lamp has asked for the wisdom of the Perl Monks concerning the following question:
I'm following no modules approach to find the day of year based on the user provided date.
#!/usr/local/bin/perl
sub dayofyear {
($day1,$month,$year)=@_;
my(@d_in_m)=(0,31,28,31,30,31,30,31,31,30,31,30,31);
$d_in_m[2]=29 if (&leap($year));
for($i=1;$i<$month;$i++) {
$k += $d_in_m[$i];
}
$k += $day1;
return $k;
}
sub leap {
$y = shift;
return 0 unless $y % 4 == 0;
return 1 unless $y % 100 == 0;
return 0 unless $y % 400 == 0;
return 1;
}
print dayofyear(10,3,2005);
It's printing the correct value. Is there any way to optimize the above code / do it in a better way?
--Binoj
Re: calculate day of year
by Corion (Patriarch) on Feb 22, 2005 at 08:26 UTC
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use strict;
use Time::Local;
sub dayofyear {
my ($d,$m,$y) = @_;
$m--;
(localtime(timelocal(0,0,0,$d,$m,$y)))[7]
};
Beware of the behaviour of timelocal for years < 1000. | [reply] [d/l] |
Re: day of year calculation
by reneeb (Chaplain) on Feb 22, 2005 at 11:48 UTC
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#! /usr/bin/perl
use strict;
use warnings;
use Date::Calc qw(Day_of_Year);
my $doy = Day_of_Year($year,$month,$day);
print $doy;
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Re: day of year calculation
by betterworld (Curate) on Feb 22, 2005 at 11:56 UTC
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Have you noticed that if you duplicate the last line (the line with "print"), you'll get two different results? First, it prints 69, which seems correct, but then it prints 138.
This is because you don't use strict. Especially you should write "my $k" to solve this issue. | [reply] |
Re: calculate day of year
by slayven (Pilgrim) on Feb 22, 2005 at 09:19 UTC
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| [reply] |
Re: day of year calculation
by mlh2003 (Scribe) on Feb 22, 2005 at 13:52 UTC
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One way to speed things up a little is to get rid of the for loop by replacing the array containing the number of days in each month with the cumulative number of days. If the year is a leap year, add 1 to the result if the month is >2.
UPDATE (for explanation): In the code I check for the month > 1 since I subtract 1 from the month in the previous line.
#!/usr/local/bin/perl
use strict;
sub dayofyear {
my ($day1,$month,$year)=@_;
my @cumul_d_in_m=(0,31,59,90,120,151,181,212,243,273,304,334,365);
my $doy=$cumul_d_in_m[--$month]+$day1;
$doy++ if (&leap($year) && $month>1);
return $doy;
}
sub leap {
my $y = shift;
return 0 unless $y % 4 == 0;
return 1 unless $y % 100 == 0;
return 0 unless $y % 400 == 0;
return 1;
}
print dayofyear(10,3,2005)."\n";
print dayofyear(10,3,2004)."\n";
print dayofyear(10,2,2005)."\n";
print dayofyear(10,2,2004)."\n";
This gives the following output:
69 70 41 41
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#!/usr/local/bin/perl
use strict;
sub dayofyear {
my ($day1,$month,$year)=@_;
my @cumul_d_in_m =
(0,31,59,90,120,151,181,212,243,273,304,334,365);
my $doy=$cumul_d_in_m[--$month]+$day1;
return $doy if $month < 2;
return $doy unless $year % 4 == 0;
return ++$doy unless $year % 100 == 0;
return $doy unless $year % 400 == 0;
return ++$doy;
}
print dayofyear(10,3,2005)."\n";
print dayofyear(10,3,2004)."\n";
print dayofyear(10,2,2005)."\n";
print dayofyear(10,2,2004)."\n";
| [reply] [d/l] |
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