Compute average file size

Question

I am looking for a handy one-liner for computing the average file size in a directory.

What I want is:

size of all files / number of files in directory

Answer 1

With GNU find:

find . -type f -printf '%s\n' | awk '{s+=$0}
  END {printf "Count: %u\nAverage size: %.2f\n", NR, s/NR}'

Or for disk usage:

find . -type f -printf '%k\n' | awk '{s+=$0}
  END {printf "Count: %u\nAverage size: %.2f\n", NR, s*1024/NR}'

Note that if there are several hardlinks of the same file in there, that will count its disk usage several time.

The above only counts regular files, not symlinks or directories or other special files. It includes hidden files.

The same with zsh builtins:

zmodload -i zsh/stat
count() {zstat -Hs -- $REPLY; ((size+=$s[size], count++, 0))}
size=0 count=0
**/*(oNDN.+count)
print -f "Count: %s\nAverage size: %.2f\n" $count $(($size./count))

Answer 2

A simplistic solution in one line:

ls -Rl -- "$DIR" | awk 'BEGIN{sum=0;count=0};/^-/{sum+=$5;++count};END{print sum/count}'

It has a syntactic cheat in it, only considering ls output lines that begin with '-', which should constitutes data for regular files.

Answer 3

for only files in a directory, ignoring sub directories:

expr $(du -Ss | cut -f1) / $(find /path/to/dir -maxdepth 1 -type f | wc -l)

Or to count all files including files in subdirectories:

expr $(du -s | cut -f1) / $(find /path/to/dir -type f | wc -l)

Answer 4

With FreeBSD / Mac OS X find, stat and awk (not exactly a handy one-liner though):

find -x . -type f -maxdepth 1 -exec sh -c 'stat -f '%z' "${@}"' _ '{}' + | 
LC_ALL=C awk -v pwd="${PWD}" '
   BEGIN{ sum=0; count=0; }
   { sum+=$1; ++count; }
   END{ 
        if (count == 0) exit;
        printf ("number of files: %d\n", count); 
        printf ("average file size in B: %.5f\n", sum/count); 
        printf ("average file size in KB: %.5f\n", (sum/count) / 1024); 
        printf ("average file size in MB: %.5f\n", (sum/count) / (1024*1024)); 
        printf ("directory: %s\n", pwd); 
   }
'