work for files and multilevel directories:

$ chmod 750 du.py
$ ./du.py
2,9M

Here is a recursive function (it recursively sums up the size of all subfolders and their respective files) which returns exactly the same bytes as when running "du -sb ." in linux (where the "." means "the current folder"):

import os

def getFolderSize(folder):
    total_size = os.path.getsize(folder)
    for item in os.listdir(folder):
        itempath = os.path.join(folder, item)
        if os.path.isfile(itempath):
            total_size += os.path.getsize(itempath)
        elif os.path.isdir(itempath):
            total_size += getFolderSize(itempath)
    return total_size

print "Size: " + str(getFolderSize("."))

Python 3.5 recursive folder size using os.scandir

def folder_size(path='.'):
    total = 0
    for entry in os.scandir(path):
        if entry.is_file():
            total += entry.stat().st_size
        elif entry.is_dir():
            total += folder_size(entry.path)
    return total

Using pathlib I came up this one-liner to get the size of a folder:

sum(file.stat().st_size for file in Path(folder).rglob('*'))

And this is what I came up with for a nicely formatted output:

from pathlib import Path


def get_folder_size(folder):
    return ByteSize(sum(file.stat().st_size for file in Path(folder).rglob('*')))


class ByteSize(int):

    _kB = 1024
    _suffixes = 'B', 'kB', 'MB', 'GB', 'PB'

    def __new__(cls, *args, **kwargs):
        return super().__new__(cls, *args, **kwargs)

    def __init__(self, *args, **kwargs):
        self.bytes = self.B = int(self)
        self.kilobytes = self.kB = self / self._kB**1
        self.megabytes = self.MB = self / self._kB**2
        self.gigabytes = self.GB = self / self._kB**3
        self.petabytes = self.PB = self / self._kB**4
        *suffixes, last = self._suffixes
        suffix = next((
            suffix
            for suffix in suffixes
            if 1 < getattr(self, suffix) < self._kB
        ), last)
        self.readable = suffix, getattr(self, suffix)

        super().__init__()

    def __str__(self):
        return self.__format__('.2f')

    def __repr__(self):
        return '{}({})'.format(self.__class__.__name__, super().__repr__())

    def __format__(self, format_spec):
        suffix, val = self.readable
        return '{val:{fmt}} {suf}'.format(val=val, fmt=format_spec, suf=suffix)

    def __sub__(self, other):
        return self.__class__(super().__sub__(other))

    def __add__(self, other):
        return self.__class__(super().__add__(other))

    def __mul__(self, other):
        return self.__class__(super().__mul__(other))

    def __rsub__(self, other):
        return self.__class__(super().__sub__(other))

    def __radd__(self, other):
        return self.__class__(super().__add__(other))

    def __rmul__(self, other):
        return self.__class__(super().__rmul__(other))   

Usage:

>>> size = get_folder_size("c:/users/tdavis/downloads")
>>> print(size)
5.81 GB
>>> size.GB
5.810891855508089
>>> size.gigabytes
5.810891855508089
>>> size.PB
0.005674699077644618
>>> size.MB
5950.353260040283
>>> size
ByteSize(6239397620)

I also came across this question, which has some more compact and probably more performant strategies for printing file sizes.