# How happy is this emoticon?

In this challenge, submissions will be programs or function which, when given an emoticon such as :-), :(, or :D, will rate their happiness from 0 to 3.

An emoticon will be one of the following:

• :(: 0
• :|: 1
• :): 2
• :D: 3

Emoticons may also have noses (a - after the :).

Test cases:

:(  -> 0
:-| -> 1
:D  -> 3
:-) -> 2
:|  -> 1


This is a code golf challenge, shortest answer per language wins.

• What about (:? – anatolyg Apr 2 '20 at 13:06
• @anatolyg I figured having to handle reversed strings as well would make the challenge too difficult – Redwolf Programs Apr 2 '20 at 15:34
• @anatolyg Don't forget ☺and ☻ either – Chronocidal Apr 2 '20 at 16:20
• @anatolyg What about D:? – user92069 Apr 3 '20 at 8:59
• It would be cool to see a solution that looked like a smiley ... – Jerry Jeremiah Jun 9 '20 at 23:45

# Python 3, 26 bytes

lambda s:ord(s[-1])*2%19%4


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$$f(x) = (2x \bmod 19) \bmod 4$$

• You can save 1 character by using the formula in the Ruby solution below : codegolf.stackexchange.com/a/202907/61997 – mr.mams Apr 3 '20 at 10:41
• @mr.mams, then it will be absolutely identical to another python answer – Dion Apr 5 '20 at 7:40

# Ruby, 20 bytes

s[-1].ord will give us the codepoint of the last character in the string, which we can then plug into this formula: $$(160 \bmod c) \bmod 7$$

->s{160%s[-1].ord%7}


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• Nice find on the formula! I tried brute-forcing for expressions in Python no longer than 160%n%7 (7 bytes) and got only that with nothing shorter. Here are the 8-byte ones: TIO – xnor Apr 2 '20 at 3:14

# APL (Dyalog), 11 8 bytes

3 bytes saved thanks to @Bubbler!

⊃'(|)'⍳⌽


Try it online!

       ⌽  reverse the string
⊃         take the last (now first) byte
'(|)'⍳    index inside '(|)' (if not found ('D') returns the length (3))

• 8 bytes – Bubbler Apr 2 '20 at 1:47
• @Bubbler Unfortunately, it says the link can’t be decode; mind repasting? – AviFS Apr 2 '20 at 3:02
• @AviF.S. Here it is. Weird though, because I can open it fine on my PC. – Bubbler Apr 2 '20 at 3:22
• Very odd! Still doesn’t work for me but I’m on the mobile app (iPhone). Though I couldn’t find anything on Meta re: bugs with TIO and mobile... Suppose I’ll just hold my excitement ‘til I can check on my computer :) – AviFS Apr 2 '20 at 3:33
• thanks @Bubbler – Uriel Apr 2 '20 at 8:48

# Python 3, 28 27 bytes

lambda e:"(|)D".find(e[-1])


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Because we don't care about the eyes/nose, we can just look at the mouth (the last character) and find its index in a string with all the mouths, ordered from saddest to happiest :) Thanks @Surculose Sputum!

• find works the same as index and saves 1 byte – Surculose Sputum Apr 1 '20 at 16:42

# Python 3, 25 bytes

lambda s:160%ord(s[-1])%7


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Math taken from @dingledooper

# C (gcc),  37  33 bytes

Saved 4 bytes on both versions thanks to @ceilingcat and @dingledooper

I overlooked the rule about the optional nose, so this is not as effective as expected.

f(char*s){s=390%~-s[*++s%9<1]&3;}


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### How?

Among the characters that we have to deal with, the hyphen is the only one whose ASCII code is congruent to $$\0\$$ modulo $$\9\$$. We use this property to decide whether we need to work on the second or the third character.

