Which Card is it?

Question

Introduction

Long ago, when I used to code card games with usual playing cards, I used to specify a number for each card and call a function with some number to get a card. This somewhat inspired me to make this challenge.

So for the people unaware of the playing cards, a deck of cards consist of 52 cards (13 in each of the four suits, i.e, Hearts, Diamonds, Spades, Clubs). In each suit, there are 13 cards - firstly the cards numbered from 2-10, then the Jack(J), Queen(Q), King(K) and the Ace(A). This is the order

Challenge

The challenge is to take an integer between 1-52 as input and display the card at that position. But, your output must be in words. Also, order must be maintained, i.e, first 13 cards will be of Hearts, then Diamonds, then Spades and finally Clubs.

For example, if someone chooses the number 30.The card would then belong to the third suit, i.e, the Spades. Also, it would be the fourth card in the suit, which means the number 5. Hence your output in words must be: five of spades and it should always follow this format, i.e, first the card, followed by an of and the name of the suit at the end, with required spaces in between.

Input And Output

The input will be an integer between 1-52 (both inclusive). Note that here counting starts from 1. You may choose to start from 0. However, you must maintain the order of the cards which is mentioned above. Your output should be the card at that position written in words. You do not need to handle invalid inputs. Also, your output may be in lower-case or in upper-case.

Given below is the list of all the possible inputs and their outputs:

1 -> two of hearts
2 -> three of hearts
3 -> four of hearts
4 -> five of hearts
5 -> six of hearts
6 -> seven of hearts
7 -> eight of hearts
8 -> nine of hearts
9 -> ten of hearts
10 -> jack of hearts
11 -> queen of hearts
12 -> king of hearts
13 -> ace of hearts
14 -> two of diamonds
15 -> three of diamonds
16 -> four of diamonds
17 -> five of diamonds
18 -> six of diamonds
19 -> seven of diamonds
20 -> eight of diamonds
21 -> nine of diamonds
22 -> ten of diamonds
23 -> jack of diamonds
24 -> queen of diamonds
25 -> king of diamonds
26 -> ace of diamonds
27 -> two of spades
28 -> three of spades
29 -> four of spades
30 -> five of spades
31 -> six of spades
32 -> seven of spades
33 -> eight of spades
34 -> nine of spades
35 -> ten of spades
36 -> jack of spades
37 -> queen of spades
38 -> king of spades
39 -> ace of spades
40 -> two of clubs
41 -> three of clubs
42 -> four of clubs
43 -> five of clubs
44 -> six of clubs
45 -> seven of clubs
46 -> eight of clubs
47 -> nine of clubs
48 -> ten of clubs
49 -> jack of clubs
50 -> queen of clubs
51 -> king of clubs
52 -> ace of clubs

Scoring

This is code-golf, so the shortest code wins.

Answer 1

Python 3,  115  90 bytes

from unicodedata import*
lambda n:name(chr(n%13+n%13//11+[6,0,4,2][-n//13]*8+127137))[13:]

An unnamed function returning the string in uppercase.

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How?

Unicode characters have names. The names of some of these are like "PLAYING CARD TWO OF SPADES", hence we can get the characters of the Unicode character representing the required card and strip off the first 13 characters to get our output.

The Unicode characters of interest are within a block like so:

            0   1   2   3   4   5   6   7   8   9   A   B   C   D   E   F
U+1F0Ax     x   As  2s  3s  4s  5s  6s  7s  8s  9s  Ts  Js  x   Qs  Ks  x
U+1F0Bx     x   Ah  2h  3h  4h  5h  6h  7h  8h  9h  Th  Jh  x   Qh  Kh  x
U+1F0Cx     x   Ad  2d  3d  4d  5d  6d  7d  8d  9d  Td  Jd  x   Qd  Kd  x
U+1F0Dx     x   Ac  2c  3c  4c  5c  6c  7c  8c  9c  Tc  Jc  x   Qc  Kc  x                           

Where the x are not characters we are after (the four in the C column are "knights"; three in F are "jokers"; one in 0 is generic; the rest are reserved characters).

