Python:os.stat().st_size提供的值与du不同 [英] Python: os.stat().st_size gives different value than du

问题描述

我正在创建一个实用程序，它将遍历目录并获取所有目录的子目录和文件的大小并存储值.但是，尺寸计算不正确.

I'm creating a utility that will walk through directories and get the sizes of child directories and files for all directories and store the value. However, the sizes aren't computed correctly.

这是我的课程，它自动遍历所有子目录:

Here's my class, which automatically recurses through all sub-directories:

class directory:
    '''
    Class that automatically traverses directories
    and builds a tree with size info
    '''
    def __init__(self, path, parent=None):

        if path[-1] != '/':
            # Add trailing /
            self.path = path + '/'
        else:
            self.path = path
        self.size = 4096
        self.parent = parent
        self.children = []
        self.errors = []
        for i in os.listdir(self.path):
            try:
                self.size += os.lstat(self.path + i).st_size
                if os.path.isdir(self.path + i) and not os.path.islink(self.path + i):
                    a = directory(self.path + i, self)
                    self.size += a.size
                    self.children.append(a)
            except OSError:
                self.errors.append(path + i)

我有一个用于测试该程序的视频目录:

I have a directory of videos that I'm testing this program with:

>>> a = directory('/var/media/television/The Wire')
>>> a.size
45289964053

但是，当我与du尝试相同时，我得到了

However, when I try the same with du, I get

$ du -sx /var/media/television/The\ Wire
44228824

目录不包含任何链接或任何特殊内容.

The directories don't contain any links or anything special.

有人可以解释为什么 os.stat()给出奇怪的尺寸读数吗?

Could someone explain why os.stat() is giving weird size readings?

平台:

• Linux(Fedora 13)
• Python 2.7

推荐答案

将此文件视为foo

-rw-rw-r-- 1 unutbu unutbu 25334 2010-10-31 12:55 foo

它由25334个字节组成.

It consists of 25334 bytes.

tune2fs告诉我foo驻留在块大小为4096字节的文件系统上:

tune2fs tells me foo resides on a filesystem with block size 4096 bytes:

% sudo tune2fs -l /dev/mapper/vg1-OS1
...
Block size:               4096
...

因此，即使文件系统的内容仅由1个字节组成，它在文件系统上的最小文件也将占用4096个字节.随着文件的增大，将以4096字节的块分配空间.

Thus, the smallest file on the filesystem will occupy 4096 bytes, even if its contents consist of just 1 byte. As the file grows larger, space is allocated in 4096-byte blocks.

du报告

% du -B1 foo
28672   foo

请注意28672/4096 =7.这意味着foo占用了文件系统上的7个4096字节块.这是容纳25334个字节所需的最小块数.

Note that 28672/4096 = 7. This is saying that foo occupys 7 4096-byte blocks on the filesystem. This is the smallest number of blocks needed to hold 25334 bytes.

% du foo
28  foo

此版本的 du 仅报告28672/1024的四舍五入.

This version of du is just reporting 28672/1024 rounded down.

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