[SOLVED] how to double the first word in a line that have 3 words using sed?

Issue

I have a file called test that have:

1 2 3
2 3
4 5 6 7
8 9 10
11 12 13 14 15 16 17
18 19 20

I want to get the lines that have 3 words in them, and then print them, while the first word is duplicated.

I cant use pipeline, and I can use >| to put it in a tmp file and read from it.

so the output in this case is:

1 1 2 3
8 8 9 10
18 18 19 20

I understand more or less what regular expr I need, but the rest im struggling, could someone please help :?

this is what I did:

sed 's/'^[^ ]*[ ]+[^ ]+[ ]+[^ ]+[ ]*$'/&&/1/ test

I know this is not the solution but please help me understand :/

Solution

You could try this,

$ sed -nr 's/^([^ ]+) +[^ ]+ +[^ ]+$/\1 &/p' file
1 1 2 3
8 8 9 10
18 18 19 20

From man sed

-n, --quiet, --silent
             suppress automatic printing of pattern space
p      Print the current pattern space.

^ Asserts that we are at the start. (..) called capturing group, which is used to capture characters. Later you could refer those captured characters by back-referencing it's index number. ([^ ]+) Captures any character but not of a space one or more times. + repeats the previous token one or more times. $ asserts that we are at the line end.

OR

$ sed -n 's/^\([^[:blank:]]\+\)\([[:blank:]]\+\)[^[:blank:]]\+[[:blank:]]\+[^[:blank:]]\+$/\1\2&/p' file
1 1 2 3
8 8 9 10
18 18 19 20

[^[:blank:]]\+ Matches one or more non-space characters. [[:blank:]]\+ matches one or more space characters. & in the replacement part will print all the matched characters.

Answered By - Avinash Raj

Answer Checked By - Senaida (WPSolving Volunteer)