open file 1 and write to 4th column a certain pattern

d.mintis · 08-02-2018, 12:16 PM

This is the format of file 1
20336 20336 20301
20337 20336 20301
20338 20336 20304
20339 20336 20304
20340 20336 20307
20341 20336 20307

I want to create a new file which will have the following format using awk in bash
20336 20336 20301 20302
20337 20336 20301 20303
20338 20336 20304 20305
20339 20336 20304 20306
20340 20336 20307 20308
20341 20336 20307 20309

scasey · 08-02-2018, 12:23 PM

Homework?
Show us what you've tried. We're happy to help.

d.mintis · 08-02-2018, 12:26 PM

it is more practicing. I tried the following awk '{$(NF+1.5)=20302+1*int(NR-1)); print}'

So
1. I set a new field
2. I need to find the right calc so as to create the following numbering pattern -> 1 2 4 5 7 8 ....

scasey · 08-02-2018, 02:42 PM

Quote:

Originally Posted by d.mintis

it is more practicing. I tried the following awk '{$(NF+1.5)=20302+1*int(NR-1)); print}'

So
1. I set a new field
2. I need to find the right calc so as to create the following numbering pattern -> 1 2 4 5 7 8 ....

Can't help much with awk, but it looks like you're skipping every third number, yes? So only print if the number is not divisible by 3. Don't have a clue how to do that.

Have you read the awk man page or done a comprehensive web search?

syg00 · 08-02-2018, 05:53 PM

You can't have fractions of field numbers - $NF+1 will do the job. A subsequent print will include it as a field of $0.
How you scale it is your decision.

allend · 08-02-2018, 08:33 PM

Quote:

I need to find the right calc so as to create the following numbering pattern -> 1 2 4 5 7 8 ....

If this pattern is to continue 1 2 4 5 7 8 10 11 ...
perhaps something like

Code:

awk '{if (NR%2) i+=1; print $0,20300+NR+i,NR+i-1}'