As such we can add some value to 0x1F0A1 = 127137 (As) to find the card we want.

The value to add is only complicated by three things:

1. We need to reorder the suits (from s,h,d,c to h,d,s,c)
2. We need to reorder the ranks from (A,2,...,K to 2,...,K,A)
3. We need to avoid the columns without cards of interest.

Using the one-indexing option allows the use of negative integer division to index into an array of row-wise offsets for the suit re-ordering with [6,0,4,2][-n//13]*8+ (effectively [48,0,32,16][-n//13]), we can then also place the aces into the correct locations with n%13+ and then avoid the knights in column C with n%13//11+ (effectively (n%13>10)+).

Answer 2

Perl6/Rakudo 70 bytes

Index 0

Using perl6 -pe, and with no dictionary compression:

chr('🂱🃁🂡🃑'.ords[$_/13]+($_+1)%13*1.091).uniname.substr(13)

It just looks up the card in Unicode (starting from the Ace), asks for the name and uses that. This is a similar route (though I didn't know it at the time!) to Jonathan Aitken's Python answer - only I index from all 4 aces rather than 4 offsets from the Ace of Spades, and I multiply by 1.091 to make the index round away from the Knight entry in Unicode.

See all the output (for input values 0 to 51) https://glot.io/snippets/ez5v2gkx83

Edited to cope with Knights in the Unicode deck, because Unicode.

Perl6 ♥ Unicode

Answer 3

05AB1E, 54 bytes

0-indexed

“»€Å‹¡Šdesž…“#“‚•„í†ìˆÈŒšï¿Ÿ¯¥Š—¿—ÉŸÄ‹ŒÁà“#âí" of "ýsè

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Explanation

“»€Å‹¡Šdesž…“#                                          # push list of suits
              “‚•„í†ìˆÈŒšï¿Ÿ¯¥Š—¿—ÉŸÄ‹ŒÁà“#             # push list of ranks
                                           â            # cartesian product
                                            í           # reverse each
                                             " of "ý    # join on " of "
                                                    sè  # index into cardlist with input

Answer 4

Emojicode, 202 bytes

🍇i🚂😀🍪🍺🐽🔫🔤two.three.four.five.six.seven.eight.nine.ten.jack.queen.king.ace🔤🔤.🔤🚮i 13🔤 of 🔤🍺🐽🔫🔤hearts.diamonds.spades.clubs🔤🔤.🔤➗i 13🍪🍉

0 indexed. Try it online!

Explanation:

🍇		start of the closure block
  i🚂		 closure takes an integer argument i
  😀		 print:
    🍪		  concatenate these strings:
      🍺🐽🔫🔤...🔤🔤.🔤🚮i 13  [a]
      🔤 of 🔤
      🍺🐽🔫🔤...🔤🔤.🔤➗i 13  [b]
    🍪
🍉

[a]:
🍺		tell Emojicode to dereference without checking
🐽		 get the nth element of the following array
  🔫		  create an array using the following string and separator
    🔤...🔤
    🔤.🔤
  🚮 i 13	n, i mod 13

[b]
🍺🐽🔫🔤...🔤🔤.🔤➗i 13
same but with ⌊i÷13⌋

Answer 5

Python 2, 167 148 bytes

n=input();print 'two three four five six seven eight nine ten jack queen king ace'.split()[n%13]+' of '+['hearts','diamonds','spades','clubs'][n/13]

Zero-indexed.

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EDIT: Bubbler made a great point using the split method (and providing a shorter answer). On the second block using split() yields the same byte count.

Answer 6

R, 154 bytes

paste(el(strsplit("Two,Three,Four,Five,Six,Seven,Eight,Nine,Ten,Jack,Queen,King,Ace",",")),"of",rep(c("Hearts","Diamonds","Spades","Clubs"),e=13))[scan()]

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Takes input (1-indexed) from STDIN and with source(...,echo=T) will print the result to console.

