So here's what I came up with (deciding to barrel foward with some rework and cleanup of the existing algorithm primarily as an exercise. I really was going to(and in fact already had) put BrowserUk's code in the main script. But the more I look at the result of the cleanup the more I like it. Computing rather than attrition makes for longer code. But it appeals more to my sense of order.
Yeah I have to put exception handling in there. The code as it stands is awfully 'happy path'. But I'll trap errors in the input data going in to the primary function for now, then worry about the support functions later on.
Thanks again to tye for the encouragement and help with the "is_deeply" and "is not ok" stuff. o/
#!/usr/bin/perl -w use strict; use warnings; use Time::Local; use Test::More "no_plan"; test(); sub test { is ( identify_quarter(1), 1); is ( identify_quarter(2), 1); is ( identify_quarter(3), 1); is ( identify_quarter(4), 2); is ( identify_quarter(5), 2); is ( identify_quarter(6), 2); is ( identify_quarter(7), 3); is ( identify_quarter(8), 3); is ( identify_quarter(9), 3); is ( identify_quarter(10), 4); is ( identify_quarter(11), 4); is ( identify_quarter(12), 4); is_deeply ( [ get_previous_quarter(2011,4)],[(2011,3)]); is_deeply ( [ get_previous_quarter(2011,4)],[(2011,3)]); is_deeply ( [ get_previous_quarter(2011,3)],[(2011,2)]); is_deeply ( [ get_previous_quarter(2011,1)],[(2010,4)]); is ( last_month_of_quarter(1), 3); is ( last_month_of_quarter(2), 6); is ( last_month_of_quarter(3), 9); is ( last_month_of_quarter(4), 12); is_deeply ( [next_month(2011,1)] , [ (2011,2) ] ); is_deeply ( [next_month(2011,12)] , [ (2012,1) ] ); is ( last_day_of_month(2011,01), 31); is ( last_day_of_month(2011,02), 28); is ( last_day_of_month(2012,02), 29); is_deeply ( [ get_last_day_of_prev_quarter(2011,1,1) ],[ (2010,12, +31) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,2,1) ],[ (2010,12, +31) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,3,1) ],[ (2010,12, +31) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,3,1) ],[ (2010,12, +31) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,4,1) ],[ (2011,3,3 +1) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,5,1) ],[ (2011,3,3 +1) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,6,1) ],[ (2011,3,3 +1) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,7,1) ],[ (2011,6,3 +0) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,8,1) ],[ (2011,6,3 +0) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,9,1) ],[ (2011,6,3 +0) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,10,1) ],[ (2011,9, +30) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,11,1) ],[ (2011,9, +30) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,12,1) ],[ (2011,9, +30) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2011,9,1) ],[ (2011,6,3 +0) ] ); is_deeply ( [ get_last_day_of_prev_quarter(2012,1,1) ],[ (2011,12, +31) ] ); is_deeply ( [ get_last_bus_day_of_prev_quarter(2011,9,1) ],[ (2011 +,6,30) ]); is_deeply ( [ get_last_bus_day_of_prev_quarter(2012,1,1) ],[ (2011 +,12,30) ]); } sub get_last_day_of_prev_quarter { my ($yyyy,$mm,$dd) = @_; my ($year, $quarter) = get_previous_quarter($yyyy,identify_quarter +($mm)); my $month = $quarter * 3; my $day = last_day_of_month($year,$month); return ($year,$month,$day); } sub get_last_bus_day_of_prev_quarter { my ($yyyy,$mm,$dd) = @_; my ($year,$month,$day) = get_last_day_of_prev_quarter($yyyy,$mm,$d +d); my $epoch = timelocal(0,0,12, $day, $month-1, $year-1900); my @timestuff = localtime($epoch); # # If the day is a Sunday (0) or a Saturday (6) # # Subtract 1 or 2 days worth of seconds, then convert # back. my $date_diff = 0; if ($timestuff[6] == 0) { $date_diff = 2; } if ($timestuff[6] == 6) { $date_diff = 1; } if ($date_diff > 0) { $epoch = $epoch - (60*60*24) * $date_diff; } @timestuff = localtime($epoch); return ($timestuff[5] + 1900, $timestuff[4] + 1,$timestuff[3]); } sub last_day_of_month { # This logic is moronic, but it works. # To get the last day of a given month, first take the first day # of the first month, convert it to epoch seconds, subtract a day # (86400 seconds), then re-convert back and pull the "day of month +" # element out of the list. my ($yyyy,$mm) = @_; my ($year,$month) = next_month($yyyy,$mm); my $epoch = timelocal(0,0,12,1, $month-1, $year-1900); $epoch = $epoch - ( 60*60*24 ); my @timestuff = localtime($epoch); return $timestuff[3]; } sub next_month { my ($yyyy,$mm) = @_; if ($mm == 12) { $mm = 1; $yyyy++; } else { $mm++; } return ($yyyy,$mm); } sub get_previous_quarter { my ($yyyy,$quarter) = @_; if ($quarter == 1) { $quarter = 4; $yyyy--; } else { $quarter--; } return ($yyyy,$quarter); } sub identify_quarter { my ($mm) = @_; return int(($mm-1) / 3) +1; }
In reply to Re: Care to simplify some working code?
by Voronich
in thread Care to simplify some working code?
by Voronich
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |