How can I insert a row using awk?
My file looks as:
1 43
2 34
3 65
4 75
I would like to insert three rows with "?" So my desire file looks as:
1 ?
2 ?
3 ?
4 43
5 34
6 65
7 75
I am trying with the below script.
awk '{if(NR<=3){print "NR ?"}} {printf" " NR $2}' file.txt
$ awk 'BEGIN {printf "1 ?\n2 ?\n3 ?\n"} {printf "%d", $1 + 3; printf " %s\n", $2}' file.txt
1 ?
2 ?
3 ?
4 43
5 34
6 65
7 75
Here's one way to do it:
$ awk 'BEGIN{s=" "; for(c=1; c<4; c++) print c s "?"}
{print c s $2; c++}' ip.txt
1 ?
2 ?
3 ?
4 43
5 34
6 65
7 75
You could also add the 3 lines before awk, e.g.:
{ seq 3; cat file.txt; } | awk 'NR <= 3 { $2 = "?" } $1 = NR' OFS='\t'
Output:
1 ?
2 ?
3 ?
4 43
5 34
6 65
7 75
I would do it following way using GNU AWK, let file.txt content be
1 43
2 34
3 65
4 75
then
awk 'BEGIN{OFS=" "}NR==1{print 1,"?";print 2,"?";print 3,"?"}{print NR+3,$2}' file.txt
output
1 ?
2 ?
3 ?
4 43
5 34
6 65
7 75
Explanation: I set output field separator (OFS) to 7 spaces. For 1st row I do print three lines which consisting of subsequent number and ? sheared by output field separator. You might elect to do this using for loop, especially if you expect that requirement might change here. For every line I print number of row plus 4 (to keep order) and 2nd column ($2). Thanks to use of OFS, you would need to make only one change if requirement regarding number of spaces will be altered. Note that construct like
{if(condition){dosomething}}
might be written in GNU AWK in more concise manner as
(condition){dosomething}
(tested in gawk 4.2.1)