It's not pretty, BUT it comes in 2 bytes shorter than the best solution I could using outer (presented below), so let this be a reminder to examine another approach!

paste(                          # concatenate together, separating by spaces,
                                # and recycling each arg to match the length of the longest
el(strsplit("Two,...",",")),    # split on commas and take the first element
"of",                           # 
 rep(c("Hearts",...),           # replicate the suits (shorter as a vector than using strsplit
               e=13)            # each 13 times
                    )[scan()]   # and take the input'th index.

R, 156 bytes

outer(el(strsplit("Two,Three,Four,Five,Six,Seven,Eight,Nine,Ten,Jack,Queen,King,Ace",",")),c("Hearts","Diamonds","Spades","Clubs"),paste,sep=" of ")[scan()]

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Essentially the same as above; however, outer will do the recycling properly, but having to set sep=" of " for the paste made this just a hair longer.

Answer 7

Excel, 156 bytes

=TRIM(MID("two  threefour five six  seveneightnine ten  jack queenking ace",1+MOD(A1,13)*5,5))&" of "&CHOOSE(1+(A1/13),"hearts","diamonds","spades","clubs")

Cards from 0-51. Unfortunately, Excel does not feature a function to convert 1 to "one"...

Using TRIM and MID is shorter than using CHOOSE for the face values, but longer than using CHOOSE for the Suit.

Answer 8

Java 8, 141 bytes

n->"two;three;four;five;six;seven;eight;nine;ten;jack;queen;king;ace".split(";")[n%13]+" of "+"hearts;diamonds;spades;clubs".split(";")[n/13]

Input is 0-indexed.

Explanation:

Try it online.

n->         // Method with integer parameter and String return-type
  "two;three;four;five;six;seven;eight;nine;ten;jack;queen;king;ace".split(";")[n%13]
            //  Take `n` modulo-13 as 0-indexed card value
   +" of "  //  append " of "
   +"hearts;diamonds;spades;clubs".split(";")[n/13]
            //  append `n` integer-divided by 13 as 0-indexed suit

Answer 9

Husk, 52 bytes

;↔!Πmw¶¨×‼sÿẋδẎ₆ṡ⁷Ḃ6‰fωθ»&⌈θƒV₆x⁵▼Ëġ`nıEṅ'jĊk⁸"eïkÄc

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I'm always happy to show off Husk's string compression system :D

Explanation

The majority of the program (from ¨ onwards) is obviously a compressed string. When uncompressed it turns into:

hearts diamonds spades clubs
of
two three four five six seven eight nine ten jack queen king ace

The program then is:

;↔!Πmw¶¨…
       ¨…    The previous string
      ¶      Split on lines
    mw       Split each line into words
             - we now have a list of lists of words
   Π         Cartesian product of the three lists
             - with this we obtain all possible combinations of suits and values
               with "of" between the two (e.g. ["spades","of","king"])
  !          Pick the combination at the index corresponding to the input
 ↔           Reverse it, so words are in the correct order
;            Wrap it in a list. Result: [["king","of","spades"]]

There are a couple of things left to explain:

• We build the cards with suits before values because of how the cartesian product Π works: if we did it the other way around, the list of cards would be ordered by value (i.e. two of hearts, two of diamonds, two of spades, two of clubs, three of hearts...). As a consequence, we have to reverse our result.

• The result of the program is a two-dimensional matrix of strings. This is automatically printed by Husk as a single string built by joining rows of the matrix with newlines and cells with spaces. The reason we build this matrix instead of using the more straightforward w (join a list of words with spaces) is that if using w the type inferencer guesses another interpretation for the program, producing a different result.

Answer 10

Kotlin, 154 152 140 bytes

i->"two,three,four,five,six,seven,eight,nine,ten,jack,queen,king,ace".split(',')[i%13]+" of ${"heart,diamond,spade,club".split(',')[i/13]}s"

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Updated to use just lambda expression.

