Shell: list directories ordered by file count (including in subdirectories)

Question

I've nearly reached my limit for the permitted number of files in my Linux home directory, and I'm curious about where all the files are.

In any directory I can use for example find . -type f | wc -l to show a count of how many files are in that directory and in its subdirectories, but what I'd like is to be able to generate a complete list of all subdirectories (and sub-subdirectories etc) each with a count of all files contained in it and its subdirectories - if possible ranked by count, descending.

Eg if my file structure looks like this:

Home/
  file1.txt
  file2.txt
  Docs/
    file3.txt
    Notes/
      file4.txt
      file5.txt
    Queries/
      file6.txt
  Photos/
    file7.jpg

The output would be something like this:

7  Home
4  Home/Docs
2  Home/Docs/Notes
1  Home/Docs/Queries
1  Home/Photos

Any suggestions greatly appreciated. (Also a quick explanation of the answer, so I can learn from this!). Thanks.

Answer 1

I use the following command

find . -xdev -type f | cut -d "/" -f 2 | sort | uniq -c | sort -n

Which produces something like:

[root@ip-***-***-***-*** /]# find . -xdev -type f | cut -d "/" -f 2 | sort | uniq -c | sort -n
      1 .autofsck
      1 stat-nginx-access
      1 stat-nginx-error
      2 tmp
     14 boot
     88 bin
    163 sbin
    291 lib64
    597 etc
    841 opt
   1169 root
   2900 lib
   7634 home
  42479 usr
  80964 var

Answer 2

This should work:

find ~ -type d -exec sh -c "fc=\$(find '{}' -type f | wc -l); echo -e \"\$fc\t{}\"" \; | sort -nr

Explanation: In the command above will run "find ~ -type d" to find all the sub-directories the home-directory. For each of them, it runs a short shell script that finds the total number of files in that sub-directory (using the "find $dir -type f | wc -l" command that you already know), and will echo the number followed by the directory name. The sort command then runs to sort by the total number of files, in a descending order.

This is not the most efficient solution (you end up scanning the same directory many times), but I am not sure you can do much better with a one liner :-)