How many days in a month?

Question

Given a textual representation (case-insensitive full name or 3 character abbreviation) of a month return the number of days in the month.

For example, december, DEC, and dec should all return 31.

February can have either 28 or 29 days.

Assume the input is a month in one of the correct forms.

Answer 1

Pyke, 9 bytes

l4C9@~%R@

Try it here!

l4        -   input.title()
    @     -  v.index(^)
  C9      -   ['PADDING', 'January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December']
        @ - v[^]
     ~%R  -  ['Padding', 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

Or 15 bytes if all input formats are required

l43<C9 3L<@~%R@

Try it here!

l43<            -   input.title()[:3]
          @     -  v.index(^)
    C9 3L<      -   ['PAD', 'Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']
              @ - v[^]
           ~%R  -  ['Padding', 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

Answer 2

JavaScript (ES6),  48 47 44 43  42 bytes

m=>31^'311'[parseInt(m[1]+m[2],34)*3%49%8]

Demo

let f =

m=>31^'311'[parseInt(m[1]+m[2],34)*3%49%8]

;(
  'JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC,' +
  'JANUARY,FEBRUARY,MARCH,APRIL,MAY,JUNE,JULY,AUGUST,SEPTEMBER,OCTOBER,NOVEMBER,DECEMBER'
).split`,`
.forEach(m => console.log(m, '->', f(m)))

How?

These operations lead to a lookup table of 8 entries, which would not be very interesting if the values were randomly distributed. But any result greater than 2 is mapped to 31 days. Therefore, only the first 3 entries need to be stored explicitly.

Month | [1:2] | Base 34 -> dec. | * 3  | % 49 | % 8 | Days
------+-------+-----------------+------+------+-----+-----
  JAN |    AN |             363 | 1089 |   11 |   3 |  31
  FEB |    EB |             487 | 1461 |   40 |   0 |  28
  MAR |    AR |             367 | 1101 |   23 |   7 |  31
  APR |    PR |             877 | 2631 |   34 |   2 |  30
  MAY |    AY |              10 |   30 |   30 |   6 |  31
  JUN |    UN |            1043 | 3129 |   42 |   2 |  30
  JUL |    UL |            1041 | 3123 |   36 |   4 |  31
  AUG |    UG |            1036 | 3108 |   21 |   5 |  31
  SEP |    EP |             501 | 1503 |   33 |   1 |  30
  OCT |    CT |             437 | 1311 |   37 |   5 |  31
  NOV |    OV |             847 | 2541 |   42 |   2 |  30
  DEC |    EC |             488 | 1464 |   43 |   3 |  31

Answer 3

Javascript (ES6), 36 33 bytes

-3 bytes thanks to @JustinMariner and @Neil

m=>31-new Date(m+31).getDate()%31

Sorry @Arnauld, abusing JavaScript weirdness is shorter than your fancy base conversions.

How it works

For some reason, JavaScript allows entering dates outside of the specified month. The code counts how many days outside the month the date is to determine how many days there are in the month. Examples:
"FEB31" → Thu Mar 02 2000 → 31 - 2 % 31 → 29
"October31" → Tue Oct 31 2000 → 31 - 31 % 31 → 31

Test cases

f=m=>31-new Date(m+31).getDate()%31
;(
 "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC,"+
 "January,February,Mars,April,May,June,July,August,September,October,November,December"
).split`,`.forEach(s=>console.log(s,"->",f(s)))

Answer 4

Python 2, 46 45 38 bytes

-1 byte thanks to @totallyhuman

lambda m:29-int(m[1:3],35)%238%36%-5/2

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Answer 5

Bash, 21 bytes

cal $1|xargs|tail -c3

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Takes input as command-line argument and outputs with a trailing newline. The day count for February depends on that of the current year

Requires the util-linux 2.29 version of cal, which is the one available on TIO. Also is locale-dependent, so LC_TIME must be changed on non-English systems (thanks @Dennis for clarification).

Idea of piping through xargs to trim cal's output is from this SO answer.

Answer 6

Proton, 50 bytes

k=>31-((e=k.lower()[1to3])in"eprunov")-3*(e=="eb")

Try it online!

