Scrabble scorer

Challenge:

Take a string of upper or lower case letters as input (optional), and calculate the score that string would get in a game of Scrabble in English.

Rules:

The score of each letter is as follows (use this even if there are other versions of the game):

1 point: E, A, I, O, N, R, T, L, S, U
2 points: D, G
3 points: B, C, M, P
4 points: F, H, V, W, Y
5 points: K
8 points: J, X
10 points: Q, Z


The score of a string is simply the sum of the scores of each of the letters used. You may assume that you have plenty of tiles available, so long words, and words with many of the same letters are valid input.

Test cases:

ABC       ->    7
PPCG      ->   11
STEWIE    ->    9
UGPYKXQ   ->   33
FIZZBUZZ  ->   49
ABCDEFGHIJKLMNOPQRSTUVWXYZ  -> 87


The shortest answer in each language wins! The input and output formats are flexible, so you may take the input as an array of characters (upper or lower case) if you want.

• I'm hoping to see a MATLAB/Octave solution. All my attempts were horribly long... =/ Apr 17, 2018 at 19:04
• I'm hoping to see a Beatnik solution. Cuz, you know, that would be the right tool for the job. Apr 17, 2018 at 19:28
• @StewieGriffin Does 85 bytes count as horribly long? Apr 17, 2018 at 21:11
• Hasn't Mathematica a built-in for it? Apr 18, 2018 at 0:30
• @manassehkatz you should definitely give it a go! I highly recommend that you post the challenge in the Sandbox to get some feedback and help with it before posting it on the main site. Complex challenges are notoriously difficult to get right without any feedback. Apr 18, 2018 at 19:37

sed 4.2.2, 81

s/[QZ]/JD/g
s/[JX]/KB/g
s/K/FE/g
s/[FHVWY]/BE/g
s/[BCMP]/DE/g
s/[DG]/EE/g
s/./1/g


Output is in unary.

Reduces each letter to a combination of lower-scoring letters until all letters are 1-scorers. Then replaces those with 1s to give a unary count.

Try it online!

Octave, 50 bytes

@(t)'				'/3*sum(65:90==t')'


Try it online!

Challenge accepted. Explanation:

@(t)             % Define anonymous function taking a single argument t.
' ... '/3    % Row vector with letter scores. Corresponds to char([1 3 3 2 ...]*3).
% The factor 3 was necessary to avoid a newline.

*                % Dot product (yes, * is dot product, .* is not. Go figure). Equivalent to sum of element-wise products.
65:90       % Alphabet
==t'   % Broadcast equality with input string.
sum(         )  % Sum this matrix. Gives the count of each letter in the alphabet
' % Transpose into column vector for dot product

• Very clever! Using unprintables was a nice touch! :) Apr 18, 2018 at 8:22
• @StewieGriffin It's only one byte compared to -47 but that's code-golfing for you! Apr 18, 2018 at 8:31
• Sigh. Well and truly out golfed. I hadn't realised you could use == like that in Octave. Doesn't work in MATLAB. Good to know. Apr 18, 2018 at 8:43
• @TomCarpenter I don't mean to rub any salt in the wound, but the 'old' way of doing this (with bsxfun) is also shorter at 61 bytes: Try it online! Apr 18, 2018 at 8:57
• WAT 50 bytes I don't even Apr 18, 2018 at 10:54

f s=length s+sum[n|x<-s,(n,y)<-zip(9:7:[1..])$words"QZ JX DG BCMP FHVWY K",xelemy]  Try it online! Explanation Most letters give a score of 1 and thus we don't need to keep track of these, instead we just decrement each score (saves 1 byte on 10 as well) and then add the length of the string to the resulting score. Thanks @nimi for -2 bytes (rearanging the words and using [1..] instead of [4,3..])! • zip[1..]$words"DG BCMP FHVWY K . . JX . QZ" gives another alternative with equal length
– Angs
Apr 17, 2018 at 21:54

Beatnik, 733 bytes

Since it really had to be done, here it is. It was a really nasty to debug and provided a few challenges.

