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Rounding up numbers in PERL

Posted on 2009-07-07
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Last Modified: 2012-06-27
I have this code:

my $avg = ($n1 + $n2 + $n3 + $n4 + $n5 + $n6 + $n7)/ 7;

which gives me an average of all the values in those variables.
The result is this:
24,36,45,67,887,33,320, 201.285714285714, with 201.285714285714 being the variable

how do I round this number to no decimal places, so it would just be 201     ?
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Question by:Europa MacDonald
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15 Comments
 
LVL 85

Expert Comment

by:ozo
ID: 24792363
use POSIX:
#round up
print ceil $avg;
#round down
print floor $avg;
#round to nearest
print floor $avg+0.5;
#or
printf "%.f",$avg;
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Author Comment

by:Europa MacDonald
ID: 24792383
using print floor $avg; gave me

456,657,432,345,677,899,322, floor 540.714285714286
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Author Comment

by:Europa MacDonald
ID: 24792392
written :

  print OUT "$n1,$n2,$n3,$n4,$n5,$n6,$n7, floor $avg\n";
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Expert Comment

by:Suhas .
ID: 24792399
print "%d",$avg gives the rounded value of an average result.
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LVL 85

Expert Comment

by:ozo
ID: 24792402
use POSIX;
print OUT "$n1,$n2,$n3,$n4,$n5,$n6,$n7,",floor $avg,"\n";
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Author Comment

by:Europa MacDonald
ID: 24792411
still getting 24,36,45,67,887,33,320, floor 201.285714285714
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LVL 85

Expert Comment

by:ozo
ID: 24792427
> print "%d",$avg gives the rounded value of an average result.
no, it gives
%d540.714285714286
printf  "%d",$avg; #gives the truncated value of $avg
printf  "%.f",$avg; #gives the rounded value of $avg
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LVL 85

Expert Comment

by:ozo
ID: 24792439
> still getting 24,36,45,67,887,33,320, floor 201.285714285714
do you still have the "floor" in  quotes?
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LVL 85

Accepted Solution

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ozo earned 2000 total points
ID: 24792444
print "$n1,$n2,$n3,$n4,$n5,$n6,$n7,",(floor $avg),"\n";
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LVL 85

Expert Comment

by:ozo
ID: 24792454
or
 print "$n1,$n2,$n3,$n4,$n5,$n6,$n7,", floor($avg) , "\n";
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Author Closing Comment

by:Europa MacDonald
ID: 31600507
thankyou :-)
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LVL 9

Expert Comment

by:Suhas .
ID: 24792466
#thanks for the correction for print to printf ozo.
# since Michael request is "how do I round this number to no decimal places, so it would just be 201"


Michael,
Is your requirement like 201.**** to 201, if so its better to use %d
or less than 201.5 to 201 and greather than 201.5 to 202 , then use %f with condition


Cheers...
Suhas
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Author Comment

by:Europa MacDonald
ID: 24792492
thankyou Suhas
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Expert Comment

by:ozo
ID: 24792510
the question title was about rounding up
but 201.285714285714 to 201 is rounding down
or it could be rounding to the nearest integer, or it could be truncating.
The difference would be seen with 201.9 and -201.9
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Expert Comment

by:Tintin
ID: 24792785
A useful feature to use is the perl FAQ.  From the command line type

perldoc -q floor

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