-14 bytes thanks to Jonathan Frech

Thirty days hath September, April, June, and November. All the rest had peanut butter. All except my grandmother; she had a little red trike, but I stole it. muahahahahaha

(I've been waiting to tell that joke (source: my math professor) for ages on this site :D :D :D)

Answer 7

C# (.NET Core), 52+13=65 38+24=62 bytes

m=>D.DaysInMonth(1,D.Parse(1+m).Month)

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+24 for using D=System.DateTime;

Acknowledgements

-3 bytes thanks to Grzegorz Puławski.

Answer 8

Python 3, 60 bytes

x=input().lower()[1:3];print(31-(x in"eprunov")-3*(x=="eb"))

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Porting my Proton solution

-10 bytes thanks to totallyhuman

Answer 9

Shell/GNU Date, 39, 26 bytes

date -d1$1+1month-1day +%d

Where $1 is the name of the month.

Try it online!

edit: Thanks Dennis for saving many bytes!

Answer 10

AWK, 45 44 bytes

L=tolower($1){$0=L~/v|p|un/?30:L~/f/?28:31}1

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Answer 11

Python 3 - 93 86 84 82 bytes

Variants of answer (showing the progression of time, and bytes for each, with TIO links):

Original Answer (93 bytes)

-7 bytes thanks to Jonathan Frech. (86 bytes)

-2 more bytes thanks to my own further testing of the monthrange results, with the second value always being the higher value. (84 bytes) ¹

-2 more by using import calendar as c and referencing it with c.monthrange. (82 bytes, current revision)

---

lambda x:c.monthrange(1,time.strptime(x[:3],'%b')[1])[1];import time,calendar as c

---

Obviously not as nice as HyperNeutrino's answer which doesn't use built-ins, but this still works.

---

Footnotes

¹: Test cases via TIO.run showing the proof for how I'm handling those monthrange values, for a varying number of month test cases.

Answer 12

Haskell, 65 63 62 bytes

f.map((`mod`32).fromEnum)
f(_:b:c:_)|c<3=28|c>13,b>3=30
f _=31

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Pattern matching approach. The first line is to handle the case-insensitivity. Then we return 28 if the third letter is smaller than C (number 3), 30 if the second letter is larger than C and the third one larger than M, or 31 otherwise.

Edit: -1 byte thanks to Leo

---

Alternative (65 64 bytes)

f s|let i#n=n<mod(fromEnum$s!!i)32=sum$29:[2|2#2]++[-1|2#13,1#3]

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Answer 13

Perl 5, 47 + 1 (-p) = 48 bytes

$_=substr$_,1,2;$_=31-("eprunov"=~/$_/i)-3*/b/i

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Answer 14

Mediawiki Template, 19 bytes

{{#time:t|{{{1}}}}}

Answer 15

APL (Dyalog), 32 bytes*

Tacit prefix function. Assumes ⎕IO (Index Origin) 0, which is default on many systems.

31 28 30⊃⍨∘⊃'.p|un|no|f'⎕S 1⍠1

Try it online!

⍠1 case insensitively

1 return the length of the

⎕S PCRE Search for

'.p|un|no|f' any-char,"p" or "un" or "no" or "f"

⊃⍨∘⊃ and use the first element of that (0 if none) to pick from

31 28 30 this list

Thus:

• Apr, Sep, Jun, and Nov will select the number at index 2, namely 30

• Feb will select the number at index 1, namely 28

• anything else will select the number at index 0, namely 31

---

* Using Classic and counting ⍠ as ⎕OPT.

Answer 16

MATL, 22 bytes

14L22Y2c3:Z)Z{kj3:)km)

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Explanation

14L    % Push numeric array of month lengths: [31 28 ... 31]
22Y2   % Push cell array of strings with month names: {'January', ..., 'December'}
c      % Convert to 2D char array, right-padding with spaces
3:Z)   % Keep first 3 columns
Z{     % Split into cell array of strings, one each row
k      % Convert to lower case
j      % Input string
3:)    % Keep first 3 characcters
k      % Convert to lower case
m      % Ismember: gives a logical index with one match
)      % Use that as index into array of month lengths. Implicit display

Answer 17

Retina, 32 31 28 bytes

i`f
28
i`p|v|un
30
\D

^$
31

Try it online! Edit: Saved 1 byte thanks to @RobertBenson. Saved 3 bytes thanks to @ovs.

