Last Friday of each month: Difference between revisions
(Racket version) |
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key=lambda x:int(x.split("-")[1])): |
key=lambda x:int(x.split("-")[1])): |
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print item</lang> |
print item</lang> |
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=={{header|Racket}}== |
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<lang racket> |
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#lang racket |
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(require srfi/19 math) |
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(define (days-in-month m y) |
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(define lengths #(0 31 #f 31 30 31 30 31 31 30 31 30 31)) |
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(define d (vector-ref lengths m)) |
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(if d d (days-in-feb y))) |
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(define (leap-year? y) |
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(and (divides? 4 y) |
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(or (not (divides? 100 y)) |
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(divides? 400 y)))) |
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(define (days-in-feb y) |
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(if (leap-year? y) 29 28)) |
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(define (last-day-in-month m y) |
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(make-date 0 0 0 0 (days-in-month m y) m y 0)) |
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(define (week-day date) |
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(define days #(sun mon tue wed thu fri sat)) |
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(vector-ref days (date-week-day date))) |
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(define (last-fridays y) |
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(for/list ([m (in-range 1 13)]) |
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(prev-friday (last-day-in-month m y)))) |
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(define 24hours (make-time time-duration 0 (* 24 60 60))) |
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(define (prev-day d) |
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(time-utc->date |
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(subtract-duration |
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(date->time-utc d) 24hours))) |
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(define (prev-friday d) |
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(if (eq? (week-day d) 'fri) |
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d |
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(prev-friday (prev-day d)))) |
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(for ([d (last-fridays 2012)]) |
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(displayln (~a (date->string d "~a ~d ~b ~Y")))) |
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</lang> |
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Output: |
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<lang racket> |
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Fri 27 Jan 2012 |
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Fri 24 Feb 2012 |
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Fri 30 Mar 2012 |
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Fri 27 Apr 2012 |
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Fri 25 May 2012 |
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Fri 29 Jun 2012 |
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Fri 27 Jul 2012 |
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Fri 31 Aug 2012 |
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Fri 28 Sep 2012 |
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Fri 26 Oct 2012 |
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Fri 30 Nov 2012 |
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Fri 28 Dec 2012 |
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</lang> |
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=={{header|REXX}}== |
=={{header|REXX}}== |
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Revision as of 15:16, 4 May 2013
You are encouraged to solve this task according to the task description, using any language you may know.
Write a program or a script that returns the last Fridays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc.).
Example of an expected output:
./last_fridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
- Cf.
Ada
Uses GNAT. Applicable to any day of the week, cf. [[1]].
<lang Ada>with Ada.Text_IO, GNAT.Calendar.Time_IO, Ada.Command_Line,
Ada.Calendar.Formatting, Ada.Calendar.Arithmetic;
procedure Last_Weekday_In_Month is
procedure Put_Line(T: Ada.Calendar.Time) is
use GNAT.Calendar.Time_IO;
begin
Ada.Text_IO.Put_Line(Image(Date => T, Picture => ISO_Date));
end Put_Line;
use Ada.Calendar, Ada.Calendar.Arithmetic;
subtype Day_Name is Formatting.Day_Name; use type Formatting.Day_Name;
T, Selected : Time;
Weekday: Day_Name := Day_Name'Value(Ada.Command_Line.Argument (1));
Year : Year_Number := Integer'Value (Ada.Command_Line.Argument (2));
begin
for Month in 1 .. 12 loop
T := Time_Of (Year => Year, Month => Month, Day => 01);
while Ada.Calendar.Month(T) = Month loop
if Formatting.Day_Of_Week (T) = Weekday then Selected := T; end if; T := T + Day_Count(1);
end loop;
Put_Line(Selected);
end loop;
end Last_Weekday_In_Month;</lang>
- Output:
>./last_weekday_in_month friday 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
AutoHotkey
<lang AHK>if 1 = ; no parameter passed { InputBox, 1, Last Fridays of year, Enter a year:, , , , , , , , %A_YYYY% If ErrorLevel ExitApp }
YYYY = %1% ; retrieve command line parameter Stmp = %YYYY%0101000000 count= 0
While count < 12 { FormatTime, ddd, %stmp%, ddd FormatTime, M, %stmp%, M If (ddd = "Fri"){ if (M-1 = count){ t := stmp stmp += 7, days } else res .= SubStr(t, 1, 4) "-" SubStr(t, 5, 2) "-" SubStr(t, 7, 2) "`n" ,count++ ,stmp := YYYY . SubStr("0" M, -1) . "01" } else stmp += 1, days } MsgBox % res</lang> Output for 2012:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
C
Doesn't work with Julian calendar (then again, you probably don't need to plan your weekends for middle ages).
