Bash Script Unix. Pattern matching

Question

How can I write the script for multiple strings in a single line with no space:

acgtttgggcccagctctccgccctcacacacaccccggggt

for visual purpose:

acg ttt ggg ccc agc tct ccg ccc tca cac aca ccc cgg ggt

and will have to match the 4th 3 letter sequence repeated 2 times. so in the above sequence we have ccc as the 4th seq. and it is repeated again after agc tct ccg.

so would I have to use grep for it?

Answer 1

Then how about:

#!/bin/bash

# add a space every three letters
str="acgtttgggcccagctctccgccctcacacacaccccggggt"
result=$(sed -e 's/\(...\)/\1 /g' <<< "$str")
echo $result

# check if the 4th sequence is repeated two times
awk '
{   ref = $4;                       # set the 4th sequence as a reference
    for (i=5; i<=NF; i++)           # iterate from 5th sequence to the end
        if (ref == $i) count++      # count the same one as the reference
    printf "4th sequence \"%s\" repeated %d times.\n", ref, count
}' <<< "$result"

which yields:

acg ttt ggg ccc agc tct ccg ccc tca cac aca ccc cgg ggt
4th sequence "ccc" repeated 2 times.  

The script is composed of two parts: 1st one to split the string with spaces, and the 2nd one to count the repetition of the 4th triplet.

• The sed script sed -e 's/\\(...\\)/\\1 /g' inserts a space after every three letters.
• The awk script loops over the sequences for the one which is same as the 4th triplet.
• If you just want to make sure the repetition is exactly two times or not, you may modify the script to compare count with 2.

Hope this helps.