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How to find a minimum value of a row?

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# 1  
Old 03-14-2013
How to find a minimum value of a row?
input

Code:
1 2 3 4 5 2 8
2 1 1 1 4 2 1
5 4 4 4 2 1 3
2 2 6 7 4 5 4
5 5 5 4 3 3 5

I woud like to print a min of each row such that my output would look like

Code:
1
1
1
2
3

Thanks!
# 2  
Old 03-14-2013
Code:
awk 'BEGIN{min=9}{for(i=1;i<=NF;i++){min=(min<$i)?min:$i}print min;min=9}' file

This User Gave Thanks to Yoda For This Post:
johnkim0806
# 3  
Old 03-14-2013
I also found that this works as well
Code:
awk '{for (i=1;i<=NF;i++) {$1=($i<$1)?$i:$1} print $1}' file.txt

Last edited by Scrutinizer; 03-14-2013 at 03:41 PM.. Reason: code tags
This User Gave Thanks to johnkim0806 For This Post:
Yoda
# 4  
Old 03-14-2013
In my opinion, you both are trying to be too clever. Throw away the unnecessary hardcoded maximum (9 in post #2) and the pointless ternary operators in favor of more readable code:
Code:
awk '{m=$1; for (i=2; i<=NF; i++) if ($i < m) m = $i; print m}' file.txt

Regards,
Alister
This User Gave Thanks to alister For This Post:
johnkim0806
# 5  
Old 03-14-2013
This is much easy to read! Thanks Smilie
# 6  
Old 03-14-2013
Aside from readability, that ternary approach is also inefficient. It requires an assignment to be made even when one is not needed. Further, assigning to a field triggers recomputing and rebuilding of $0.

Regards,
Alister
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