<lang c>#include <stdio.h>
- include <stdlib.h>
int main(int c, char *v[]) { int days[] = {31,29,31,30,31,30,31,31,30,31,30,31}; int m, y, w;
if (c < 2 || (y = atoi(v[1])) <= 1700) return 1;
days[1] -= (y % 4) || (!(y % 100) && (y % 400));
w = y * 365 + (y - 1) / 4 - (y - 1) / 100 + (y - 1) / 400 + 6;
for(m = 0; m < 12; m++) { w = (w + days[m]) % 7; printf("%d-%02d-%d\n", y, m + 1, days[m] + (w < 5 ? -2 : 5) - w); }
return 0; }</lang>
C#
<lang csharp>using System; using System.Collections.Generic; using System.Globalization; using System.Linq;
namespace RosettaCode.LastFridaysOfYear {
internal static class Program
{
private static IEnumerable<DateTime> LastFridaysOfYear(int year)
{
for (var month = 1; month <= 12; month++)
{
var date = new DateTime(year, month, 1).AddMonths(1).AddDays(-1);
while (date.DayOfWeek != DayOfWeek.Friday)
{
date = date.AddDays(-1);
}
yield return date;
}
}
private static void Main(string[] arguments)
{
int year;
var argument = arguments.FirstOrDefault();
if (string.IsNullOrEmpty(argument) || !int.TryParse(argument, out year))
{
year = DateTime.Today.Year;
}
foreach (var date in LastFridaysOfYear(year))
{
Console.WriteLine(date.ToString("d", CultureInfo.InvariantCulture));
}
}
}
}</lang> Output:
01/27/2012 02/24/2012 03/30/2012 04/27/2012 05/25/2012 06/29/2012 07/27/2012 08/31/2012 09/28/2012 10/26/2012 11/30/2012 12/28/2012
C++
called with ./last_fridays 2012
<lang cpp>#include <boost/date_time/gregorian/gregorian.hpp>
- include <iostream>
- include <cstdlib>
int main( int argc , char* argv[ ] ) {
using namespace boost::gregorian ;
greg_month months[ ] = { Jan , Feb , Mar , Apr , May , Jun , Jul ,
Aug , Sep , Oct , Nov , Dec } ;
greg_year gy = atoi( argv[ 1 ] ) ;
for ( int i = 0 ; i < 12 ; i++ ) {
last_day_of_the_week_in_month lwdm ( Friday , months[ i ] ) ;
date d = lwdm.get_date( gy ) ;
std::cout << d << std::endl ;
}
return 0 ;
}</lang> Output:
2012-Jan-27 2012-Feb-24 2012-Mar-30 2012-Apr-27 2012-May-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Oct-26 2012-Nov-30 2012-Dec-28
CoffeeScript
<lang coffeescript> last_friday_of_month = (year, month) ->
# month is 1-based, JS API is 0-based, then we use
# non-positive indexes to work backward relative to the
# first day of the next month
i = 0
while true
last_day = new Date(year, month, i)
if last_day.getDay() == 5
return last_day.toDateString()
i -= 1
print_last_fridays_of_month = (year) ->
for month in [1..12] console.log last_friday_of_month year, month
do ->
year = parseInt process.argv[2] print_last_fridays_of_month year
</lang> output <lang> > coffee last_friday.coffee 2012 Fri Jan 27 2012 Fri Feb 24 2012 Fri Mar 30 2012 Fri Apr 27 2012 Fri May 25 2012 Fri Jun 29 2012 Fri Jul 27 2012 Fri Aug 31 2012 Fri Sep 28 2012 Fri Oct 26 2012 Fri Nov 30 2012 Fri Dec 28 2012 </lang>
D
<lang d>import std.stdio, std.datetime, std.traits;
void lastFridays(in uint year) {
auto date = Date(year, 1, 1);
foreach (_; [EnumMembers!Month]) {
date.day(date.daysInMonth);
date.roll!"days"(-(date.dayOfWeek + 2) % 7);
writeln(date);
date.add!"months"(1, AllowDayOverflow.no);
}
}
void main() {
lastFridays(2012);
}</lang>
2012-Jan-27 2012-Feb-24 2012-Mar-30 2012-Apr-27 2012-May-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Oct-26 2012-Nov-30 2012-Dec-28
Fortran
Algorithm: compute day of week for last day of month, then subtract just enough to get to the preceding friday. Do this for each month. To simplify computations further, we only need to compute day of week of january 1st (the others are found by adding month lengths). Since day of week need only be known modulo 7, we do not compute modulo at all except once when subtracting. <lang fortran>program fridays
implicit none
integer :: days(1:12) = (/31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31/)
integer :: year, k, y, m
read *, year
if (mod(year, 400) == 0 .or. (mod(year, 4) == 0 .and. mod(year, 100) /= 0)) days(2) = 29
y = year - 1
k = 44 + y + y/4 + 6*(y/100) + y/400
do m = 1, 12
k = k + days(m)
print "(I4,A1,I2.2,A1,I2)", year, '-', m, '-', days(m) - mod(k, 7)
end do
end program </lang>
Go
<lang go>package main
import (
"fmt" "os" "strconv" "time"
)
func main() {
y := time.Now().Year()
if len(os.Args) == 2 {
if i, err := strconv.Atoi(os.Args[1]); err == nil {
y = i
}
}
for m := time.January; m <= time.December; m++ {
d := time.Date(y, m+1, 1, 0, 0, 0, 0, time.UTC).Add(-24 * time.Hour)
d = d.Add(-time.Duration((d.Weekday() + 2) % 7) * 24 * time.Hour)
fmt.Println(d.Format("2006-01-02"))
}
}</lang> Output:
> ./fridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
Icon and Unicon
This will write the last fridays for every year given as an argument. There is no error checking on the year.