Given the ASCII code $$\n\$$ of the relevant smiley character, we apply the following formula to get the happiness:

$$h(n)=(390\bmod (n-1))\bmod 4$$

# C (gcc),  36  32 bytes

Using Uriel's formula is 1 byte shorter.

f(char*s){s=s[*++s%9<1]*2%19&3;}


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• -1 byte for both, by recursively searching for the end of the string. – Bubbler Apr 2 '20 at 1:43

# Retina 0.8.2, 16 12 bytes

T-:(|)D__d


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T


Enter transliteration mode

(|)Dd


Replace the mouth character with the corresponding digit (0 to 3)

-:__


Remove each of - and : from the string.

Implicitly output the result, which will be a single digit.

# Jelly,  8  7 bytes

⁽$9 - base 250 literal = 10058 ,4 - pair with four ḥ - Jelly hash A using 10058 as a salt and [1,2,3,4] as the domain ’ - decrement  Previous 8 byter: “|)D”iⱮS  Try it online! ### How? “|)D”iⱮS - Link: list of characters, A e.g. ":-)" OR ":-(" Ɱ - map across c in A with: i - first index of c in (or 0 if not found): “|)D” - list of characters = "|)D" [0,0,2] [0,0,0] S - sum 2 0  # Raku, 19 bytes {TR:d/(|)D:-/0123/}  Try it online! • If raku -p is acceptable, you could reduce to tr:d/(|)D:-/0123/ – Sebastian Apr 4 '20 at 4:18 # Keg, -hr, 18 bytes ?^⑵¦P0|\🄃1|R2|\¦3™  Try it online! The same switch statement format, but with a different character checking criteria. # Bash + GNU utilities, 17 bytes tr '(|)D:-' 0-3\  Try it online! Note: There's a space character after the backslash. Input on stdin, output on stdout. The challenge doesn't specify the format of the output. Depending on the input, this program may print the output in either %2d or %3d format (that is, with one or two spaces before the 0, 1, 2, or 3). If that's not acceptable, then tr -s '(|)D:-' 0-3\ (20 bytes) always prints the digit in %2d format. (There's a space after the backslash here too.) Or tr '(|)D:-' 0-4|tr -d 4 (23 bytes) prints just the digit (with no spaces). ## Excel, 25 bytes =FIND(RIGHT(A1),"(|)D")-1  # Mornington Crescent, 1820 bytes Try it online! A port of @dingledooper's fantastic Ruby answer, which just so happens to work seamlessly in Mornington Crescent because of a convenient 7 in it. Props to them! Take Northern Line to Leicester Square Take Northern Line to Leicester Square Take Piccadilly Line to Turnpike Lane Take Piccadilly Line to Turnpike Lane Take Piccadilly Line to Leicester Square Take Northern Line to Leicester Square Take Northern Line to Charing Cross Take Northern Line to Charing Cross Take Bakerloo Line to Baker Street Take Bakerloo Line to Paddington Take Bakerloo Line to Charing Cross Take Bakerloo Line to Charing Cross Take Northern Line to Moorgate Take Circle Line to Moorgate Take Metropolitan Line to Chalfont & Latimer Take Metropolitan Line to King's Cross St. Pancras Take Victoria Line to Seven Sisters Take Victoria Line to Victoria Take Circle Line to Victoria Take Circle Line to Bank Take Circle Line to Hammersmith Take Circle Line to Cannon Street Take Circle Line to Hammersmith Take Circle Line to Cannon Street Take Circle Line to Bank Take Circle Line to Hammersmith Take District Line to Upminster Take District Line to Hammersmith Take District Line to Upminster Take District Line to Victoria Take Circle Line to Victoria Take Circle Line to Aldgate Take Circle Line to Aldgate Take Metropolitan Line to Chalfont & Latimer Take Metropolitan Line to Preston Road Take Metropolitan Line to Baker Street Take Metropolitan Line to Preston Road Take Metropolitan Line to Pinner Take Metropolitan Line to Preston Road Take Metropolitan Line to King's Cross St. Pancras Take Victoria Line to Seven Sisters Take Victoria Line to King's Cross St. Pancras Take Circle Line to King's Cross St. Pancras Take Metropolitan Line to Pinner Take Metropolitan Line to Preston Road Take Metropolitan Line to Pinner Take Metropolitan Line to Preston Road Take Metropolitan Line to King's Cross St. Pancras Take Circle Line to King's Cross St. Pancras Take Northern Line to Mornington Crescent  The only major thing I feel I need to point out is how I got 160 in the program. To do this, I took the first letter of Paddington, which gives an ASCII value of 80 when run through Charing Cross, then multiplied it by 2 in Chalfont & Latimer to get 160. Subsequently, the program computes 160 % (ASCII of last character) % 7 via Preston Road, the formula found by @dingledooper. # Perl 5-p, 24 bytes s/.*(.)/ord($1)*2%19%4/e


