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Loops & Variable

 
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Top Forums UNIX for Dummies Questions & Answers Loops & Variable
# 15  
Old 07-01-2015
Quote:
(yes, the values differ for each execution of the loop, but the schema stays the same). You could - if you don't want to convolute your "awk" statement too much, which is a good idea, always do one thing in a step, but do it good - put it in a variable:
Code:
 eval replacement=\${TMPArray${y}[$x]}
awk -v "REP=${replacement}"[...]

How you use the variable replacement in the script?
Code:
eval replacement="\${PrimerSet${PSet}[$oligo]}" | awk -v "REP=${replacement}" '{gsub("^.*" REP ".*$","",$0)} 1' $PSet.$oligo > $PSet.Result$oligo

the above does not work
Last edited by Xterra; 07-01-2015 at 01:46 PM..
# 16  
Old 07-01-2015
If you could tell us WHAT doesn't work? On first sight I can comment that you're using parentheses instead of braces for the REP assignment.
# 17  
Old 07-01-2015
Quote:
If you could tell us WHAT doesn't work? On first sight I can comment that you're using parentheses instead of braces for the REP assignment.
Sorry, I meant to use braces. Still, that does not work. The outfiles are empty which makes me think I am not assigning the variable replacement correctly.
# 18  
Old 07-01-2015
Quote:
Originally Posted by Xterra
How you use the variable replacement in the script?
Sigh. Exactly in the way i wrote it in the line immediately following the assignment and precisely in the way you quoted it.

Quote:
Originally Posted by Xterra
Code:
eval replacement="\${PrimerSet${PSet}[$oligo]}" | awk -v "REP=$(replacement)" '{gsub("^.*" REP ".*$","",$0)} 1' $PSet.$oligo > $PSet.Result$oligo

the above does not work
Yes. Because it is wrong for the reasons explained at length in my last post. I hate quoting myself but you probably overread the 12 paragraphs following

Quote:
First, the pipe is not necessary
You might want to read them now.

/PS: i just now saw you posted in the meantime and corrected the braces/parentheses confusion. If you think the variable is not correctly assigned why don't you just "echo" it?

Code:
[....]
    replacement=....
    echo $replacement
    [....]

This will immediately show you if the variable is correctly assigned or not, no?

I hope this helps.

bakunin
Last edited by bakunin; 07-01-2015 at 02:18 PM..
# 19  
Old 07-01-2015
Quote:
Originally Posted by Xterra
Sorry, I meant to use braces. Still, that does not work. The outfiles are empty which makes me think I am not assigning the variable replacement correctly.
This is exactly what I meant in my earlier post:

Quote:
Originally Posted by RudiC
.
.
.
Are you sure the gsub with the regex anchored on both ends will do what you want?
Rethink the use of gsub with that anchored regex.
# 20  
Old 07-01-2015
Maybe something like this?
Code:
#!/bin/bash
PrimerSet1=( NULL "GGGGGG|TTTTT" "AAAAAA|CCCCCC" "CCAAAA|CCCCAA" )
PrimerSet2=( NULL "XXXXX|YYYYYY" "ZZZZZZZ|BBBBBBB" "KKKKKK|LLLLLL" )
PrimerSet3=( NULL "SSSSS|FFFFFF" "EEEEEE|JJJJJJ" "PPPPPP|UUUUUU" )
for PSet in {1..3}
do
    for oligo in {1..3}
    do
        eval replacement=\${PrimerSet${PSet}[$oligo]}
        awk -v "REP=${replacement}" '{gsub("^.*" REP ".*$","",$0)} 1' $PSet.$oligo > $PSet.Result$oligo
    done
done

# 21  
Old 07-01-2015
Quote:
Originally Posted by Xterra
Maybe something like this?
Well, fame at last.

Yes, like this. Now, if you enter a "echo"line between the line starting with "eval" and the one starting with "awk" you can control easily what exactly is in the variable (and, if it is not to your liking, do something to correct it).

This is another advantage of well-written programs: they can be easily debugged if there is a suspicion they don't exactly do what they are supposed to.

If you now get rid of the "NULL" data it would be perfect.

I hope this helps.

bakunin
This User Gave Thanks to bakunin For This Post:
Xterra
 
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