<lang Icon>procedure main(A) every write(lastfridays(!A)) end
procedure lastfridays(year) every m := 1 to 12 do {
d := case m of {
2 : if IsLeapYear(year) then 29 else 28
4|6|9|11 : 30
default : 31
} # last day of month
z := 0
j := julian(m,d,year) + 1 # first day of next month
until (j-:=1)%7 = 4 do z -:=1 # backup to last friday=4
suspend sprintf("%d-%d-%d",year,m,d+z)
}
end
link datetime, printf</lang>
printf.icn provides formatting datetime.icn provides julian and IsLeapYear
Output:
last_fridays.exe 2012 2012-1-27 2012-2-24 2012-3-30 2012-4-27 2012-5-25 2012-6-29 2012-7-27 2012-8-31 2012-9-28 2012-10-26 2012-11-30 2012-12-28
J
<lang j>require'dates' last_fridays=: 12 {. [:({:/.~ }:"1)@(#~ 5 = weekday)@todate (i.366) + todayno@,&1 1</lang>
In other words, start from January 1 of the given year, and count forward for 366 days, keeping the fridays. Then pick the last remaining day within each represented month (which will be a friday because we only kept the fridays). Then pick the first 12 (since on a non-leap year which ends on a thursday we would get an extra friday).
Example use:
<lang j> last_fridays 2012 2012 1 27 2012 2 24 2012 3 30 2012 4 27 2012 5 25 2012 6 29 2012 7 27 2012 8 31 2012 9 28 2012 10 26 2012 11 30 2012 12 28</lang>
Java
<lang java5>import java.text.*; import java.util.*;
public class LastFridays {
public static void main(String[] args) throws Exception {
int year = Integer.parseInt(args[0]);
GregorianCalendar c = new GregorianCalendar(year, 0, 1);
for (String mon : new DateFormatSymbols(Locale.US).getShortMonths()) {
if (!mon.isEmpty()) {
int totalDaysOfMonth = c.getActualMaximum(Calendar.DAY_OF_MONTH);
c.set(Calendar.DAY_OF_MONTH, totalDaysOfMonth);
int daysToRollBack = (c.get(Calendar.DAY_OF_WEEK) + 1) % 7;
int day = totalDaysOfMonth - daysToRollBack;
c.set(Calendar.DAY_OF_MONTH, day);
System.out.printf("%d %s %d\n", year, mon, day);
c.set(year, c.get(Calendar.MONTH) + 1, 1);
}
}
}
}</lang>
Output (for java LastFridays 2012):
2012 Jan 27 2012 Feb 24 2012 Mar 30 2012 Apr 27 2012 May 25 2012 Jun 29 2012 Jul 27 2012 Aug 31 2012 Sep 28 2012 Oct 26 2012 Nov 30 2012 Dec 28
Mathematica
<lang Mathematica>FridaysOfYear[Y_] :=
NestWhile[(DaysPlus[#, - 1]) &, #, (DateString[#, "DayName"] != "Friday") &] & /@
Most@Reverse@NestList [DaysPlus[# /. {x_, y_, X_} -> {x, y, 1}, - 1] &, {Y + 1, 1, 1}, 12]
Column@FridaysOfYear[2012]</lang> Output:
{2012,1,27}
{2012,2,24}
{2012,3,30}
{2012,4,27}
{2012,5,25}
{2012,6,29}
{2012,7,27}
{2012,8,31}
{2012,9,28}
{2012,10,26}
{2012,11,30}
{2012,12,28}
MATLAB / Octave
<lang Matlab> function t = last_fridays_of_year(y)
t1 = datenum([y,1,1,0,0,0]);
t2 = datenum([y,12,31,0,0,0]);
t = datevec(t1:t2);
t = t(strmatch('Friday', datestr(t,'dddd')), :); % find all Fridays
t = t([find(diff(t(:,2)) > 0); end], :); % find Fridays before change of month
end;
datestr(last_fridays_of_year(2012),'yyyy-mm-dd')
</lang>
Output:
ans = 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
Maxima
<lang maxima>weekday(year, month, day) := block([m: month, y: year, k],
if m < 3 then (m: m + 12, y: y - 1),
k: 1 + remainder(day + quotient((m + 1)*26, 10) + y + quotient(y, 4)
+ 6*quotient(y, 100) + quotient(y, 400) + 5, 7),
['monday, 'tuesday, 'wednesday, 'thurdsday, 'friday, 'saturday, 'sunday][k]
)$
leapyearp(year) := is(mod(year, 4) = 0 and (mod(year, 100) # 0 or mod(year, 400) = 0))$