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Steals the math from @Uriel's Python answer

# MATL, 9 bytes

'|)D'jmfs


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### Explanation

'|)D' % Push this string
j     % Read input as unevaluated string
m     % ismember: true for chars of the first string that are present in the second
f     % find: (1-based) indices of true entries. The result will have length 0 or 1
s     % sum. This is needed to transform an empty array into 0
% Implicitly display


# naz, 120 bytes

8a5m2x1v1a2x2v4a2x3v9a9a5a2x4v1x1f2r3x3v1e2f0x1x2f3x1v3e3x2v5e3x4v6e4f0x1x3f0m1o0x1x4f0m1a1o0x1x5f0m2a1o0x1x6f0m3a1o0x1f


Explanation (with 0x commands removed)

8a5m2x1v                 # Set variable 1 equal to 40 ("(")
1a2x2v                   # Set variable 2 equal to 41 (")")
4a2x3v                   # Set variable 3 equal to 45 ("-")
9a9a5a2x4v               # Set variable 4 equal to 68 ("D")
1x1f                     # Function 1
2r                   # Read the second byte in the input string, removing it
3x3v1e             # Jump back to the start of the function if it equals variable 3
1x2f                     # Function 2
3x1v3e               # Jump to function 3 if the register equals variable 1
3x2v5e         # Jump to function 5 if the register equals variable 2
3x4v6e   # Jump to function 6 if the register equals variable 4
1x3f                     # Function 3
0m1o                 # Output 0
1x4f                     # Function 4
0m1a1o               # Output 1
1x5f                     # Function 5
0m2a1o               # Output 2
1x6f                     # Function 6
0m3a1o               # Output 3
1f                       # Call function 1


# 05AB1E, 8 bytes

θÇx19%4%


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### Explanation

I used Uriel's formula.

θ        # Get mouth (last char)
Ç       # ASCII value
x      # Multiply by 2
19%   # Modulo 19
4% # Modulo 4

• If you don't want to have the answer wrapped into a list, you could swap the θÇ to Çθ. (PS: x actually pushes 2n without popping, but there is also · which pops and pushes 2n. Doesn't matter here, but figured I'd let you know in case you're using it in future answers.) Here also another 8-bytes alternative based on dingledooper's Ruby answer: ƵxIÇθ%7%. Try it online. – Kevin Cruijssen Apr 2 '20 at 10:10
• @Kevin Cruijssen I knew that x did that, just that I looked for * 2 and not 2 * in info.txt. I also made a solution based on dingledooper's formula, but it was 9 bytes. – PkmnQ Apr 2 '20 at 10:14

# C# (Visual C# Interactive Compiler), 17 14 bytes

Turns out Ranges finally started to actually exist and work! Thanks to an anonymous user for pointing this out (by proposing an edit, but anonymous users cannot comment...)

s=>160%s[^1]%7


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# Pyth, 8 bytes

x"(|)D"e


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### Exalanation

x"(|)D"e
: Implicit evaluated input
e  : Last element of input
"(|)D"   : The string "(|)D"
x         : First occurrence of the last element of input in "(|)D"


# sed, 20 bytes

y/(|)D/0123/;s/:-*//


Pretty self explanatory. Replaces the the "mouth" characters with numbers using the y/// transform command and then strips off the eyes and the nose if they exist.

Try it online!

