Bash sort by regexp

Question

I have something about 100 files with the following syntax

ahfsdjfhdfhj_EPI_34_fdsafasdf
asdfasdf_EPI_2_fdsf
hfdjh_EPI_8_dhfffffffffff
ffffffffffasdfsdf_EPI_         


        

Answer 1

If Perl is okay you can:

print sort foo <>;    
sub foo {
        ($x = $a) =~s/.*EPI_(\d+).*/$1/;
        ($y = $b) =~s/.*EPI_(\d+).*/$1/;
        return $x <=> $y;
}

and use it as:

perl prg.pl inputfile

See it

Answer 2

This might work for you:

 ls | sed 's/.*EPI_\([0-9]*\)/\1 &/' | sort -n | sed 's/\S* //'

Answer 3

From your example it appears that delimiter is _ and text EPI_nnn comes at the same position after delimiter _. If that is always the case then you can use following command to sort the file:

sort -n -t "_" -k 3 file.txt

UPDATE:

If position of EPI_ text is not fixed then use following shell command:

sed 's/^\(.*EPI_\)\(.*\)$/\2##\1/' file.txt | sort -n -t "_" -k1 | sed 's/^\(.*\)##\(.*\)$/\2\1/'

Answer 4

 sed -e 's/EPI_/EPI /' file1 file2 ...|sort -n -k 2 -t ' '

Pipe that to sed -e 's/ /_/' to get back the original form.