lastfridays(year) := block(
[m: [31, if leapyearp(year) then 29 else 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31], v: [ ]],
for month thru 12 do v: endcons(sconcat(year, "-", month, "-",
lmax(sublist(makelist(i, i, 1, m[month]), lambda([day], weekday(year, month, day) = 'friday)))), v),
v
)$
lastfridays(2012); ["2012-1-27", "2012-2-24", "2012-3-30", "2012-4-27", "2012-5-25", "2012-6-29", "2012-7-27","2012-8-31", "2012-9-28", "2012-10-26", "2012-11-30", "2012-12-28"]</lang>
OCaml
Using the module Unix from the standard OCaml library:
<lang ocaml>#load "unix.cma" open Unix
let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0); exit 1
let print_date t =
Printf.printf "%d-%02d-%02d\n" (t.tm_year + 1900) (t.tm_mon + 1) t.tm_mday
let is_date_ok tm t =
(tm.tm_year = t.tm_year && tm.tm_mon = t.tm_mon && tm.tm_mday = t.tm_mday)
let () =
let _year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let year = _year - 1900 in
let fridays = Array.make 12 (Unix.gmtime 0.0) in
for month = 0 to 11 do
for day_of_month = 1 to 31 do
let tm = { (Unix.gmtime 0.0) with
tm_year = year;
tm_mon = month;
tm_mday = day_of_month;
} in
let _, t = Unix.mktime tm in
if is_date_ok tm t (* check for months that have less than 31 days *)
&& t.tm_wday = 5 (* is a friday *)
then fridays.(month) <- t
done;
done;
Array.iter print_date fridays</lang>
Output:
$ ocaml last_fridays.ml 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
With a dedicated library
<lang ocaml>open CalendarLib
let usage() =
Printf.eprintf "%s <year>\n" Sys.argv.(0); exit 1
let print_date (year, month, day) =
Printf.printf "%d-%02d-%02d\n" year month day
let () =
let year =
try int_of_string Sys.argv.(1)
with _ -> usage()
in
let fridays = ref [] in
for month = 1 to 12 do
let num_days = Date.days_in_month (Date.make_year_month year month) in
let rec aux day =
if Date.day_of_week (Date.make year month day) = Date.Fri
then fridays := (year, month, day) :: !fridays
else aux (pred day)
in
aux num_days
done;
List.iter print_date (List.rev !fridays)</lang>
Run this script with the command:
ocaml unix.cma str.cma -I +calendar calendarLib.cma last_fridays.ml 2012
Perl
<lang Perl>#!/usr/bin/perl -w use strict ; use DateTime ; use feature qw( say ) ;
foreach my $month ( 1..12 ) {
my $dt = DateTime->last_day_of_month( year => $ARGV[ 0 ] , month => $month ) ;
while ( $dt->day_of_week != 5 ) {
$dt->subtract( days => 1 ) ;
}
say $dt->ymd ;
}</lang> Output:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
Perl 6
<lang perl6>sub MAIN (Int $year = Date.today.year) {
my @fri;
for Date.new("$year-01-01") .. Date.new("$year-12-31") {
@fri[.month] = .Str if .day-of-week == 5;
}
.say for @fri[1..12];
}</lang>
Example:
$ ./lastfri 2038 2038-01-29 2038-02-26 2038-03-26 2038-04-30 2038-05-28 2038-06-25 2038-07-30 2038-08-27 2038-09-24 2038-10-29 2038-11-26 2038-12-31
A solution without a result array to store things in:
<lang perl6>sub MAIN (Int $year = Date.today.year) {
say ~.value.reverse.first: *.day-of-week == 5
for classify *.month, Date.new("$year-01-01") .. Date.new("$year-12-31");
}</lang>
Here, classify sorts the dates into one bin per month (but preserves the order in each bin). We then take the list inside each bin (.value) and find the last (.reverse.first) date which is a Friday.
