在我重新发明这个特殊的轮子之前,有没有人有一个很好的用Python计算目录大小的例程?如果该例程能以Mb/Gb等格式格式化大小,那就太好了。


当前回答

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

其他回答

问题的第二部分

def human(size):

    B = "B"
    KB = "KB" 
    MB = "MB"
    GB = "GB"
    TB = "TB"
    UNITS = [B, KB, MB, GB, TB]
    HUMANFMT = "%f %s"
    HUMANRADIX = 1024.

    for u in UNITS[:-1]:
        if size < HUMANRADIX : return HUMANFMT % (size, u)
        size /= HUMANRADIX

    return HUMANFMT % (size,  UNITS[-1])

到目前为止,建议的一些方法实现了递归,其他方法使用shell或不会生成格式整齐的结果。当您的代码对于Linux平台是一次性的,您可以像往常一样获得格式化,包括递归,作为一行程序。除了最后一行的输出,它将适用于当前版本的python2和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('.'))

简单,高效,将工作于文件和多级目录:

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

Chris的回答很好,但可以通过使用set来检查已看到的目录来使其更加惯用,这也避免了对控制流使用异常:

def directory_size(path):
    total_size = 0
    seen = set()

    for dirpath, dirnames, filenames in os.walk(path):
        for f in filenames:
            fp = os.path.join(dirpath, f)

            try:
                stat = os.stat(fp)
            except OSError:
                continue

            if stat.st_ino in seen:
                continue

            seen.add(stat.st_ino)

            total_size += stat.st_size

    return total_size  # size in bytes

要获取一个文件的大小,可以使用os.path.getsize()

>>> import os
>>> os.path.getsize("/path/file")
35L

它以字节为单位报告。

import os

def get_size(path):
    total_size = 0
    for dirpath, dirnames, filenames in os.walk(path):
        for f in filenames:
            if os.path.exists(fp):
                fp = os.path.join(dirpath, f)
                total_size += os.path.getsize(fp)

    return total_size   # in megabytes

谢谢monkut & troex!