How to cut the last 5 strings in one particular column for all rows

Question

I have a file with multiple columns and rows, and I need to cut the last 5 strings of my third column and print it in an extra column. For example:

1 info1 AAAAAAAAAAA
2 info23 TTTTTTTTTT
3 ingo45  ATATATATATATAT
4 info67 GGGYYYYUUUUUUU

My output will be:

1 info1 AAAAAAAAAAA AAAAA
2 info23 TTTTTTTTTT TTTTT
3 ingo45  ATATATATATATAT TATAT
4 info67 GGGYYYYUUUUUUU UUUUU

I was able to cut just the third column only by:

cat input_file | cut -f3 | grep -o '.\{5\}$' > output_file

But I want to maintain my original table. Any suggestion?

Answer 1

Short AWK approaches:

awk '{ $(NF+1)=substr($NF, length($NF)-4) }1' input-file

• $NF - the last field value

• $(NF+1) - becomes the last field (following the initial last field $NF)

• length($NF)-4 - obtaining the starting position/offset for the 5-character slice

---

The output:

1 info1 AAAAAAAAAAA AAAAA
2 info23 TTTTTTTTTT TTTTT
3 ingo45 ATATATATATATAT TATAT
4 info67 GGGYYYYUUUUUUU UUUUU

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The same can be reached via string substitution:

awk '{ sub(/.{5}$/, "& &") }1' input-file

• /.{5}$/ - match the last 5 chars at the end of the record

• & - points to the whole matched sequence

Answer 2

Using two instances of the sed & replacement string:

$ sed 's/.\{5\}$/& &/' file
1 info1 AAAAAAAAAAA AAAAA
2 info23 TTTTTTTTTT TTTTT
3 ingo45  ATATATATATATAT TATAT
4 info67 GGGYYYYUUUUUUU UUUUU