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

python3.5 +

from pathlib import Path

def get_size(folder: str) -> int:
    return sum(p.stat().st_size for p in Path(folder).rglob('*'))

用法::

In [6]: get_size('/etc/not-exist-path')
Out[6]: 0
In [7]: get_size('.')
Out[7]: 12038689
In [8]: def filesize(size: int) -> str:
   ...:     for unit in ("B", "K", "M", "G", "T"):
   ...:         if size < 1024:
   ...:             break
   ...:         size /= 1024
   ...:     return f"{size:.1f}{unit}"
   ...:

In [9]: filesize(get_size('.'))
Out[9]: '11.5M'

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])

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

下面是一个递归函数(它递归地总结所有子文件夹及其各自文件的大小),返回的字节与在linux中运行“du -sb .”时完全相同(其中“。”表示“当前文件夹”):

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("."))
import os
def get_size(path = os.getcwd()):
    print("Calculating Size: ",path)
    total_size = 0
    #if path is directory--
    if os.path.isdir(path):
      print("Path type : Directory/Folder")
      for dirpath, dirnames, filenames in os.walk(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)
    #if path is a file---
    elif os.path.isfile(path):
      print("Path type : File")
      total_size=os.path.getsize(path)
    else:
      print("Path Type : Special File (Socket, FIFO, Device File)" )
      total_size=0
    bytesize=total_size
    print(bytesize, 'bytes')
    print(bytesize/(1024), 'kilobytes')
    print(bytesize/(1024*1024), 'megabytes')
    print(bytesize/(1024*1024*1024), 'gegabytes')
    return total_size


x=get_size("/content/examples")

我相信这很有帮助!文件夹和文件!

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