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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.
m=>31^'311'[parseInt(m[1]+m[2],34)*3%49%8]
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)))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
.substr(0,3) was not required. So, on second thought, this may not be the best approach.
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-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.
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
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)))
31. It seems to work in Chrome as new Date("feb31") for example.
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Oct 30, 2017 at 7:56
+31 saving three bytes overall. None of this works in Firefox though.
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k=>31-((e=k.lower()[1to3])in"eprunov")-3*(e=="eb")
-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)
'sepaprjunnov' instead of a list of strings.
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Oct 28, 2017 at 17:55
cal $1|xargs|tail -c3
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.
-2 thanks to ngn
Tacit prefix function. Assumes ⎕IO (Index Origin) 0, which is default on many systems.
32-1⌈∘⊃'.p|un|no|feb'⎕S 1⍠1
1⌈∘⊃ and use the maximum of one and the first element of that (0 if none) to pick from
32- thirty-two minus
1⌈∘ the maximum of one and
⊃ the first element (0 if none) of
'.p|un|no|f'⎕S the PCRE *Search for any-char,"p" or "un" or "no" or "f"
1 returning the length of the matches
⍠1 and done case insensitively
Apr, Sep, Jun, and Nov will give 32−2=30
Feb will give 32−3=29 (allowed by OP)
anything else will give 32−1=31
l4C9@~%R@
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]
l43<C9 3L<@~%R@
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]
Apr, Jun, Sep, and Nov) but also think that it requires a little clarification in the OP (see my question).
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Oct 29, 2017 at 11:20
m=>D.DaysInMonth(1,D.Parse(1+m).Month)
+24 for using D=System.DateTime;
-3 bytes thanks to Grzegorz Puławski.
using System;? Or can you excluse that from the byte count?
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using D=System.DateTime; and m=>D.DaysInMonth(1,D.Parse(1+m).Month) like here: tio.run/##jc5BSwMxEAXgs/…
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Nov 10, 2017 at 9:58
x=input().lower()[1:3];print(31-(x in"eprunov")-3*(x=="eb"))
Porting my Proton solution
-10 bytes thanks to totallyhuman
date -d1$1+1month-1day +%d
Where $1 is the name of the month.
edit: Thanks Dennis for saving many bytes!
($_)=/.(..)/; instead of $_=substr$_,1,2; and () around "eprunov"=~/$_/i can be removed.
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Nov 3, 2017 at 8:33
f.map((`mod`32).fromEnum)
f(_:b:c:_)|c<3=28|c>13,b>3=30
f _=31
Pattern matching approach. The first line is to handle the case-insensitivity. Then we return 28 if the third letter is smaller than C (number 3), 30 if the second letter is larger than C and the third one larger than M, or 31 otherwise.
Edit: -1 byte thanks to Leo
f s|let i#n=n<mod(fromEnum$s!!i)32=sum$29:[2|2#2]++[-1|2#13,1#3]
c<3 instead of a==6 (February is the first month if you order them by their third letter, followed by December)
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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
{{#time:t|{{{1}}}}}
14L22Y2c3:Z)Z{kj3:)km)
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
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 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.
import ...,calendar as c so as not having to type 'calendar' twice.
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Oct 28, 2017 at 19:38
#~NextDate~"Month"~DayCount~#&
Will give either 28 or 29 for February depending on whether the current year is a leap year.
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.
m->31-new java.util.Date(m+"31 1").getDate()%31
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.
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.
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
d=DateValue(["1 "&A1&" 1"]):?DateAdd("m",1,d)-d
Saved 5 bytes thanks to Titus
<?=date(t,strtotime("$argn 1"));
Run as pipe with -nF
.' 1', it seems to work on TIO without it!
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Nov 2, 2017 at 16:51
<?=date(t,strtotime($argn)); (run as pipe with -nF)
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.' 1', but it wasn't working. After seeing your comment, I tried to figure out what I had done wrong. Because I was running it on the 31st of the month, it was taking the 31st (current) day for any month I put in, which would put it beyond the current month. Feb 31st turns into March 3rd, so the code returns 31 (the number of days in March). Because of this, every month was returning 31. So, it works without the .' 1' on any day <= 28th of the month.
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't' -> t. Also, I had to do a bunch of searching to figure out how to "run as pipe with -nF" but I got it figured out (I think). :)
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s->{for(java.time.Month m:java.time.Month.values())if(m.name().startsWith(s.toUpperCase()))System.out.print(m.length(false));}
false to a boolean expression like 1<0 to save a couple bytes.
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Oct 31, 2017 at 0:53
?31-(instr(@aprjunsepnov feb`,;)%3)
Significantly shorter with some trickery.
? 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)
->m{((Date.parse(m)>>1)-1).day}
require'date'
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.
val d={m:String->arrayOf(0,31,30,30,31,30,31,28,31,0,30)[(m[1].toInt()+m[2].toInt())%32%11]}
(z,v=n=e=b=r=y=u=c=0,t=s=a=l=1,f=p=29,m=j=g=o=30,d=31)=>eval([z[0],z[1],z[2]].join`+`.toLowerCase())
Uses Herman's method.
´H-ÐU+H f uH
Try it (includes all test cases, footer resets the H constant for the next case)
december,DEC, anddecshould all return 31" - Is that the intention? \$\endgroup\$