Using and passing arguments to shuf within awk


 
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# 1  
Old 03-26-2013
Using and passing arguments to shuf within awk

Hello all,

I would like to output a random number within a range for every line using awk and shuf. I think I'm almost there, but I don't know how to pass arguments to shuf within my awk script:

Input
Code:
1 12190 12227
1 12595 12721
1 13403 13639
1 14362 14829
1 14970 15038

awk:
Code:
awk '{ a = (shuf -i $2"-"$3 -n 1)}{print $1,a}' Input

Desired Output (with second column being a random number between field 2 and 3 of input file):

Code:
1 12211
1 12659
1 13411
1 14705
1 15021

Al I get so far is

Code:
1 012190-122271
1 012595-127211
1 013403-136391
1 014362-148291
1 014970-150381

Any help would be greatly appreciated!

Seb
# 2  
Old 03-26-2013
Perl, ok?
Code:
perl -lane 'print "$F[0] ", int(rand($F[2]-$F[1])) + $F[1]' file

This User Gave Thanks to balajesuri For This Post:
# 3  
Old 03-26-2013
I'm not used to perl, but this works nicely. Thanks!
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