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例程来使用Python计算目录的大小?如果例程以Mb / Gb等格式很好地格式化大小,那将是非常好的。
tree
* nix系统上的命令免费提供所有这些功能。tree -h -d --du /path/to/dir
。
du -sh /path/to/dir/*
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例程来使用Python计算目录的大小?如果例程以Mb / Gb等格式很好地格式化大小,那将是非常好的。
tree
* nix系统上的命令免费提供所有这些功能。tree -h -d --du /path/to/dir
。
du -sh /path/to/dir/*
Answers:
这遍历所有子目录;总结文件大小:
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')
还有一个使用os.listdir(不包含子目录)来娱乐的oneliner :
import os
sum(os.path.getsize(f) for f in os.listdir('.') if os.path.isfile(f))
参考:
已更新 要使用os.path.getsize,这比使用os.stat()。st_size方法更清晰。
感谢ghostdog74指出这一点!
os.stat - st_size给出大小(以字节为单位)。也可以用于获取文件大小和其他文件相关信息。
import os
nbytes = sum(d.stat().st_size for d in os.scandir('.') if d.is_file())
更新2018
如果您使用Python 3.4或更早版本,则可以考虑使用walk
第三方scandir
软件包提供的更有效的方法。在Python 3.5和更高版本中,此软件包已合并到标准库中,os.walk
并获得了相应的性能提升。
更新2019
最近,我一直在使用pathlib
越来越多的pathlib
解决方案:
from pathlib import Path
root_directory = Path('.')
sum(f.stat().st_size for f in root_directory.glob('**/*') if f.is_file())
st_size
如果您不想使用符号链接,则必须使用,因为您应该使用lstat
。
到目前为止建议的某些方法实现了递归,其他方法则使用了shell或不会产生整齐的格式化结果。如果您的代码一次性用于Linux平台,则可以像往常一样进行格式设置,包括递归。除print
最后一行,它会为当前版本的工作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,并且适用于文件和多级目录:
$ chmod 750 du.py
$ ./du.py
2,9M
这是一个递归函数(它递归求和所有子文件夹及其各自文件的大小),返回的字节数与运行“ du -sb”时完全相同。在linux中(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("."))
使用Python 3.5递归文件夹的大小 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
sum([entry.stat().st_size for entry in os.scandir(file)])
。注意输出以字节为单位,/ 1024以字节为单位,/(1024 * 1024)以MB为单位。
sum(entry.stat().st_size for entry in os.scandir(file))
。无需列出列表,因为也sum
需要迭代器。
monknut的答案很好,但是在符号链接断开时失败,因此您还必须检查该路径是否确实存在
if os.path.exists(fp):
total_size += os.stat(fp).st_size
lstat
。
接受的答案未考虑硬链接或软链接,并且会将这些文件计数两次。您想跟踪已看到的inode,而不要增加这些文件的大小。
import os
def get_size(start_path='.'):
total_size = 0
seen = {}
for dirpath, dirnames, filenames in os.walk(start_path):
for f in filenames:
fp = os.path.join(dirpath, f)
try:
stat = os.stat(fp)
except OSError:
continue
try:
seen[stat.st_ino]
except KeyError:
seen[stat.st_ino] = True
else:
continue
total_size += stat.st_size
return total_size
print get_size()
克里斯的回答很好,但可以通过使用一组检查可见目录的方式使它更加惯用,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
df -sb
。
递归一线:
def getFolderSize(p):
from functools import partial
prepend = partial(os.path.join, p)
return sum([(os.path.getsize(f) if os.path.isfile(f) else getFolderSize(f)) for f in map(prepend, os.listdir(p))])
对于问题的第二部分
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])
使用pathlib
我,我得出了以下这种单线来获取文件夹的大小:
sum(file.stat().st_size for file in Path(folder).rglob('*'))
这就是我想出的一种格式正确的输出:
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))
用法:
>>> 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)
我也遇到了这个问题,它有一些更紧凑,性能可能更高的打印文件大小的策略。
您可以执行以下操作:
import commands
size = commands.getoutput('du -sh /path/').split()[0]
在这种情况下,如果需要,我可以在返回结果之前未测试结果,可以使用commands.getstatusoutput进行检查。
os.walk
递归检查子文件夹大小相比,性能如何?