Answer 11

JavaScript ES6, 124 118 Bytes, 0-index

F= x=>(h=btoa`O
?NÞ{ñhº¿Å÷¿J,IëÞñ"6)Þý7§üô.yéÿ*)àüÿÿÿæ«·÷bjj'wû)i׿r[`.split`/`)[x%13]+` of ${h[x/13|16]}s`

console.log (F(51))

Base64 version

eD0+KGg9YnRvYWBPCj9OGt578Wi6v8WK979KLH9J696f8SKCG382Kd79N6f8lpyT9C556f8qKeD8Bx7///+F5qu392Jqaid3+ylp179yW5tgLnNwbGl0YC9gKVt4JTEzXStgIG9mICR7aFt4LzEzfDE2XX1zYA==

Answer 12

Stax, 58 57 56 bytes

î↑à■?R╢8«E▄¡╔ÿ•L╫<<⌠ï∞∟⌡♪Ös1"TàLα╥▀¢¡◄└%≈δñM;;}'░o=⌡»╬í√

Run and debug it

Here's the commented ungolfed representation of the same program. It uses stax's compressed literals heavily. The input is 0-indexed. It's Emigna's 05AB1E algorithm.

`SsUI'S~pTU5T@T^Ez+`j   suits
`fV:l7eTkQtL*L2!CZb6u[&YNO4>cNHn;9(`j   ranks
|*  cross-product
@   index with input
r   reverse pair
`%+(`*  join with " of "

Run this one

Answer 13

mIRCScript, 157 bytes

c echo $token(ace two three four five six seven eight nine ten jack queen king,$calc(1+$1% 13),32) of $token(clubs spades diamonds hearts,$calc(-$1// 13),32)

Load as an alias, then use: /c N. mIRC is 1-indexed, so floor division (//) on the negative value of the input produces -1 to -4 as required.

Answer 14

C (gcc), 148 bytes

f(n){printf("%.5s of %.7ss","two\0 threefour\0five\0six\0 seveneightnine\0ten\0 jack\0queenking\0ace"+n%13*5,"heart\0 diamondspade\0 club"+n/13*7);}

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0-based.

Answer 15

Haskell, 132 bytes

(!!)[v++" of "++s|s<-words"hearts diamonds spades clubs",v<-words"two three four five six seven eight nine ten jack queen king ace"]

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An anonymous function, using list comprehension to build all the combinations of suit and value, and indexing into the resulting list with the input.

Answer 16

F#, 174 168 bytes

Removed some extra whitespace as noted by Manish Kundu. Thanks!

let c x=["two";"three";"four";"five";"six";"seven";"eight";"nine";"ten";"jack";"queen";"king";"ace"].[(x-1)%13]+" of "+["hearts";"diamonds";"spades";"clubs"].[(x-1)/13]

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I'll be honest - I'm new at code golf, so I don't know if it's more appropriate to answer with a pure function like this (with parameters, but no I/O) or with a working code block with user I/O.

Answer 17

Octave, 155 153 151 150 bytes

@(x)[strsplit(' of ,s,heart,diamond,spade,club,ace,two,three,four,five,six,seven,eight,nine,ten,jack,queen,king',','){[mod(x,13)+7,1,ceil(2+x/13),2]}]

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This creates a string starting with ' of ' and 's', then all the suits followed by all the ranks. This string is split at commas into separate strings. The suits are before the ranks, because since that saves a byte when creating the indices. After this, we index it using square brackets with the following indices:

{[mod(x,13)+7,1,ceil(2+x/13),2]}

which is the rank, followed by the first element ' of ', followed by the suit, followed by 's'.

Having the 's' as part of the suits (hearts,diamonds,spades,clubs) instead of a separate string is the exact same length but less fun.

Splitting on the default separator would save 4 bytes in the strsplit-call, but the spaces around ' of ' would be removed and would have to be added manually, costing more bytes.