Answer 18

Wolfram Language (Mathematica), 46 30 bytes

#~NextDate~"Month"~DayCount~#&

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Will give either 28 or 29 for February depending on whether the current year is a leap year.

How it works

All date commands in Mathematica will interpret input such April, APR, ApRiL, and so on as the first day of the corresponding month in the current year. (As a bonus, input such as "February 2016" or {2016,2} also works as expected.)

#~NextDate~"Month" gives the first day of the month after that, and DayCount gives the number of days between its two arguments. The number of days between April 1st and May 1st is 30, the number of days in April.

Answer 19

Java 8, 47 bytes

m->31-new java.util.Date(m+"31 1").getDate()%31

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Ended up using the same idea as Herman Lauenstein's JS answer, where setting the date to the 31st pushed into the next month. Java does require a year, so that has been set to 1.

Answer 20

PHP, 38 33+1 32+1 bytes

Saved 5 bytes thanks to Titus

<?=date(t,strtotime("$argn 1"));

Run as pipe with -nF

Try it online!

Answer 21

q/kdb+, 36 bytes

Solution:

28 30 31@2^1&(*)"ebeprunov"ss(_)1_3#

Examples:

q)28 30 31@2^1&(*)"ebeprunov"ss(_)1_3#"January"
31
q)28 30 31@2^1&(*)"ebeprunov"ss(_)1_3#"FEB"
28
q)28 30 31@2^1&(*)"ebeprunov"ss(_)1_3#"jun"
30

Explanation:

There are a million ways to skin a cat. I think is slightly different to the others. Take the 2nd and 3rd letters of the input, lowercase them, then look them up in the string "ebeprunov". If they are at location 0, then this is February, if they are at a location >0 they are a 30-dayer, if they are not in the string, they are a 31-dayer.

28 30 31@2^1&first"ebeprunov"ss lower 1_3# / ungolfed solution
                                        3# / take first 3 items from list, January => Jan
                                      1_   / drop the first item from the list, Jan => an
                                lower      / lower-case, an => an
                  "ebeprunov"ss            / string-search in "ebeprunov", an => ,0N (enlisted null)
             first                         / take the first, ,0N => 0N
           1&                              / take max (&) with 1, 0N => 0N
         2^                                / fill nulls with 2, 0N => 2
        @                                  / index into
28 30 31                                   / list 28,30,31

Answer 22

Excel VBA, ₄₇ 43 Bytes

Anonymous VBE immediate window function that takes input, as month name, abbreviation, or number, from range [A1] and outputs the length of that month in the year 2001 to the VBE immediate window function.

?31-Day(DateValue("1 "&[A1]&" 1")+30)Mod 31

Old Version

d=DateValue(["1 "&A1&" 1"]):?DateAdd("m",1,d)-d

Answer 23

QBIC, 49 35 bytes

?31-(instr(@aprjunsepnov feb`,;)%3)

Significantly shorter with some trickery.

Explanation

?                          PRINT
31-(                       31 minus
  instr(                   the position of
                      ,;   our input string
    @aprjunsepnov feb`  )  in the string cntaining all non-31 months                                
    %3)                    modulo 3 (this yields a 1 for each month except feb=2)

Answer 24

Java (OpenJDK 8), 126 bytes

s->{for(java.time.Month m:java.time.Month.values())if(m.name().startsWith(s.toUpperCase()))System.out.print(m.length(false));}

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Answer 25

Perl 5, 24 bytes

23 bytes code + 1 for -p.

$_=31-/p|un|no/i-/f/i*3

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Answer 26

Ruby, 45 bytes

->m{((Date.parse(m)>>1)-1).day}
require'date'

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Ruby's Date.parse accepts a month name on its own. What would normally be a right-shift (>>) actually adds to the month of the Date object. Subtraction affects the day of the month, which will wrap backwards to the last day of the previous month.

Answer 27

Kotlin, 92 bytes

val d={m:String->arrayOf(0,31,30,30,31,30,31,28,31,0,30)[(m[1].toInt()+m[2].toInt())%32%11]}

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