Get the day of the year of a given date

Question

Challenge

Given two positive integers \$1 \le m \le 12\$ and \$1\le d \le 31\$, representing a month and days into the month, output the amount of days that have passed since January 1st, on a non-leap year. You can assume that the number of days passed will always be constrained by the number of days in the month (so \$m = 2, d = 31\$ will never be an input)

This is code-golf so the shortest code in bytes wins.

Examples

For example, \$m = 2, d = 11\$ represents February 11th. This is the 42nd day of the year (31 days in January + 11 days), so the output is 42.

Answer 1

Haskell, 27 bytes

m%d=div(275*m)9-30+d-mod 2m

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This is a “closed-form” answer (in a C-like language it would be 275*m/9-30+d-2%m).

Answer 2

JavaScript (ES6),  40 37  36 bytes

Expects (day)(month).

d=>g=m=>--m?31-(4/m&2)+~m%9%2+g(m):d

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How?

This is a recursive function that computes the number of days in each full month (i.e. lower than \$m\$), sums them all together and finally adds \$d\$ on the last iteration.

We use 4 / m & 2 to distinguish between February and all other months:

• For January, 4 / 1 & 2 is 4 & 2, which is \$0\$
• For February, 4 / 2 & 2 is 2 & 2, which is \$2\$
• For \$m>2\$, 4 / m & 2 is \$0\$ because \$0<4/m<2\$

We use ~m % 9 % 2 to subtract \$1\$ for months that do not have \$31\$ days:

    m |   1 |   2 |   3 |   4 |   5 |   6 |   7 |   8 |   9 |  10 |  11
   ~m |  -2 |  -3 |  -4 |  -5 |  -6 |  -7 |  -8 |  -9 | -10 | -11 | -12
mod 9 |  -2 |  -3 |  -4 |  -5 |  -6 |  -7 |  -8 |   0 |  -1 |  -2 |  -3
mod 2 |   0 |  -1 |   0 |  -1 |   0 |  -1 |   0 |   0 |  -1 |   0 |  -1

(This also works for December, but we never have to compute the total number of days in this month.)

Answer 3

Gaia, 6 bytes

A dyadic function taking the month above the day. Gaia has a nice collection of date/time builtins.

(∂k<Σ+

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(       # decrement the month
 ∂k     # push list [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] (builtin)
   <    # take the first month-1 elements from this list
    Σ   # sum them
     +  # add the day

Answer 4

JavaScript, 33 bytes

m=>d=>((m+9)%12*51+99)*.6%365+d|0

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Answer 5

JavaScript (Node.js), 35 32 bytes

m=>d=>--m*31+d-'003344555667'[m]

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Thanks to A username for pro golf tips!

Answer 6

C (gcc), ⁴⁷ 27 bytes

f(d,m){m=275*m/9-30+d-2%m;}

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Uses formula from Lynn's Haskell answer.

Answer 7

Bash, 16

date -d$1/1 +%-j

Input is given on the command-line as slash-separated integers, with month first.

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Answer 8

Excel formula, 18 Bytes

=DATE(1,A1,B1)-366

• Cell A1 contains the input month.
• Cell B1 contains the input day.

Saved 8 bytes from answer by Crissov by evaluating the offset as 366, then added 2 bytes by converting named ranges m and d into cell references.

If leap years were allowed, then it could be shortened to a single function =DATE(0,A1,B1)

Answer 9

Python 3, 60 55 51 bytes, 1-based month

lambda function that accepts the month (1-based) and the day.

-5 bytes: used a "31-days" default month, and the list s accounts for the cumulate (absolute) difference between the actual days of months and 31
-4 bytes: the list is removed from the function parameters, thanks to @Hunaphu and @mic_e

lambda m,d,a=[3,0,3,2,3,2,3,3,2,3,2]:28*(m-1)+sum(a[:m-1])+d  # original version

lambda m,d,s=[0,0,3,3,4,4,5,5,5,6,6,7]:31*m-31-s[m-1]+d

lambda m,d:31*m-31-[0,0,3,3,4,4,5,5,5,6,6,7][m-1]+d

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Explanation (original version):

• m,d: input month and day
• a[...]: list of days to add to the "base" 28-day month, from january to novemeber, passed as optional argument
• 28*(m-1): the total number of days in the previous "28-days" months
• +sum(a[:m-1]): add the remaining days of the previous months (as difference actual days - 28)
• +d: add the days of the selected month

---

Python 3, 56 50 46 bytes, 0-based month

-4 bytes (compared to 1-based month number): if 0-based month number is allowed as input
-6 bytes: used the "31-days" approach
-4 bytes: the list is removed from the function parameters, thanks to @Hunaphu and @mic_e

lambda m,d,a=[3,0,3,2,3,2,3,3,2,3,2]:28*(m)+sum(a[:m])+d  # original version

lambda m,d,s=[0,0,3,3,4,4,5,5,5,6,6,7]:31*m-s[m]+d

lambda m,d:31*m-[0,0,3,3,4,4,5,5,5,6,6,7][m]+d

Answer 10

Japt, 16 bytes

Well, this ain't right :\ Normally Japt would be ruling the roost in a date based challenge but not this time, despite all the tricks I can muster. Which makes me seriously worried that, after 16 long months, I've lost my edge when it comes to golfing on me phone over a few pints down the boozer!

