os.path.getsize 返回不正确的值? [英] os.path.getsize Returns Incorrect Value?

问题描述

def size_of_dir(dirname):
    print("Size of directory: ")
    print(os.path.getsize(dirname))

是有问题的代码.dirname 是一个包含 130 个文件的目录，每个文件大约 1kb.当我调用这个函数时，它返回4624，这是不是目录的大小......为什么会这样?

is the code in question. dirname is a directory with 130 files of about 1kb each. When I call this function, it returns 4624, which is NOT the size of the directory...why is this?

推荐答案

此值 (4624B) 表示描述该目录的文件的大小.目录被描述为 inode (http://en.wikipedia.org/wiki/Inode)保存有关其包含的文件和目录的信息.

This value (4624B) represents the size of the file that describes that directory. Directories are described as inodes (http://en.wikipedia.org/wiki/Inode) that hold information about the files and directories it contains.

要获取该路径内的文件/子目录的数量，请使用:

To get the number of files/subdirectories inside that path, use:

len(os.listdir(dirname))

要获取总数据量，您可以在此问题中使用代码/a>，即(如@linker 发布的那样)

To get the total amount of data, you could use the code in this question, that is (as @linker posted)

 sum([os.path.getsize(f) for f in os.listdir('.') if os.path.isfile(f)]).

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