目录的平均和最大大小 [英] Average and maximum size of directories

问题描述

我有一个目录和一堆这样的子目录：
- directory1 （ sub-dir1 ， sub-dir2 ，sub-dir3 ， sub-dir4 ， sub-dir5 ...........等等，数百人...）

I have a directory and a bunch of sub-directories like this: - directory1 (sub-dir1, sub-dir2, sub-dir3, sub-dir4, sub-dir5...........and so on, hundreds of them...)

如何查看子目录的平均大小？
如何查找子目录的最大大小？

How do I find out what is average size of the sub-directories? And how do I find what is the maximum size of the sub-directories?

所有使用Unix命令...

All using Unix commands...

谢谢。

推荐答案

如果您只有 directory1中的目录而不是文件，那么以下两个命令应分别给出最大目录的大小（以字节为单位）和其大小的平均值（以字节为单位）。

If you only have directories and not files in directory1, then the following two "commands" should give you the size (in bytes) and name of the largest directory and the average of their sizes (in bytes), respectively.

$ du -sb directory1/* | sort -n | tail -n 1
$ du -sb directory1/* | awk ' { sum+=$1; ++n } END { print sum/n } '

如果 directory1 ，这些将与上面的示例一并计数。如果普通文件不应被计算在内，以下内容可能更合适。

If there is also ordinary files within directory1, these will be counted as well with the examples above. If ordinary files should not be counted, the following might be more appropriate.

$ find directory1/ -mindepth 1 -maxdepth 1 -type d -exec du -sb {} \; | sort -n | tail -n 1
$ find directory1/ -mindepth 1 -maxdepth 1 -type d -exec du -sb {} \; | awk ' { sum+=$1; ++n } END { print sum/n } '

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