Input must be uppercase letters only. Output is unary (hope that is OK?)

J K ZZZZZZK Z ZD ZB ZZZZZZZZZZZZZZZZZA K A Z ZD ZB ZZZZZZZZZZZZZZZKF K A Z ZD ZB ZZZZZZZZZZZZZZZB K A Z ZD ZB ZZZZZZZZZZZZZZZ K A Z ZD ZB ZZZZZZZZZZZZZZKD K A Z ZD ZB ZZZZZZZZZZZZZD K A Z ZD ZB ZZZZZZZZZZZZZD K A Z ZD ZB ZZZZZZZZZZZZ K A Z ZD ZB ZZZZZZZZZZZZB K A Z ZD ZB ZZZZZZZZZKA K A Z ZD ZB ZZZZZZZZZKF K A Z ZD ZB ZZZZZZZZZZK K A Z ZD ZB ZZZZZZZZZB K A Z ZD ZB ZZZZZZZZZB K A Z ZD ZB ZZZZZZZZKD K A Z ZD ZB ZZZZZZZK K A Z ZD ZB ZZZZKB K A Z ZD ZB ZZZZZZKF K A Z ZD ZB ZZZZZZB K A Z ZD ZB ZZZZZFB K A Z ZD ZB ZZZZZA K A Z ZD ZB ZZZAK K A Z ZD ZB ZZZ K A Z ZD ZB ZD K A Z ZD ZB ZKB K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K ZZZZKF KF K A ZKA ZZZZZZZZZZZZZZZZZZY


Try it online!

General process is:

• get character from input
• subtract 65
• check if result is 0
• if 0 jump specified amount of words.
• otherwise subtract 1 and repeat check.
• the jump targets are push print operations followed be a loop back to beginning of program.

Ends with an error.

A more complete explanation:

J K ZZZZZZK Z ZD               # Get input and subtract 65
ZB ZZZZZZZZZZZZZZZZZA K A Z ZD # Character A - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZZZKF K A Z ZD  # Character B - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZZZB K A Z ZD   # Character C - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZZZ K A Z ZD    # Character D - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZZKD K A Z ZD   # Character E - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZD K A Z ZD     # Character F - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZZD K A Z ZD     # Character G - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZ K A Z ZD       # Character H - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZZZB K A Z ZD      # Character I - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZKA K A Z ZD        # Character J - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZKF K A Z ZD        # Character K - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZZK K A Z ZD        # Character L - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZB K A Z ZD         # Character M - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZZB K A Z ZD         # Character N - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZZKD K A Z ZD         # Character O - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZZK K A Z ZD           # Character P - if 0 jump to print, otherwise subtract 1
ZB ZZZZKB K A Z ZD             # Character Q - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZKF K A Z ZD           # Character R - if 0 jump to print, otherwise subtract 1
ZB ZZZZZZB K A Z ZD            # Character S - if 0 jump to print, otherwise subtract 1
ZB ZZZZZFB K A Z ZD            # Character T - if 0 jump to print, otherwise subtract 1
ZB ZZZZZA K A Z ZD             # Character U - if 0 jump to print, otherwise subtract 1
ZB ZZZAK K A Z ZD              # Character V - if 0 jump to print, otherwise subtract 1
ZB ZZZ K A Z ZD                # Character W - if 0 jump to print, otherwise subtract 1
ZB ZD K A Z ZD                 # Character X - if 0 jump to print, otherwise subtract 1
ZB ZKB                         # Character Y - if 0 jump to print, otherwise subtract 1
K ZZZZKF KF                    # Jump Point for print 1111111111
K ZZZZKF KF                    #
K ZZZZKF KF                    # Jump Point for print 11111111
K ZZZZKF KF                    #
K ZZZZKF KF                    #
K ZZZZKF KF                    # Jump Point for print 11111
K ZZZZKF KF                    # Jump Point for print 1111
K ZZZZKF KF                    # Jump Point for print 111
K ZZZZKF KF                    # Jump Point for print 11
K ZZZZKF KF                    # Jump Point for print 1
K A ZKA ZZZZZZZZZZZZZZZZZZAAAA # Jump back to start