PHP
PHP is generally used for web apps, so I am not implementing the command-line component of this task.
<lang PHP><?php function last_friday_of_month($year, $month) {
$day = 0;
while(True) {
$last_day = mktime(0, 0, 0, $month+1, $day, $year);
if (date("w", $last_day) == 5) {
return date("Y-m-d", $last_day);
}
$day -= 1;
}
}
function print_last_fridays_of_month($year) {
foreach(range(1, 12) as $month) {
echo last_friday_of_month($year, $month), "
";
}
}
date_default_timezone_set("GMT"); $year = 2012; print_last_fridays_of_month($year); ?></lang>
Output in browser:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
PicoLisp
<lang PicoLisp>(de lastFridays (Y)
(for M `(range 1 12)
(prinl
(dat$
(find '((D) (= "Friday" (day D)))
(mapcar '((D) (date Y M D)) `(range 31 22)) )
"-" ) ) ) )</lang>
Test: <lang PicoLisp>: (lastFridays 2012) 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28</lang>
Pike
<lang Pike>int(0..1) last_friday(object day) {
return day->week_day() == 5 &&
day->month_day() > day->month()->number_of_days()-7;
}
int main(int argc, array argv) {
array days = filter(Calendar.Year((int)argv[1])->months()->days()[*], last_friday);
write("%{%s\n%}", days->format_ymd());
return 0;
}</lang>
PL/I
<lang PL/I> Fridays: procedure (year) options (main); /* 8 January 2013 */
declare year character (4) varying; declare start fixed binary (31); declare months fixed decimal (2) initial (0); declare (current_month, month_one_week_hence) character (2);
put list ('Last Fridays in each month for the year ' || year || ':' );
start = days('0101' || year, 'DDMMYYYY');
/* Find first Friday */
do while (weekday(start) ^= 6); start = start + 1; end;
do until (months=12);
current_month = substr (daystodate(start, 'MMDDYYYY'), 1, 2 );
month_one_week_hence = substr (daystodate(start+7, 'MMDDYYYY'), 1, 2 );
if current_month ^= month_one_week_hence then
do;
months = months + 1;
put skip list (daystodate(start, 'DDMmmYYYY'));
end;
start = start + 7;
end;
end Fridays; </lang> The command: FRIDAYS /2008 produces:
Last Fridays in each month for the year 2008: 25Jan2008 29Feb2008 28Mar2008 25Apr2008 30May2008 27Jun2008 25Jul2008 29Aug2008 26Sep2008 31Oct2008 28Nov2008 26Dec2008
Output for 2013:
Last Fridays in each month for the year 2013: 25Jan2013 22Feb2013 29Mar2013 26Apr2013 31May2013 28Jun2013 26Jul2013 30Aug2013 27Sep2013 25Oct2013 29Nov2013 27Dec2013
Python
<lang python>import calendar c=calendar.Calendar() fridays={} year=raw_input("year") for item in c.yeardatescalendar(int(year)):
for i1 in item:
for i2 in i1:
for i3 in i2:
if "Fri" in i3.ctime() and year in i3.ctime():
month,day=str(i3).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
Using reduce
<lang python>import calendar c=calendar.Calendar() fridays={} year=raw_input("year") add=list.__add__ for day in reduce(add,reduce(add,reduce(add,c.yeardatescalendar(int(year))))):
if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
using itertools
<lang python>import calendar from itertools import chain f=chain.from_iterable c=calendar.Calendar() fridays={} year=raw_input("year") add=list.__add__
for day in f(f(f(c.yeardatescalendar(int(year))))):
if "Fri" in day.ctime() and year in day.ctime():
month,day=str(day).rsplit("-",1)
fridays[month]=day
for item in sorted((month+"-"+day for month,day in fridays.items()),
key=lambda x:int(x.split("-")[1])):
print item</lang>
Racket
<lang racket>
- lang racket
(require srfi/19 math)
(define (days-in-month m y)
(define lengths #(0 31 #f 31 30 31 30 31 31 30 31 30 31)) (define d (vector-ref lengths m)) (if d d (days-in-feb y)))
(define (leap-year? y)
(and (divides? 4 y)
(or (not (divides? 100 y))
(divides? 400 y))))
(define (days-in-feb y)
(if (leap-year? y) 29 28))
(define (last-day-in-month m y)
(make-date 0 0 0 0 (days-in-month m y) m y 0))
(define (week-day date)
(define days #(sun mon tue wed thu fri sat)) (vector-ref days (date-week-day date)))
(define (last-fridays y)
(for/list ([m (in-range 1 13)]) (prev-friday (last-day-in-month m y))))
(define 24hours (make-time time-duration 0 (* 24 60 60)))
(define (prev-day d)
(time-utc->date (subtract-duration (date->time-utc d) 24hours)))
(define (prev-friday d)
(if (eq? (week-day d) 'fri)
d
(prev-friday (prev-day d))))
(for ([d (last-fridays 2012)])
(displayln (~a (date->string d "~a ~d ~b ~Y"))))
</lang> Output: <lang racket> Fri 27 Jan 2012 Fri 24 Feb 2012 Fri 30 Mar 2012 Fri 27 Apr 2012 Fri 25 May 2012 Fri 29 Jun 2012 Fri 27 Jul 2012 Fri 31 Aug 2012 Fri 28 Sep 2012 Fri 26 Oct 2012 Fri 30 Nov 2012 Fri 28 Dec 2012 </lang>
REXX
This REXX program will find the last day-of-week (for any day) of all the months for any year.