ÒÐBì)nÐUi¹z864e5

Try it

Answer 11

Factor, 29 bytes

[ 1 -rot <date> day-of-year ]

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• 1 -rot <date> Create a timestamp from the year 1 and the two inputs. (1 is not a leap year according to Factor's calendar vocabulary.) This is the shortest way I know of to create a timestamp object from a month and day.
• day-of-year Factor has a builtin for this once you have a timestamp.

Answer 12

APL (Dyalog Unicode), 11 bytes

Full program. Prompts for [month,day]

1⎕DT⊂1900,⎕

Try it on TryAPL! (⎕ is stdin, but is emulated with the variable ∆ since TryAPL doesn't allow stdin)

⎕ prompt for [month,day]; [2,11]

1900, prepend 1900; [1900,2,11]

⊂ enclose to represent as scalar time stamp; [[1900,2,11]]

1⎕DT convert DateTime to days since 1899-12-31; 42

Answer 13

Red, 36 bytes

func[m d][pick to now reduce[d m]11]

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Answer 14

Vyxal s, 14 bytes

‹»HȦð9E»fẎ28+p

Try it Online! Port of @caird's Jelly solution

‹»HȦð9E»fẎ28+p    
 »HȦð9E»          Push compressed integer 303232332323
        f         Convert to digits 
‹        Ẏ        Slice [0:m-1]
          28+     Add 28 to each list element
             p    Prepend to the day of the month

s flag: Sums the top of the stack and outputs the sum.

Answer 15

Jelly, 16 15 bytes

“LɓịNH’D+28⁸;ḣS

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Takes \$d\$ then \$m\$ on the command line

“LɓịNH’D+28⁸;ḣS - Main link. Takes d on the left and m on the right
“LɓịNH’         - Compressed integer; 303232332323
       D        - Convert to digits; [3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3]
        +28     - Plus 28 to each; [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
           ⁸;   - Prepend d; [d, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
             ḣ  - Take the first m
              S - Sum

Answer 16

Pip, 24 bytes

b+$+A*""@<Da

The code contains unprintable characters. Here's a hexdump:

00000000: 622b 242b 412a 221f 1c1f 1e1f 1e1f 1f1e  b+$+A*".........
00000010: 1f1e 1f22 403c 4461                      ..."@<Da

If you put that into xxd -r, save the results in a file, and then run the file as Pip code with the two inputs as command-line arguments, you can try it here! Or, here's a 25-byte version in Pip Classic: Try it online!

Explanation

                          a (month) and b (day) are command-line arguments
      ""      String containing characters with codes 31, 28, 31, 30, etc.
    A*                    Get the ASCII code of each character
                      Da  Decrement a
                    @<    The first (^ that many) items of the list of charcodes
  $+                      Sum
b+                        Add b

Answer 17

Vyxal, 17 bytes

»∇ė{»4τ28+¦0p?‹i+

Try it Online! Works now, and -2.

Answer 18

Desmos, 90 82 59 bytes

f(m,d)=total([0,31,28,31,30,31,30,31,31,30,31,30][1...m])+d

Saved 23 bytes 'cause I was so dumb... I've been trying to do some clever list manipulation to save bytes, but it never came to my mind to just put the list itself.

Very Brief Explanation:

[0,31,28,31,30,31,30,31,31,30,31,30]: The number of days in each month(excluding December), with an extra 0 element at the beginning.

total( ... [1...m])+d: Sum of the first m elements of the list explained above, then add d.

Try It On Desmos!

Try It On Desmos! - Prettified

Solution using Lynn's "closed form" formula, 34 bytes

f(m,d)=floor(275m/9)-30+d-mod(2,m)

Answer 19

This works on my PowerShell 5 on my computer; I can't get it working on TIO:

PowerShell 5, 97 bytes

function s($m,$d){1+(New-Timespan -st (Get-Date -day 1 -mo 1) -e (Get-Date -day $d -mo $m)).Days}

Call as s 2 11 for the example date (month before day-of-month).

