Percentage of work days in a month

Question

Given a year and a month, find out the percentage of work days in said month. Work days are Monday through Friday with no regard to holidays or other special things. The Gregorian calendar is used.

Input

A year and month in ISO 8601 format (YYYY-MM). The year always has four digits, the month always has two digits. The given year will not be before 1582.

Output

Output is the percentage of work days (according to above definition) in the given month, rounded to a whole number. No percent sign or fractional digits follow.

Sample 1

Input                Output

2010-05              68

Sample 2

Input                Output

2010-06              73

Sample 3

Input                Output

1920-10              68

Sample 4

Input                Output

2817-12              68

A week has passed, an answer has been accepted. For the curious, the sizes of the submissions we got in our contest:

129 – Z shell
174 – VB.NET
222 – C
233 – C
300 – C

As well as our own (unranked) solutions:

  75 – PowerShell
  93 – Ruby
112 – Bourne shell

Answer 1

64-bit Perl, 67 68

Perl 5.10 or later, run with perl -E 'code here' or perl -M5.010 filename

map{$d++,/^S/||$w++if$_=`date -d@ARGV-$_`}1..31;say int.5+100*$w/$d

Concessions to code size:

• locale-sensitive: it counts as work days the days whose date output don't start with a capital S. Run under LC_ALL=C if in doubt.
• output is pure and well-formatted, but there's "garbage" on stderr on months shorter than 31. 2> /dev/null if upset.
• for some reason, my version of date considers 2817-12 an invalid month. Who knew, GNU new apocalypse is due! Requires a 64 bit build of date for dates after 2038. (Thanks Joey)

Answer 2

PHP - 135

I made it in PHP because I had a similar problem to treat a few days ago.

<?php $a=array(2,3,3,3,2,1,1);$t=strtotime($argv[1]);$m=date(t,$t);echo round((20+min($m-28,$a[date(w,strtotime('28day',$t))]))/$m*100)

(Somewhat) More legibly, and without notices about constants being used as strings:

<?php
date_default_timezone_set('America/New_York');
$additionalDays = array(2, 3, 3, 3, 2, 1, 1);
$timestamp = strtotime($argv[1]);
$daysInMonth = date('t', $timestamp);
$limit = $daysInMonth - 28;
$twentyNinthDayIndex = date('w', strtotime("+28 days", $timestamp));
$add = $additionalDays[$twentyNinthDayIndex];
$numberOfWorkDays = 20 + min($limit, $add);
echo round($numberOfWorkDays / $daysInMonth * 100);
?>

This is made possible by a very simple algorithm to compute the number of work days in a month: check for the weekdayness of the 29th, 30th and 31st (if those dates exist), and add 20.

Answer 3

Python 152 Characters

from calendar import*
y,m=map(int,raw_input().split('-'))
c=r=monthrange(y,m)[1]
for d in range(1,r+1):
 if weekday(y,m,d)>4:c-=1
print '%.f'%(c*100./r)

Answer 4

Windows PowerShell, 80

$x=$args;1..31|%{"$x-$_"|date -u %u -ea 0}|%{$a++
$b+=!!($_%6)}
[int]($b*100/$a)

Answer 5

Bash + coreutils, 82 bytes

f()(cal -NMd$1|sed -n "s/^$2.//p"|wc -w)
dc -e`f $1 "[^S ]"`d`f $1 S`+r200*r/1+2/p

Answer 6

PHP 5.2, 88 bytes

Although I already golfed zneak´s solution down to 85 bytes (I just found one more), here´s my own:
_{I doubt that I can squeeze another three bytes out here.}

$a=_4444444255555236666304777411;echo$a[date(t,$t=strtotime($argn))%28*7+date(N,$t)]+67;

takes input from STDIN: Run with echo <yyyy>-<mm> | php -nR '<code>'.

The string $a maps days per month (date(t)) and week day of the first day of the month (date(N): Monday=1, Sunday=7) to the percentage of work days-67; strtotime converts the input to a UNIX timestamp; the rest of the code does the hashing.

+1 byte for older PHP 5: Replace N with w and $a=_...; with $a="...".
another +3 bytes for PHP 4: insert .-1 after $argn.

-5 bytes for PHP 5.5 or later (postdates the challenge):
Remove everything before echo and replace $a with "4444444255555236666304777411".

Answer 7

D: 186 Characters

auto f(S)(S s){auto d=Date.fromISOExtendedString(s~"-28"),e=d.endOfMonth;int n=20;while(1){d+=dur!"days"(1);if(d>e)break;int w=d.dayOfWeek;if(w>0&&w<6)++n;}return rndtol(n*100.0/e.day);}

More Legibly:

auto f(S)(S s)
{
    auto d = Date.fromISOExtendedString(s ~ "-28"), e = d.endOfMonth;
    int n = 20;

    while(1)
    {
        d += dur!"days"(1);

        if(d > e)
            break;

        int w = d.dayOfWeek;

        if(w > 0 && w < 6)
            ++n;
    }

    return rndtol(n * 100.0 / e.day);
}

Answer 8

Python - 142

from calendar import*
y,m=map(int,raw_input().split('-'))
f,r=monthrange(y,m)
print'%.f'%((r-sum(weekday(y,m,d+1)>4for d in range(r)))*100./r)

Thanks to fR0DDY for the calendar bit.