It wasn't optimized just to find a particular day-of-week.
<lang rexx>/*REXX program displays dates of last Fridays of each month for any year*/
parse arg yyyy
do j=1 for 12
say lastDOW('Friday',j,yyyy)
end /*j*/
exit /*stick a fork in it, we're done.*/ /*┌────────────────────────────────────────────────────────────────────┐
│ lastDOW: procedure to return the date of the last day-of-week of │ │ any particular month of any particular year. │ │ │ │ The day-of-week must be specified (it can be in any case, │ │ (lower-/mixed-/upper-case) as an English name of the spelled day │ │ of the week, with a minimum length that causes no ambiguity. │ │ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... │ │ │ │ The month can be specified as an integer 1 ──► 12 │ │ 1=January 2=February 3=March ... 12=December │ │ or the English name of the month, with a minimum length that │ │ causes no ambiguity. I.E.: Jun for June, D for December. │ │ If omitted [or an asterisk(*)], the current month is used. │ │ │ │ The year is specified as an integer or just the last two digits │ │ (two digit years are assumed to be in the current century, and │ │ there is no windowing for a two-digit year). │ │ If omitted [or an asterisk(*)], the current year is used. │ │ Years < 100 must be specified with (at least 2) leading zeroes.│ │ │ │ Method used: find the "day number" of the 1st of the next month, │ │ then subtract one (this gives the "day number" of the last day of │ │ the month, bypassing the leapday mess). The last day-of-week is │ │ then obtained straightforwardly, or via subtraction. │ └────────────────────────────────────────────────────────────────────┘*/
lastdow: procedure; arg dow .,mm .,yy . /*DOW = day of week*/ parse arg a.1,a.2,a.3 /*orig args, errmsg*/ if mm== | mm=='*' then mm=left(date('U'),2) /*use default month*/ if yy== | yy=='*' then yy=left(date('S'),4) /*use default year */ if length(yy)==2 then yy=left(date('S'),2)yy /*append century. */
/*Note mandatory leading blank in strings below.*/
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday" !=" JAnuary February MARch APril MAY JUNe JULy AUgust September",
" October November December"
upper $ ! /*uppercase strings*/ if dow== then call .er "wasn't specified",1 if arg()>3 then call .er 'arguments specified',4
do j=1 for 3 /*any plural args ?*/ if words(arg(j))>1 then call .er 'is illegal:',j end
dw=pos(' 'dow,$) /*find day-of-week*/ if dw==0 then call .er 'is invalid:',1 if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1
if datatype(mm,'month') then /*if MM is alpha...*/
do
m=pos(' 'mm,!) /*maybe its good...*/
if m==0 then call .er 'is invalid:',1
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
mm=wordpos(word(substr(!,m),1),!)-1 /*now, use true Mon*/
end
if \datatype(mm,'W') then call .er "isn't an integer:",2 if \datatype(yy,'W') then call .er "isn't an integer:",3 if mm<1 | mm>12 then call .er "isn't in range 1──►12:",2 if yy=0 then call .er "can't be 0 (zero):",3 if yy<0 then call .er "can't be negative:",3 if yy>9999 then call .er "can't be > 9999:",3
tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
/*day# of last dom.*/
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1 ?=_//7 /*calc. DOW, 0──►6*/ if ?\==tdow then _=_-?-7+tdow+7*(?>tdow) /*not DOW? Adjust.*/ return date('weekday',_,"B") date(,_,'B') /*return the answer*/
.er: arg ,_;say; say '***error!*** (in LASTDOW)';say /*tell error, and */
say word('day-of-week month year excess',arg(2)) arg(1) a._
say; exit 13 /*... then exit. */</lang>
output when using the following input: 2012 or 12
Friday 27 Jan 2012 Friday 24 Feb 2012 Friday 30 Mar 2012 Friday 27 Apr 2012 Friday 25 May 2012 Friday 29 Jun 2012 Friday 27 Jul 2012 Friday 31 Aug 2012 Friday 28 Sep 2012 Friday 26 Oct 2012 Friday 30 Nov 2012 Friday 28 Dec 2012
Ruby
<lang ruby>require 'date'
def last_friday(year, month)
# Last day of month: Date.new interprets a negative number as a relative month/day from the end of year/month. d = Date.new(year, month, -1) d -= (d.wday - 5) % 7 # Subtract days after Friday.
