Issue
I need to extract from the file the words that contain certain letters in a certain amount.
I apologize if this question has been resolved in the past, I just did not find anything that fits what I am looking for.
File:
wab 12aaabbb abababx ab ttttt baaabb zabcabc
baab baaabb cbaab ab ccabab zzz
For example
1. If I chose the letters a and the number is 1 the output should be:
wab
ab
ab
//only the words that contains a and the char appear in the word 1 time
2. If I chose the letters a,b and the number is 3, the output should be:
12aaabbb
abababx
baaabb
//only the word contains a,b, and both chars appear in the word 3 times
3. If I chose the letters a,b,c and the number 2, the output should be:
ccabab
zabcabc
//only the words that contains a,b,c and the chars appear in the word 3 times
Is it possible to find 2 letters in the same script? I was able to find in a single letter but I get only the words where the letters appear in sequence and I do not want to find only these words, that's what I did:
egrep '([a])\1{N-1}' file
And another problem I can not get only the specific words, I get all file and the letter I am looking for "a" in red. I tried using -w but it does not display anything.
::: EDIT :::
try to edit what you did to a for
i=$1
fileName=$2
letters=${@: 3}
tr -s '[:space:]' '\n' < $fileName* |
for letter in $letters; do
grep -E "^[^$letter]*($letter[^$letter]*){$i}$"
done | uniq
Solution
There are various ways to split input so that grep sees a single word per line. tr
is most common. For example:
tr -s '[:space:]' '\n' file | ...
We can build a function to find a specific number of a particular letter:
NofL(){
num=$1
letter=$2
regex="^[^$letter]*($letter[^$letter]*){$num}$"
grep -E "$regex"
}
Then:
# letter=a number=1
tr -s '[:space:]' '\n' file | NofL 1 a
# letters=a,b number=3
tr -s '[:space:]' '\n' file | NofL 3 a | NofL 3 b
# letters=a,b,c number=2
tr -s '[:space:]' '\n' file | NofL 2 a | NofL 2 b | NofL 2 c
Answered By - jhnc Answer Checked By - Pedro (WPSolving Volunteer)