[SOLVED] stop counting if diff character in the file from the provided patterns

Issue

I am trying to create a script to count certain patterns such as aac tgt ttg. Generally, the DNA file should be a text file containing a valid DNA string with no newline characters or white space characters of any kind within it. (It will be terminated with a newline character.) This DNA file should contain nothing but a sequence of the bases a, c, g, and t in any order.

#! /bin/bash
   diffchar=$(grep -cv 'aac\|gtt\|tgt\|cag' $1 )
if  [[ $diffchar -eq 1 ]]
then
echo "error"

elif  [[  $diffchar -ne 1 ]]
then

         
        count=$(grep -o 'aac\|gtt\|tgt\|cag' $1 | sort -k1,1nr -k2,2 | uniq -c)
#       newcount=$(tr -d '\n' $count | awk -f histogram.awk | sort | uniq -c)

#       echo "$newcount"
        echo $count 
fi

Solution

Since all your search patterns are 3 characters long you can split your file into 3-character blocks and use the idiom sort | uniq -c to create a histogram.

sed '1s/.../\n&/g' yourFile | sed 1d | sort | uniq -c

Above command prints something like

  1 aac
  5 cag
  1 gtt
  3 tgt
  1 XYZ

For everything that isn't one of your patterns (XYZ in this case) you can automatically throw an error using

histogram=$(sed '1s/.../\n&/g' yourFile | sed 1d | sort | uniq -c)
if echo "$histogram" | grep -vEq '^[0-9 ]+(aac|gtt|tgt|cag)$'; then
  echo error
else
  echo "$histogram"
fi

Answered By - Socowi

Answer Checked By - Senaida (WPSolving Volunteer)