Replace 4th occurrence of regex in line

Question

This is the output I have now. How can I get from Input to Output?

Input:

1.2.3.4.5
2.3.4.5.6

Output:

1.2.3.4-5
2.3.4.5-6

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Bonus: I have the other 80% of the script done but this is my ideal output. Is there a super simple way to do it?

Input:

1.2.3.4-1.2.3.10

Output:

1.2.3.4-10

Answer 1

With awk:

awk -F '[-.]' '{printf "%s.%s.%s.%s-%s\n", $1, $2, $3, $4, $NF}'

Since it looks like you want fields 1-4 and the last field from input separated by either . or -, and the output formatted with all but the last field separated by ..

More concisely:

awk -F '[-.]' -v OFS=. '{$4 = $4"-"$NF; NF=4}1'

• Add the last field to the fourth field and set the number of fields to 4.

Answer 2

First part:

$ sed 's/\./-/4' file.txt

Which replaces the 4th occurrence of . with a -.

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For your Bonus section:

sed -r 's/(.*)-.*\.(.*)$/\1-\2/' file.txt

input:

1.2.3.4-1.2.3.10

output:

1.2.3.4-10

Answer 3

We can use grouping with greedy match with sed like so:

$ sed -r 's/(.*)\.(.*)$/\1-\2/' input.txt                                
1.2.3.4-5
2.3.4.5-6

Same idea for perl:

$ perl -pe's/(.*)\.(.*)$/\1-\2/' input.txt                               
1.2.3.4-5
2.3.4.5-6

But we can play around with . as column separator in awk:

$ awk -F '.' 'BEGIN{OFS=FS}{$NF=-$NF;sub(/\.-/,"-");print}' input.txt        
1.2.3.4-5
2.3.4.5-6