This is one byte longer, but perhaps it gets bonus points for having an emoticon in the code itself? :-\

y/(|)D/0123/;s/:-\?//


The shorter solution (16 bytes)

y/(|)D:-/0123  /


also works, but adds extra whitespace to each line of output. It also has two emoticons D: and :-/. :)

Try it online!

# batch, 109 bytes 99 bytes

@For %%G in ("(=0",")=2","D=3","|=1")Do @Set %%G
@Set "T=%~1"
@<Nul Call Call Set/P"=%%%T:~-1%%%"


• 106 Bytes when lines are terminated by LF instead of CR LF, using Shift+CTRL+J – T3RR0R Apr 2 '20 at 16:24
• the space in front of do is optional – HaxAddict1337 Apr 25 '20 at 13:34

# Charcoal, 10 bytes

Ｉ⌕(|)D§Ｓ±¹


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

       Ｓ    Input string
§     Cyclically indexed by
¹  Literal 1
±   Negated
⌕          Find index in
(|)D      Literal string of mouths
Ｉ           Cast to string
Implicitly print


# Io, 31 bytes

Io's strings are made of integers, so character-converting is unneccecary. Although Io doesn't allow us to index the last item of a sequence using last()...

method(x,160%x reverse at(0)%7)


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# Red, 32 bytes

func[s][select"(0|1)2D3"last s]



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# J, 10 9 bytes

-1 byte thanks to FrownyFrog ang Bubbler

'(|)'i.{:


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• You can drop D as bubbler pointed out. – FrownyFrog Apr 3 '20 at 19:34
• @FrownyFrog Yes, indeed. Thanks! – Galen Ivanov Apr 3 '20 at 20:35

# PowerShell, 16 bytes

2*\$args[-1]%19%4


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PowerShell port of the modulo formula going around. Takes input via splatting

# Wolfram Language (Mathematica), 38 bytes

Mod[2Last@ToCharacterCode@#,19]~Mod~4&


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This borrows Uriel's formula, which happens to be the same length as dingledooper's, but I can't help but wonder if another formula would be shorter. I will investigate.

For comparison, here's the string indexing solution (51 bytes):

("(|)D"~StringPosition~Last@Characters@#)[[1, 1]]-1&


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# JavaScript, 3532 29 bytes

I based the logic off of the Java solution submitted April 2 and edited April 8, by branboyer. I suppose it would be referred to as a port of their answer, but I don’t know exactly how that’s supposed to be marked.

a=>"(|)D".indexOf(a[2]||a[1])


An anonymous function, taking a string and returning an integer. Please help me shorten it further, I’m new to golfing code. Just the code, without the way of inputting (no idea how to get input in try it online) is at this Try it online link

• Welcome to the site! As this is an anonymous function, you can remove the b= from the byte count. You can just link to a third party test area, such as Try It Online! if you don't want to include the HTML/full script, which is definitely not required in your answer. As for tips on how to shorten it, I'd recommend you check out the Tips for golfing in Javascript and look at other JS answers around the site. – caird coinheringaahing Jun 9 '20 at 22:31
• You can also drop the ; – math junkie Jun 9 '20 at 22:32
• You don't need to add any HTML for javascript answers. If you want to shorten it, you can replace a[a.length-1] with a[2]||a[1] – Redwolf Programs Jun 9 '20 at 22:39
• Here is a TryItOnline link of your answer, which verifies all test cases. – dingledooper Jun 9 '20 at 22:56
• Thanks for the Try it online link, @dingledooper! I will edit that in instead of my link. – Bardic Wizard Jun 9 '20 at 22:57

# International Phonetic Esoteric Language, 11 bytes

iʁ2f{J}ⱱ4ⱱo


Based on Uriel's formula.

iʁ2f{J}ⱱ4ⱱo
iʁ          (split the face into it's ord values)
2f        (mouth * 2)
{J}ⱱ    (mod 19) (base 26)
4ⱱ  (mod 4)
o (print)


# MAWP, 27 bytes

|2W!!92W1MP92W1MWA!!4P4WA:.
`

Try it!