您说的单线...这是一种线:
sum([sum(map(lambda fname: os.path.getsize(os.path.join(directory, fname)), files)) for directory, folders, files in os.walk(path)])
尽管我可能会将其拆分出来,但它不执行任何检查。
要转换为kb,请参阅可重用库以获取文件大小的可读版本?并在其中工作
如果您已经安装了glob2并进行了人性化设置,则聚会晚了一点,但只一行一行。请注意,在Python 3中,默认设置iglob
具有递归模式。对于读者来说,如何修改Python 3的代码是一项琐碎的练习。
>>> import os
>>> from humanize import naturalsize
>>> from glob2 import iglob
>>> naturalsize(sum(os.path.getsize(x) for x in iglob('/var/**'))))
'546.2 MB'
glob
支持递归。您可以使用:glob.glob('/var/**', recursive=True)
以下脚本打印指定目录的所有子目录的目录大小。它还尝试(如果可能)从缓存递归函数的调用中受益。如果省略参数,则脚本将在当前目录中运行。输出按目录大小从最大到最小排序。因此,您可以根据需要对其进行调整。
PS我已经使用配方578019以人类友好的格式显示目录大小(http://code.activestate.com/recipes/578019/)
from __future__ import print_function
import os
import sys
import operator
def null_decorator(ob):
return ob
if sys.version_info >= (3,2,0):
import functools
my_cache_decorator = functools.lru_cache(maxsize=4096)
else:
my_cache_decorator = null_decorator
start_dir = os.path.normpath(os.path.abspath(sys.argv[1])) if len(sys.argv) > 1 else '.'
@my_cache_decorator
def get_dir_size(start_path = '.'):
total_size = 0
if 'scandir' in dir(os):
# using fast 'os.scandir' method (new in version 3.5)
for entry in os.scandir(start_path):
if entry.is_dir(follow_symlinks = False):
total_size += get_dir_size(entry.path)
elif entry.is_file(follow_symlinks = False):
total_size += entry.stat().st_size
else:
# using slow, but compatible 'os.listdir' method
for entry in os.listdir(start_path):
full_path = os.path.abspath(os.path.join(start_path, entry))
if os.path.isdir(full_path):
total_size += get_dir_size(full_path)
elif os.path.isfile(full_path):
total_size += os.path.getsize(full_path)
return total_size
def get_dir_size_walk(start_path = '.'):
total_size = 0
for dirpath, dirnames, filenames in os.walk(start_path):
for f in filenames:
fp = os.path.join(dirpath, f)
total_size += os.path.getsize(fp)
return total_size
def bytes2human(n, format='%(value).0f%(symbol)s', symbols='customary'):
"""
(c) http://code.activestate.com/recipes/578019/
Convert n bytes into a human readable string based on format.
