当计算子目录的大小时，它应该更新其父目录的文件夹大小，这将一直进行下去，直到它到达根父目录。

下面的函数计算文件夹及其所有子文件夹的大小。

import os

def folder_size(path):
    parent = {}  # path to parent path mapper
    folder_size = {}  # storing the size of directories
    folder = os.path.realpath(path)

    for root, _, filenames in os.walk(folder):
        if root == folder:
            parent[root] = -1  # the root folder will not have any parent
            folder_size[root] = 0.0  # intializing the size to 0

        elif root not in parent:
            immediate_parent_path = os.path.dirname(root)  # extract the immediate parent of the subdirectory
            parent[root] = immediate_parent_path  # store the parent of the subdirectory
            folder_size[root] = 0.0  # initialize the size to 0

        total_size = 0
        for filename in filenames:
            filepath = os.path.join(root, filename)
            total_size += os.stat(filepath).st_size  # computing the size of the files under the directory
        folder_size[root] = total_size  # store the updated size

        temp_path = root  # for subdirectories, we need to update the size of the parent till the root parent
        while parent[temp_path] != -1:
            folder_size[parent[temp_path]] += total_size
            temp_path = parent[temp_path]

    return folder_size[folder]/1000000.0

蒙纳特的答案是很好的，但它失败的符号链接，所以你还必须检查这个路径是否真的存在

if os.path.exists(fp):
    total_size += os.stat(fp).st_size

我使用带有scandir的python 2.7.13，这里是我的一行递归函数，以获得文件夹的总大小:

from scandir import scandir
def getTotFldrSize(path):
    return sum([s.stat(follow_symlinks=False).st_size for s in scandir(path) if s.is_file(follow_symlinks=False)]) + \
    + sum([getTotFldrSize(s.path) for s in scandir(path) if s.is_dir(follow_symlinks=False)])

>>> print getTotFldrSize('.')
1203245680

https://pypi.python.org/pypi/scandir

Python 3.6+递归文件夹/文件大小使用os.scandir。和@blakev的回答一样强大，但更短，采用EAFP python风格。

import os

def size(path, *, follow_symlinks=False):
    try:
        with os.scandir(path) as it:
            return sum(size(entry, follow_symlinks=follow_symlinks) for entry in it)
    except NotADirectoryError:
        return os.stat(path, follow_symlinks=follow_symlinks).st_size

你说一句俏皮话… 这里有一句话:

sum([sum(map(lambda fname: os.path.getsize(os.path.join(directory, fname)), files)) for directory, folders, files in os.walk(path)])

尽管我可能会把它分开，它不执行检查。

要转换为kb，请参阅可重用库以获得人类可读版本的文件大小?然后算进去

我在这里有点晚(和新)，但我选择使用subprocess模块和Linux中的'du'命令行来检索文件夹大小的准确值，单位为MB。我必须使用if和elif用于根文件夹，否则子进程会由于返回的非零值而引发错误。

import subprocess
import os

#
# get folder size
#
def get_size(self, path):
    if os.path.exists(path) and path != '/':
        cmd = str(subprocess.check_output(['sudo', 'du', '-s', path])).\
            replace('b\'', '').replace('\'', '').split('\\t')[0]
        return float(cmd) / 1000000
    elif os.path.exists(path) and path == '/':
        cmd = str(subprocess.getoutput(['sudo du -s /'])). \
            replace('b\'', '').replace('\'', '').split('\n')
        val = cmd[len(cmd) - 1].replace('/', '').replace(' ', '')
        return float(val) / 1000000
    else: raise ValueError