end
year = Integer(ARGV.shift) (1..12).each {|month| puts last_friday(year, month)}</lang>
Friday is d.wday == 5; the expression (d.wday - 5) % 7 counts days after Friday.
Using the ActiveSupport library for some convenience methods
<lang ruby>require 'rubygems' require 'activesupport'
def last_friday(year, month)
d = Date.new(year, month, 1).end_of_month until d.wday == 5 d = d.yesterday end d
end</lang>
Run BASIC
<lang runbasic>input "Year:";yr dayOne$ = "01-01-";yr n1 = date$(dayOne$) for i = 1 to 12
n1 = n1 + 26 m1$ = left$(date$(n1),2) while m1$ = left$(date$(n1),2) ' find end of month n1 = n1 + 1 wend n1 = n1 -1 while (n1 Mod 7) <> 3 ' find Friday n1 = n1 - 1 wend print date$(n1) ' print last Friday's date
next i</lang>
Year:?2013 01/25/2013 02/22/2013 03/29/2013 04/26/2013 05/31/2013 06/28/2013 07/26/2013 08/30/2013 09/27/2013 10/25/2013 11/29/2013 12/27/2013
Scala
<lang scala>import java.util.Calendar import java.text.SimpleDateFormat
object Fridays {
def lastFridayOfMonth(year:Int, month:Int)={
val cal=Calendar.getInstance
cal.set(Calendar.YEAR, year)
cal.set(Calendar.MONTH, month)
cal.set(Calendar.DAY_OF_WEEK, Calendar.FRIDAY)
cal.set(Calendar.DAY_OF_WEEK_IN_MONTH, -1)
cal.getTime
}
def fridaysOfYear(year:Int)=for(month <- 0 to 11) yield lastFridayOfMonth(year, month)
def main(args:Array[String]){
val year=args(0).toInt
val formatter=new SimpleDateFormat("yyyy-MMM-dd")
fridaysOfYear(year).foreach{date=>
println(formatter.format(date))
}
}
}</lang> Output:
2012-Jan-27 2012-Feb-24 2012-Mrz-30 2012-Apr-27 2012-Mai-25 2012-Jun-29 2012-Jul-27 2012-Aug-31 2012-Sep-28 2012-Okt-26 2012-Nov-30 2012-Dez-28
Seed7
Uses the libraries time.s7i and duration.s7i. Applicable to any day of the week, cf. [[2]].
<lang seed7>$ include "seed7_05.s7i";
include "time.s7i"; include "duration.s7i";
const proc: main is func
local
var integer: weekday is 1; # 1 for monday, 2 for tuesday, and so on up to 7 for sunday.