symbols can be either "customary", "customary_ext", "iec" or "iec_ext",
see: http://goo.gl/kTQMs
>>> bytes2human(0)
'0.0 B'
>>> bytes2human(0.9)
'0.0 B'
>>> bytes2human(1)
'1.0 B'
>>> bytes2human(1.9)
'1.0 B'
>>> bytes2human(1024)
'1.0 K'
>>> bytes2human(1048576)
'1.0 M'
>>> bytes2human(1099511627776127398123789121)
'909.5 Y'
>>> bytes2human(9856, symbols="customary")
'9.6 K'
>>> bytes2human(9856, symbols="customary_ext")
'9.6 kilo'
>>> bytes2human(9856, symbols="iec")
'9.6 Ki'
>>> bytes2human(9856, symbols="iec_ext")
'9.6 kibi'
>>> bytes2human(10000, "%(value).1f %(symbol)s/sec")
'9.8 K/sec'
>>> # precision can be adjusted by playing with %f operator
>>> bytes2human(10000, format="%(value).5f %(symbol)s")
'9.76562 K'
"""
SYMBOLS = {
'customary' : ('B', 'K', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'),
'customary_ext' : ('byte', 'kilo', 'mega', 'giga', 'tera', 'peta', 'exa',
'zetta', 'iotta'),
'iec' : ('Bi', 'Ki', 'Mi', 'Gi', 'Ti', 'Pi', 'Ei', 'Zi', 'Yi'),
'iec_ext' : ('byte', 'kibi', 'mebi', 'gibi', 'tebi', 'pebi', 'exbi',
'zebi', 'yobi'),
}
n = int(n)
if n < 0:
raise ValueError("n < 0")
symbols = SYMBOLS[symbols]
prefix = {}
for i, s in enumerate(symbols[1:]):
prefix[s] = 1 << (i+1)*10
for symbol in reversed(symbols[1:]):
if n >= prefix[symbol]:
value = float(n) / prefix[symbol]
return format % locals()
return format % dict(symbol=symbols[0], value=n)
############################################################
###
### main ()
###
############################################################
if __name__ == '__main__':
dir_tree = {}
### version, that uses 'slow' [os.walk method]
#get_size = get_dir_size_walk
### this recursive version can benefit from caching the function calls (functools.lru_cache)
get_size = get_dir_size
for root, dirs, files in os.walk(start_dir):
for d in dirs:
dir_path = os.path.join(root, d)
if os.path.isdir(dir_path):
dir_tree[dir_path] = get_size(dir_path)
for d, size in sorted(dir_tree.items(), key=operator.itemgetter(1), reverse=True):
print('%s\t%s' %(bytes2human(size, format='%(value).2f%(symbol)s'), d))
print('-' * 80)
if sys.version_info >= (3,2,0):
print(get_dir_size.cache_info())
样本输出:
37.61M .\subdir_b
2.18M .\subdir_a
2.17M .\subdir_a\subdir_a_2
4.41K .\subdir_a\subdir_a_1
----------------------------------------------------------
CacheInfo(hits=2, misses=4, maxsize=4096, currsize=4)
编辑:将null_decorator移至上方,建议使用user2233949
很方便:
import os
import stat
size = 0
path_ = ""
def calculate(path=os.environ["SYSTEMROOT"]):
global size, path_
size = 0
path_ = path
for x, y, z in os.walk(path):
for i in z:
size += os.path.getsize(x + os.sep + i)
def cevir(x):
global path_
print(path_, x, "Byte")
print(path_, x/1024, "Kilobyte")
print(path_, x/1048576, "Megabyte")
print(path_, x/1073741824, "Gigabyte")
calculate("C:\Users\Jundullah\Desktop")
cevir(size)
Output:
C:\Users\Jundullah\Desktop 87874712211 Byte
C:\Users\Jundullah\Desktop 85815148.64355469 Kilobyte
C:\Users\Jundullah\Desktop 83803.85609722137 Megabyte
C:\Users\Jundullah\Desktop 81.83970321994275 Gigabyte
我正在使用带有scandir的 python 2.7.13,这是我的一线递归函数来获取文件夹的总大小:
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
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,并一直进行到到达根父目录为止。
以下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
如果您使用的是Windows操作系统,则可以执行以下操作:
通过启动以下命令来安装模块pywin32:
点安装pywin32
然后编写以下代码:
import win32com.client as com
def get_folder_size(path):
try:
fso = com.Dispatch("Scripting.FileSystemObject")
folder = fso.GetFolder(path)
size = str(round(folder.Size / 1048576))
print("Size: " + size + " MB")
except Exception as e:
print("Error --> " + str(e))
该脚本告诉您哪个文件是CWD中最大的文件,还告诉您该文件位于哪个文件夹中。该脚本在win8和python 3.3.3 shell上对我有用
import os
folder=os.cwd()
number=0
string=""
for root, dirs, files in os.walk(folder):
for file in files:
pathname=os.path.join(root,file)
## print (pathname)
## print (os.path.getsize(pathname)/1024/1024)
if number < os.path.getsize(pathname):
number = os.path.getsize(pathname)
string=pathname
## print ()
print (string)
print ()
print (number)
print ("Number in bytes")
诚然,这是一种小技巧,仅适用于Unix / Linux。
du -sb .