Golfed by @mazzy, 82 bytes

PowerShell, 82 bytes

param($m,$d)1+(New-Timespan -st (Date -day 1 -mo 1) -e (Date -day $d -mo $m)).Days

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The golfing relies on an alias or implementation of command (Date) that does not exist in a default Windows 10 installation of PowerShell 5. The TIO PowerShell is PowerShell 6 on Linux.

Answer 20

R, 57 bytes

(my own attempt)

function(m,d){F[c(8:14,2:8)]=30:31;F[3]=28;sum(F[1:m],d)}

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An alternative R solution to pajonk's answer, without using any date built-ins.

---

R, 34 bytes

(port of Lynn's answer)

function(m,d)(275*m)%/%9-30+d-2%%m

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Answer 21

K (ngn/k), 25 20 bytes

Solution:

{+/y,x#28+4\3390446}

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Explanation:

Assumes months are 0-indexed!

Naive approach, there are likely better ones out there.

{+/y,x#28+4\3390446} / the solution
{                  } / lambda taking implicit x, y args
          4\3390446  / creates 3 0 3 2 3 2 3 3 2 3 2 from base-4           
       28+           / add 28 to each item (vectorised)
     x#              / take 'month' items from this list (January = 0)
   y,                / prepend 'days'
 +/                  / sum up

Edits:

• -5 bytes; using deltas against 28 instead of days-per-month

Extra:

• +2 bytes for 1-indexed month {+/y,x#0,28+4\3390446}

Answer 22

Charcoal, 20 bytes

Ｉ⁺↨Ｅ…”)“M⟧₂”⊖Ｎ⁻³¹ι¹Ｎ

Try it online! Link is to verbose version of code. Explanation:

     ”)“M⟧₂”            Compressed string `03010100101`
    …                    Truncated to length
             Ｎ          Month as an integer
            ⊖           Decremented
   Ｅ                     Map over characters
                ³¹       Literal integer `31`
               ⁻         Subtract
                  ι      Current value
  ↨                ¹     Take the sum
 ⁺                       Plus
                    Ｎ    Day as an integer
Ｉ                        Cast to string
                         Implicitly print

Base 1 conversion is used in case the list is empty (i.e. January).

Answer 23

Perl 5 (-p), 36 bytes

/ /;$_=strftime"%-j",(0)x3,$',$`-1,0

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Answer 24

MATL, 9 bytes

14Liq:)hs

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Explanation

14L % Push [31 28 31 30 31 30 31 31 30 31 30 31]: month lengths (predefined literal)
i   % Input: month, m
q   % Subtract 1
:   % Inclusive range from 1 to that
)   % Index into the array of month lengths: gives its first m-1 terms
h   % Implicit input: day, d. Concatenate with previous array
s   % Sum of array. Implicit display

Answer 25

R, 40 bytes

function(m,d)format(ISOdate(1,m,d),"%j")

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Using buildins, returning as string.

---

Returning as a number (also using buildins):

R, 44 bytes

function(m,d)as.POSIXlt(ISOdate(1,m,d))$yd+1

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Answer 26

PHP, 42 bytes

fn($m,$d)=>date(z,mktime(0,0,0,$m,$d,1))+1

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Pretty much builtin. The year parameter could be omitted, but the current year would then be used. To avoid a leap year 1 was used (it actually corresponds to 2001 according to the doc). 1 is added because the result is zero indexed.

This is satisfying, the byte count is The Answer, which is also the answer to the test case!

1 byte shorter but less satisfying:

PHP, 41 bytes

fn($m,$d)=>date(z,strtotime("1-$m-$d"))+1

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Answer 27

Python 3, 41 bytes

lambda m,d:31*~-m-(539785049600>>3*m&7)+d

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Python 3, 41 chars, 50 bytes

lambda m,d:d+ord('0\x00>v´ðĮŪƨǦȢɠʜ'[m])/2

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Answer 28

Excel, 30 24 bytes

• =DATE(1,m,d)-DATE(1,1,0) [24]
• =DAYS(DATE(1,m,d),DATE(1,1,0)) [30]

Edit: I used the year 2001 before, because I do not trust Excel with dates before 1905.

Answer 29

Mathematica, 30 bytes

{1}~DateDifference~{1,#,#2+1}&

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Returns as a quantity object:

Using the DateDifference builtin and the fact that {y} is treated as January 1st on year y. I thought of {1}~DateDifference~{##}& at first, but sadly DateDifference starts from 0 days instead of 1.

Answer 30

Java, 141 bytes

int d(int d,int m){for(int i=1;i<m;i++){switch(i){case(2):d+=28;break;case(4):case(6):case(9):case(11):d+=30;break;default:d+=31;}}return d;}