var integer: year is 0;
var integer: month is 1;
var time: aDate is time.value;
var time: selected is time.value;
begin
if length(argv(PROGRAM)) <> 2 then
writeln("usage: lastWeekdayInMonth weekday year");
writeln(" weekday: 1 for monday, 2 for tuesday, and so on up to 7 for sunday.");
else
weekday := integer parse (argv(PROGRAM)[1]);
year := integer parse (argv(PROGRAM)[2]);
for month range 1 to 12 do
aDate := date(year, month, 1);
while aDate.month = month do
if dayOfWeek(aDate) = weekday then
selected := aDate;
end if;
aDate +:= 1 . DAYS;
end while;
writeln(strDate(selected));
end for;
end if;
end func;</lang>
Output when called with s7 rosetta/lastWeekdayInMonth 5 2013:
2013-01-25 2013-02-22 2013-03-29 2013-04-26 2013-05-31 2013-06-28 2013-07-26 2013-08-30 2013-09-27 2013-10-25 2013-11-29 2013-12-27
Tcl
<lang tcl>package require Tcl 8.5 set year [lindex $argv 0] foreach dm {02/1 03/1 04/1 05/1 06/1 07/1 08/1 09/1 10/1 11/1 12/1 12/32} {
# The [clock scan] code is unhealthily clever; use it for our own evil purposes set t [clock scan "last friday" -base [clock scan $dm/$year -gmt 1] -gmt 1] # Print the interesting part puts [clock format $t -format "%Y-%m-%d" -gmt 1]
}</lang> Sample execution:
$ tclsh8.5 lastfri.tcl 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT year=2012 LOOP month=1,12
LOOP day=31,22,-1 dayofweek=DATE (number,day,month,year,nummer) IF (dayofweek==5) THEN PRINT year,"-",month,"-",day EXIT ENDIF ENDLOOP
ENDLOOP </lang> Output:
2012-1-27 2012-2-24 2012-3-30 2012-4-27 2012-5-25 2012-6-29 2012-7-27 2012-8-31 2012-9-28 2012-10-26 2012-11-30 2012-12-28
UNIX Shell
Using ncal. Will switch to Julian calender as ncal sees fit, and will not calculate past year 9999 (chances are you'll be too dead by then to worry about weekends anyway).
<lang bash>#!/bin/sh
if [ -z $1 ]; then exit 1; fi
- weed out multiple erros due to bad year
ncal 1 $1 > /dev/null && \ for m in 01 02 03 04 05 06 07 08 09 10 11 12; do echo $1-$m-`ncal $m $1 | grep Fr | sed 's/.* \([0-9]\)/\1/'` done</lang>
Using date --date from GNU date??? This code is not portable.
<lang bash>#!/bin/sh
- Free code, no limit work
- $Id: lastfridays,v 1.1 2011/11/10 00:48:16 gilles Exp gilles $
- usage :
- lastfridays 2012 # prints last fridays of months of year 2012
debug=${debug:-false}
- debug=true
epoch_year_day() { #set -x x_epoch=`expr ${2:-0} '*' 86400 + 43200` date --date="${1:-1970}-01-01 UTC $x_epoch seconds" +%s }
year_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" +%Y } day_of_epoch() { LC_ALL=C date --date="1970-01-01 UTC ${1:-0} seconds" +%A } date_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" "+%Y-%m-%d" } month_of_epoch() { date --date="1970-01-01 UTC ${1:-0} seconds" "+%m" }
last_fridays() { year=${1:-2012}
next_year=`expr $year + 1`
$debug && echo "next_year $next_year"
current_year=$year
day=0
previous_month=01
while test $current_year != $next_year; do
$debug && echo "day $day"
current_epoch=`epoch_year_day $year $day`
$debug && echo "current_epoch $current_epoch"
current_year=`year_of_epoch $current_epoch`
current_day=`day_of_epoch $current_epoch`
$debug && echo "current_day $current_day"
test $current_day = 'Friday' && current_friday=`date_of_epoch $current_epoch`
$debug && echo "current_friday $current_friday"
current_month=`month_of_epoch $current_epoch`
$debug && echo "current_month $current_month"
# Change of month => previous friday is the last of month
test "$previous_month" != "$current_month" \
&& echo $previous_friday
previous_month=$current_month
previous_friday=$current_friday
day=`expr $day + 1`
done
}
- main
last_fridays ${1:-2012}</lang>
Sample execution:
lastfridays 2012 2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations
func WeekDay(Year, Month, Day); \Return day of week (0=Sun, 1=Mon ... 6=Sat) int Year, Month, Day; \works for years from 1583 onward [if Month<=2 then [Month:= Month+12; Year:= Year-1]; return rem((Day-1 + (Month+1)*26/10 + Year + Year/4 + Year/100*6 + Year/400)/7); ];
int Year, Month, LastDay, WD; [Year:= IntIn(8); \from command line for Month:= 1 to 12 do
[LastDay:= WeekDay(Year, Month+1, 1) - WeekDay(Year, Month, 28); if LastDay < 0 then LastDay:= LastDay + 7; LastDay:= LastDay + 27; \ = number of days in Month WD:= WeekDay(Year, Month, LastDay); WD:= WD - 5; if WD < 0 then WD:= WD + 7; LastDay:= LastDay - WD; IntOut(0, Year); ChOut(0, ^-); if Month < 10 then ChOut(0, ^0); IntOut(0, Month); ChOut(0, ^-); IntOut(0, LastDay); CrLf(0); ];
]</lang>
- Output:
2012-01-27 2012-02-24 2012-03-30 2012-04-27 2012-05-25 2012-06-29 2012-07-27 2012-08-31 2012-09-28 2012-10-26 2012-11-30 2012-12-28
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