之所以匹配,是因为实际上这是一个运行du -sb .
命令的Python bash包装器。
import subprocess
def system_command(cmd):
""""Function executes cmd parameter as a bash command."""
p = subprocess.Popen(cmd,
stdout=subprocess.PIPE,
stderr=subprocess.PIPE,
shell=True)
stdout, stderr = p.communicate()
return stdout, stderr
size = int(system_command('du -sb . ')[0].split()[0])
我在这里有点迟(又是新来的),但是我选择在Linux上使用子进程模块和'du'命令行来检索以MB为单位的文件夹大小的准确值。我必须对根文件夹使用if和elif,因为否则子进程会由于返回非零值而引发错误。
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
解决方案的属性:
du
做st.st_blocks
的磁盘空间,因此仅适用于类Unix系统代码:
import os
def du(path):
if os.path.islink(path):
return (os.lstat(path).st_size, 0)
if os.path.isfile(path):
st = os.lstat(path)
return (st.st_size, st.st_blocks * 512)
apparent_total_bytes = 0
total_bytes = 0
have = []
for dirpath, dirnames, filenames in os.walk(path):
apparent_total_bytes += os.lstat(dirpath).st_size
total_bytes += os.lstat(dirpath).st_blocks * 512
for f in filenames:
fp = os.path.join(dirpath, f)
if os.path.islink(fp):
apparent_total_bytes += os.lstat(fp).st_size
continue
st = os.lstat(fp)
if st.st_ino in have:
continue # skip hardlinks which were already counted
have.append(st.st_ino)
apparent_total_bytes += st.st_size
total_bytes += st.st_blocks * 512
for d in dirnames:
dp = os.path.join(dirpath, d)
if os.path.islink(dp):
apparent_total_bytes += os.lstat(dp).st_size
return (apparent_total_bytes, total_bytes)
用法示例:
>>> du('/lib')
(236425839, 244363264)
$ du -sb /lib
236425839 /lib
$ du -sB1 /lib
244363264 /lib
解决方案的属性:
代码:
def humanized_size(num, suffix='B', si=False):
if si:
units = ['','K','M','G','T','P','E','Z']
last_unit = 'Y'
div = 1000.0
else:
units = ['','Ki','Mi','Gi','Ti','Pi','Ei','Zi']
last_unit = 'Yi'
div = 1024.0
for unit in units:
if abs(num) < div:
return "%3.1f%s%s" % (num, unit, suffix)
num /= div
return "%.1f%s%s" % (num, last_unit, suffix)
用法示例:
>>> humanized_size(236425839)
'225.5MiB'
>>> humanized_size(236425839, si=True)
'236.4MB'
>>> humanized_size(236425839, si=True, suffix='')
'236.4M'
使用pathlib在Python 3.6上有效的解决方案。
from pathlib import Path
sum([f.stat().st_size for f in Path("path").glob("**/*")])
使用的Python 3.6+递归文件夹/文件大小os.scandir
。与@blakev 的答案一样强大,但更短,并且采用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
def recursive_dir_size(path):
size = 0
for x in os.listdir(path):
if not os.path.isdir(os.path.join(path,x)):
size += os.stat(os.path.join(path,x)).st_size
else:
size += recursive_dir_size(os.path.join(path,x))
return size
我写了这个函数,它给了我准确的目录总大小,我尝试了其他的os.walk循环解决方案,但是我不知道为什么最终结果总是小于实际大小(在ubuntu 18 env上)。我一定做错了什么,但谁在乎,写这篇文章就可以了。