How to loop over files in natural order in Bash?

readarray -d '' entries < <(printf '%s\0' *.fas | sort -zV)
for entry in "${entries[@]}"; do
  # do something with $entry
done

where printf '%s\0' *.fas yields a NUL separated list of directory entries with the extension .fas, and sort -zV sorts them in natural order.

Note that you need GNU sort installed in order for this to work.

With option sort -g it compares according to general numerical value

 for FILE in `ls ./raw/ | sort -g`; do echo "$FILE"; done

0.log 1.log 2.log ... 10.log 11.log

This will only work if the name of the files are numerical. If they are string you will get them in alphabetical order. E.g.:

 for FILE in `ls ./raw/* | sort -g`; do echo "$FILE"; done

raw/0.log raw/10.log raw/11.log ... raw/2.log

You will get the files in ASCII order. This means that vvchr10* comes before vvchr2*. I realise that you can not rename your files (my bioinformatician brain tells me they contain chromosome data, and we simply don't call chromosome 1 "chr01"), so here's another solution (not using sort -V which I can't find on any operating system I'm using):

ls *.fas | sed 's/^\([^0-9]*\)\([0-9]*\)/\1 \2/' | sort -k2,2n | tr -d ' ' |
while read filename; do
  # do work with $filename
done

This is a bit convoluted and will not work with filenames containing spaces.

Another solution: Suppose we'd like to iterate over the files in size-order instead, which might be more appropriate for some bioinformatics tasks:

du *.fas | sort -k2,2n |
while read filesize filename; do
  # do work with $filename
done

To reverse the sorting, just add r after -k2,2n (to get -k2,2nr).

You mean that files with the number 10 comes before files with number 3 in your list? Thats because ls sorts its result very simple, so something-10.whatever is smaller than something-3.whatever.

One solution is to rename all files so they have the same number of digits (the files with single-digit in them start with 0 in the number).