How to print the amount of files in a folder (recursively) seperated by extensions? - bash

Question

For example, I have a folder containing files of different types (.jpg, .png, .txt, ..) and would like to know how many files of each extensions there is in my folder separatly.
The output would be something like this:
.jpg : 255
.png : 123
.txt : 12
No extension : 1
For now, I only know how to find how many files exist for one given extension using this command:
find /folderpath -type f -name '*.jpg' | wc -l
However I would like it to be able to find by itself the files extensions.
Thanks for your help.

Answer 1

You can do this for a single directory with:
ls | grep '\.' | sed 's/.*\././' | sort | uniq -c
(I'm ignoring files with no . - tweak if you want something else)
I'd suggest fleshing this out into a script (say, extension_counts) that takes a list of directories, and for each one outputs the path followed by the report in the format you wish.
Quick and dirty version:
#!/bin/sh
for dir in $*; do
echo $dir
(cd $dir && ls | grep '\.' | sed 's/.*\././' | sort | uniq -c)
done
... but you should consider hardening this.
Then for the recursive part, you can use find and xargs:
find . -type d | xargs extension_counts
You could be a bit smarter and do it all in one script file by defining extension_counts as a function, but that's an optimisation.
There are some pitfalls to parsing the output of ls (or find). In this case the only potential issue I can think of is filenames containing a newline (yes, this is possible). You could just accept that you're using a tool not designed for weird filenames, or you could write something more robust in a language with firmer data structures, such as Python, Perl, Ruby, Go, etc.

Answer 2

This could be done with a quick awk one liner as well:
find /folderpath -type f -name '*.*' | awk -F"." 'BEGIN{OFS=" : "}{extensions[$NF]++}END{for (ext in extensions) { print ext, extensions[ext]}};'
That awk script will split each line by a period -F"."
Set the OFS (Output Field Separator) by " : " BEGIN{OFS=" : "}
Load an array using the file extension for the key extensions[$NF] where $NF is the last field in the record. The value of the array will be a count ++.
When all the lines are processed we iterate the array for (ext in extensions) and print out the index and value {print ext, extensions[ext]}

Answer 3

I would proceed this way :
list the file names (rather than their paths produced by find) :
find . -type f | rev | cut -d/ -f1 | rev
We reverse each line so that we can easily address the last field
reduce to their extension :
sed -E 's/^.*\././;t end;s/.*/No extension/;:end'
Here we remove everything up to the first dot, or if the substitution could not be done (because there was no dot) we replace everything by "No extension".
sort the result :
sort
group by extension and add the count :
uniq -c
For a complete command as follows :
find . -type f | rev | cut -d/ -f1 | rev | sed -E 's/^.*\././;t end;s/.*/No extension/;:end' | sort | uniq -c
Note that the presentation differs from yours, which could be easily fixed with an additional sed :
2 .119
1 .147
[...]
1 .Xauthority
1 .xml
1 .xsession-errors
2 .zip
1 .zshrc
48 No extension