Answer 9

Ruby, 124 119 111

require 'date'
e=Date.civil *$*[0].split(?-).map(&:to_i),-1
p ((e+1<<1..e).count{|d|d.cwday<6}*1e2/e.day).round

Requires Ruby 1.9 due to splatting the year and month before the -1 "day" argument and ?- for "-". For Ruby 1.8, we must add 2 characters:

require 'date'
e=Date.civil *$*[0].split('-').map(&:to_i)<<-1
p ((e+1<<1..e).count{|d|d.cwday<6}*1e2/e.day).round

Edit: Shave five characters based on @Dogbert's help.
Edit: Shave eight more characters based on @steenslag's help.

Answer 10

C#, 158 bytes

s=>{var d=DateTime.Parse(s);int i=0,t=DateTime.DaysInMonth(d.Year,d.Month),w=0;while(i<t)w-=-(int)d.AddDays(i++).DayOfWeek%6>>31;return Math.Round(1e2*w/t);};

Anonymous method which returns the required percentage.

Full program with ungolfed, commented method and test cases:

using System;

class WorkingDayPercentage
{
    static void Main()
    {
        Func <string, double> f =
        s =>
        {
            var d = DateTime.Parse(s);                      // extracts a DateTime object from the supplied string
            int i = 0,                                      // index variable
                t = DateTime.DaysInMonth(d.Year, d.Month),  // number of total number of days in the specified month
                w = 0;                                      // number of working days in the month

            while (i < t)                                   // iterates through the days of the month
                w -= -(int)d.AddDays(i++).DayOfWeek%6 >> 31;// d.AddDays(i) is the current day
                                                            // i++ increments the index variable to go to the next day
                                                            // .DayOfWeek is an enum which hold the weekdays
                                                            // (int)..DayOfWeek gets the days's index in the enum
                                                            // note that 6 is Saturday, 0 is Sunday, 1 is Monday etc.
                                                            // (int)DayOfWeek % 6 converts weekend days to 0
                                                            // while working days stay strictly positive
                                                            // - changes the sign of the positive numbers
                                                            // >> 31 extracts the signum
                                                            // which is -1 for negative numbers (working days)
                                                            // weekend days remain 0
                                                            // w -= substracts the negative numbers
                                                            // equivalent to adding their modulus

            return Math.Round(1e2 * w / t);                 // the Math.round function requires a double or a decimal
                                                            // working days and total number of days are integers
                                                            // also, a percentage must be returned
                                                            // multiplying with 100.0 converts the expression to a double
                                                            // however, 1e2 is used to shorten the code
        };

        // test cases:
        Console.WriteLine(f("2010-05")); // 68
        Console.WriteLine(f("2010-06")); // 73
        Console.WriteLine(f("1920-10")); // 68
        Console.WriteLine(f("2817-12")); // 68
    }
}

Alternative function, which adds negative values to the number of working days, changing the sign in the return with no extra byte cost:

s=>{var d=DateTime.Parse(s);int i=0,t=DateTime.DaysInMonth(d.Year,d.Month),w=0;while(i<t)w+=-(int)d.AddDays(i++).DayOfWeek%6>>31;return-Math.Round(1e2*w/t);};

Answer 11

APL (Dyalog Unicode), 55 bytes^SBCS

Anonymous tacit prefix function.

⌊.5+100×2÷/(2 5)2{≢⍎⍕↓⍺↓¯2⌽' ',cal⍎' '@5⊢⍵⊣⎕CY'dfns'}¨⊂

⊂ enclose date to treat it as a whole

(2 5)2{…}¨ apply the following function on that, but with the left arguments [2,5] and 2:

 ⎕CY'dfns' copy the "dfns" library

 ⍵⊣ discard the report in favour of the date

 ' '@5⊢ replace the 5th character (-) with a space

 ⍎ execute that to get two-element list

 cal call the calendar function on that

 ' ', prepend a column of spaces to that

 ¯2⌽ rotate the last two columns (Saturday) to the front

 ⍺↓ drop left-argument number of rows (2, headers) and columns (if specified; 5=Sat+Sun)

 ↓ split matrix into list of lines

 ⍕ format (flattens with insertion of double-spacing)

 ⍎ execute (turns remaining day numbers into a flat numeric list)

 ≢ tally those

2÷/ divide each pair (there's only one)

100× multiply by a hundred

.5+ add a half

⌊ floor

Try it online!

Answer 12

Oracle SQL, 110 bytes

select round(100*sum(1-trunc(to_char(x+level-1,'d')/6))/sum(1))from dual,t connect by level<=add_months(x,1)-x

It works with an assumption that input data is stored in a table in using date data type, e.g.

with t as (select to_date('2010-06','yyyy-mm') x from dual)

Answer 13

Perl 6, 78 bytes

{round 100*@_.grep(6>*.day-of-week)/@_}o{Date.new("$_-01")...^*.month-*.month}

Try it online!

Answer 14

Rebol - 118 113

w: b: 0 o: d: do join input"-01"while[d/2 = o/2][if d/7 < 6[++ w]++ b d: o + b]print to-integer round w / b * 100

Ungolfed:

w: b: 0 
o: d: do join input "-01"
while [d/2 = o/2] [
    if d/7 < 6 [++ w]
    ++ b
    d: o + b
]
print to-integer round w / b * 100