Answer 18

V, 154 147 144 142 Bytes

-7 Bytes thanks to DJMcMayhem

13i1heart
2diamond
3spade
4club
ÚGxCtwo
three
four
five
six
seven
eight
nine
ten
jack
queen
king
aceH$A of 012j$d4ñ13jPñÍ «/ 
{ÀjYHVGpAs

Try it online!

Hexdump:

00000000: 3133 6931 6865 6172 740a 3264 6961 6d6f  13i1heart.2diamo
00000010: 6e64 0a33 7370 6164 650a 3463 6c75 620a  nd.3spade.4club.
00000020: 1bda 1647 7843 7477 6f0a 7468 7265 650a  ...GxCtwo.three.
00000030: 666f 7572 0a66 6976 650a 7369 780a 7365  four.five.six.se
00000040: 7665 6e0a 6569 6768 740a 6e69 6e65 0a74  ven.eight.nine.t
00000050: 656e 0a6a 6163 6b0a 7175 6565 6e0a 6b69  en.jack.queen.ki
00000060: 6e67 0a61 6365 1b16 4824 4120 6f66 201b  ng.ace..H$A of .
00000070: 3016 3132 6a24 6434 f131 336a 50f1 cd20  0.12j$d4.13jP.. 
00000080: ab2f 200a 7bc0 6a59 4856 4770 4173       ./ .{.jYHVGpAs

Answer 19

C#, 219 207 202 197 bytes (0 indexed)

static string O(int i){string[]s={"two","three","four","five","six","seven","eight","nine","ten","jack","queen","king","ace","hearts","diamonds","spades","clubs"};return s[i%13]+" of "+s[i/14+13];}

thanks to input from @Ciaran_McCarthy and @raznagul

Takes an input of int I, subtracts 1 to match 0 indexing of the string array and outputs the number based on I mod 13 and the suit based on i/14+13.

works pretty well for my second code golf, just wondering if i could get it shorter using LINQ or something else.

Answer 20

PowerShell, 207 192 182 174 165 163 161 157 bytes

0-Indexed

$args|%{(-split'two three four five six seven eight nine ten jack queen king ace')[$_%13]+' of '+('hearts','diamonds','spades','clubs')[$_/13-replace'\..*']}

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4 bytes saved thanks to AdmBorkBork in the comments

Answer 21

dc, 161 160 158 142 bytes

[TWO][THREE][FOUR][FIVE][SIX][SEVEN][EIGHT][NINE][TEN][JACK][QUEEN][KING][ACE][HEART][DIAMOND][SPADE][CLUB][z2-:rz1<A]dsAxD~;rP[ OF ]PD+;rP83P

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Over a year later, I was looking for something in my old golfs and found 16 bytes of inefficiency in the previous code:

[ACE][KING][QUEEN][JACK][TEN][NINE][EIGHT][SEVEN][SIX][FIVE][FOUR][THREE][TWO]0[d1+si:rlidD>A]dsAxr[CLUB]3:s[SPADE]2:s[DIAMOND]1:s[HEART]0:sD~;rP[ OF ]P;sP83P

Try that chunkier version online!

Top of stack used as input. 0-indexed.

In the old version, the array of ranks (r) was created using an iterator to run through the strings on the stack (0[d1+si:rlidD>A]dsAxr), but with only four items, it's shorter to manually assign the suits (3:s, etc.). The new version simply uses one array for all of it and also doesn't check against a fixed stack size. [z2-:rz1<A]dsAx replaces the above stack-crawler, popping everything into r at the position (stack depth)-2; it stops when there's one value left.

After all that setup is done, we divide by 13 leaving remainder & quotient on stack (D~). In the old version, we used these values to pick from the two arrays. In the new version, without a separate array of suits, we add 14 (E+) to the suit value, since our suits are after our ranks in array r. Print the strings, and print OF in the middle (;rP[ OF ]PE+;rP).

