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alter code to print only average and group

Posted on 2009-07-12
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Last Modified: 2012-05-07
I have tried changing this code to print to file only the average and the grouping to file, like this

23.43,02231
21.33,11002

could you show me how to do this please ?

I know it is altering the print OUT lines but what I am doing isnt working
and keep the time stamp at the end
#!/usr/bin/perl

use POSIX;

use strict;

use warnings;

use Data::Dumper;

use List::Util qw(sum);

 

my $startTime = time();

 

##### Open files

open(my $IN,"invim2.vim") or die "Could not open input: $!\n";

open(my $OUT,">outvim2.vim") or die "Could not open output: $!\n";

 

my @avgs1;

my @avgs2;

my ($maxavg1, $minavg1, $maxavg2, $minavg2);

 

        ##### Step 1: read a line, and split it on comma

        while(<$IN>){

        chomp;

        my @arr = split /,/;

        next unless @arr;

 

 

        # discard 1st and 2nd last and last item

        # shift @arr;

        # pop @arr;

        # pop @arr;

        # pop @arr;

 

 

        ##### Step 2: Sort data into ascending order

        @arr = sort {$a <=> $b} @arr;

        # print "arr @arr\n";

 

 

        ##### Step 3: Calculate difference between neighboring pairs

        my @diff = (map {$arr[$_] - $arr[$_-1]} (1..$#arr));

        #  print "diff @diff\n";

 

 

        ##### Step 4: Calculate amount in each group

        my @groups = (0)x5;

        foreach my $a (@arr) {

                $groups[$a/10]++;

        }

        #   print "group @groups\n";

 

 

        ##### Step 5: Calculate average

 

        my $avg1 = sum(@arr)/@arr;

        # printf "arr avg %.2f\n", $avg1;

        if(!$maxavg1 || $avg1 > $maxavg1) { $maxavg1 = $avg1; }

        if(!$minavg1 || $avg1 < $minavg1) { $minavg1 = $avg1; }

        push @avgs1, $avg1;

 

        my $avg2 = sum(@diff)/@diff;

        # printf "diff avg %.2f\n", $avg2;

        if(!$maxavg2 || $avg2 > $maxavg2) { $maxavg2 = $avg2; }

        if(!$minavg2 || $avg2 < $minavg2) { $minavg2 = $avg2; }

        push @avgs2, $avg2;

 

        ##### Step 6: Print results

        print $OUT join("\t", @arr) . "\tavg " . sprintf("%.2f\t", $avg1) . "group " . join("", @groups) . "\n";

        print $OUT join("\t", @diff) . "\t\tavg " . sprintf("%.2f\t", $avg2) . "\n";

        print $OUT "\n";

}

 

my $stopTime = time();

my $totalTime = $stopTime - $startTime;

 

#  printf "input: maxavg %d minavg %d avgavg %d\n", $maxavg1, $minavg1, sum(@avgs1)/@avgs1;

printf $OUT "For data values:\t\t avg avg %.2f\t maximum avg %.2f\t minimum avg %.2f \n",

                        sum(@avgs1)/@avgs1, $maxavg1, $minavg1 ;

#  printf "diff: maxavg %d minavg %d avgavg %d\n", $maxavg2, $minavg2, sum(@avgs2)/@avgs2;

printf $OUT "For data difference values:\t avg avg %.2f\t maximum avg %.2f\t minimum avg %.2f \n",

                        sum(@avgs2)/@avgs2, $maxavg2, $minavg2, ;

 

printf "Time started: %s   Time ended: %s   Total time:  %d minutes\n",

        strftime("%H:%M", localtime($startTime)), strftime("%H:%M", localtime($stopTime)), $totalTime/60;

printf $OUT "Time started: %s   Time ended: %s   Total time:  %d minutes\n",

        strftime("%H:%M", localtime($startTime)), strftime("%H:%M", localtime($stopTime)), $totalTime/60;

 

 

close($IN);

close($OUT);

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Question by:MichaelGlancy
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Expert Comment

by:mrjoltcola
ID: 24836449
In which step?
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Accepted Solution

by:
mrjoltcola earned 500 total points
ID: 24836452
Something like this should do:
printf $OUT "%.2f,%s\n", $avg1, join("", @groups);

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Author Comment

by:MichaelGlancy
ID: 24836457
at the print out step 6
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Author Closing Comment

by:MichaelGlancy
ID: 31602661
thankyou
0
 
LVL 40

Expert Comment

by:mrjoltcola
ID: 24836474
Remember printf is a special routine that comes in handy

printf <format string>, arg1, arg2, ..., argN

It correlates to the C version of printf, so see: perldoc -f printf
or see here for format specifiers. C/C++/Perl all share pretty much the same format since it originated from C.

http://www.cplusplus.com/reference/clibrary/cstdio/printf/


I like it because its easy to format the output in the format string rather than concatenating a lot of stuff together with regular print.
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Author Comment

by:MichaelGlancy
ID: 24836480
ok well I can see some of that point

I have noticed

printf <....
and
printf OUT

the first is printing to the screen ? and the second to the text file ?
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Expert Comment

by:mrjoltcola
ID: 24836484
Yes. If the first argument to print or printf has no comma separating it and the next argument, then the argument is assumed to be a file stream.
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