(First two golfs: -1 byte by outputting uppercase; this allows me to remove the 4 ses from the end of the suits, and instead print it via its ASCII value with the 3-byte 83P at the end. -2 bytes because I repeatedly misread the challenge as requiring 1-indexing. Now 0-indexed.)

Answer 22

CJam, 114 bytes

riDmd"two three four five six seven eight nine ten jack queen king ace"S/=" of "@"hearts diamonds spades clubs"S/=

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Zero-indexed. Will probably be beaten by languages with dictionary compression, but oh well...

Answer 23

Jelly, 61 bytes

d13Uị"“¢¶/CŻ)Gụ+Ḷ}ċ<ʂḤaỴ£k7Ỵ€^ḥæ]¿9Ụ“¡¢|hḂṗƬßĖ,$ðĿȧ»Ḳ€¤j“ of 

0-indexing. Try it online!

Answer 24

Julia 0.6, 156 bytes

f(n)=print(split(" of ,hearts,diamonds,spades,clubs,two,three,four,five,six,seven,eight,nine,ten,jack,queen,king,ace",',')[[(n-1)%13+6,1,div(n-1,13)+2]]...)

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-2 bytes thanks to @Stewie Griffin

Answer 25

Haskell, 144 bytes

f n=words"two three four five six seven eight nine ten jack queen king ace"!!(n`mod`13)++" of "++words"hearts diamonds spades clubs"!!(n`div`13)

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This hits all kinds of Haskell's pain points.

Answer 26

SOGL V0.12, 53 bytes

ķζ≡◄Τū┌p9P7šH□≡⅝'╗ΦqΖ▒ƨM.‘θ.wo"C█}y1+►ΚqΚ‘θJ¼ o.'⁰/wp

Try it Here!

Answer 27

Javascript 149 143 140 bytes

a=_=>"two three four five six seven eight nine ten jack queen king ace".split` `[_%13]+' of '+["hearts","diamonds","spades","clubs"][_/13|0]

-3 bits thanks to @rick hitchcock

a=_=>"two three four five six seven eight nine ten jack queen king ace".split` `[_%13]+' of '+["hearts","diamonds","spades","clubs"][_/13|0]
console.log(a(14))
console.log(a(34))
console.log(a(51))
console.log(a(8))
console.log(a(24))

Answer 28

Perl 5 -p, 119 bytes

0-based

$_=(TWO,THREE,FOUR,FIVE,SIX,SEVEN,EIGHT,NINE,TEN,JACK,QUEEN,KING,ACE)[$_%13].' OF '.(HEART,DIAMOND,SPADE,CLUB)[$_/13].S

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Answer 29

Japt, 91 86 bytes

0-indexed.

I used a tool written by @Shaggy to generate the compressed lists.

`{`twodÈ(‚fÆfivÀ£xç P ightdÍÂdȈjackdquÁÈkˆg»­`qd gU}  {`Ê#tsk¹aÚˆäi£kclubs`qk gUzD

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Explanation:

The first compressed string contains the card values delimited by d. The second compressed string contains the card ranks delimited by k.

These chars were picked using Shaggy's tool, which generates a string delimited by a char that is optimally compressed using shoco (the compression that Japt uses). This allows us to create a list of card values and ranks.

We use backticks ` to decompress these strings, then we split the string using q, followed by the char to split on.

Once we have the lists, we map through the card values, then we get the index of the input. It is important to note that Japt wraps its indexes, so we don't have to modulo by 13.

At each item, we loop through the card ranks. We get the index by dividing the input by 13.

Once we have both items, we concatenate them with " of ", which produces the final string.

Answer 30

Jelly, 58 55 bytes

“ıĿ⁽⁷lg3e/“ẏ“£Ṣ¢÷yḊ⁾⁶ƭ⁼ẎẇḄṡ¿Onṃ⁶ḂḂfṠȮ⁻ẉÑ%»Ḳ€p/⁸ịṚj“ of 

Try it online!