• Golfed. May 31, 2019 at 22:55
• @jimmy23013 very nice, you should post that one up. Jun 3, 2019 at 18:19

Pyth, 40 bytes

sm+2x.e}dbc." zØÍ jÙ¹>;%OG5§"\ 1


Try it here

Explanation

sm+2x.e}dbc." zØÍ jÙ¹>;%OG5§"\ 1
m                              Q  For each character in the (implicit) input...
x.e  b                     1   ... find the first index in...
c." zØÍ jÙ¹>;%OG5§"\     ['dg','bcmp','fhvwy','k','','','jx','','qz']
}d                          ... containing the character...
+2                               ... 2-indexed.
s                                  Take the sum.


Brain-Flak, 210, 204, 198, 184, 170 bytes

({<([{}]<>(({}{}))(([][][][][])<((([]())<([][])>))((((()))))>)[](((()()())<((()))>)((())()()()()))((())()()())((()())()())[]((((())())()))(())){({}<{}>())}>{}{}<{{}}><>})


Try it online!

Thanks to @JoKing for saving 14 bytes!

({              # For each character

# Push array of letter scores
# Also adjust character to 1-indexing
<([{}]<>
(({}{}))    # Push 2 0s
(([][][][][])   # 10
<((([]())   # 4
<([][])>    # 8
))      # 4,4
((((()))))> # 1,1,1,1
)       # 10
[]      # Add 12 to difference
(((()()())  # 3
<((()))>    # 1,1
)       # 3
((())()()()())) # 1, 5
((())()()())    # 1, 4
((()())()())    # 2, 4
[]      # Add 22 to difference
((((())())()))  # 1,2,3
(())        # 1
)   # Push 65-char

{({}<{}>())} # Index character into score array
>
{}{}         # Add score to running total
<{{}}><>     # Clear the stack

})               # Implicit print of total score

• 170 bytes
– Jo King
Apr 18, 2018 at 1:47
• For a given definition of 'readable' :) Apr 19, 2018 at 11:59
• I made an edit to try and clarify the explanation, feel free to rollback if you find an issue. Apr 24, 2018 at 20:31

Python 2, 78 bytes

lambda x:sum(map(('ABCEIKLMNOPRSTU'+'BCDGMPQZ'*2+'FHJJKQQVWXXYZZ'*4).count,x))


Try it online!

Shorter version, port of DanielIndie's answer, 71 70 bytes

-1 byte thanks to Sunny Patel

lambda x:sum(int('02210313074020029000033739'[ord(c)-65])+1for c in x)


Try it online!

JavaScript (Node.js), 716663 62 bytes

• @Arnauld awesome as always reducing 7 bytes
• thanks to l4m2 for rducing by 1 byte
s=>Buffer(s).map(x=>s="02210313074020029000033739"[x-65]-~s)|s


Try it online!

Java 8, 7571 70 bytes

s->s.chars().map(c->"\n\n".charAt(c-65)).sum()

-1 byte by changing "02210313074020029000033739".charAt(c-65)-47 to unprintables (and two \n) so the -47 can be removed. Inspired by @Sanchises' Octave answer.

Try it online.

s->          // Method with String parameter and integer return-type
s.chars()  //  Loop over the characters as IntStream
.map(c->"\n\n".charAt(c-65))
//   Convert the character to its value
.sum()    //   And sum it all together


Octave / MATLAB, 85 bytes

@(x)sum([1:4 7 9]*any(reshape(char(strsplit('DG BCMP FHVWY K JX QZ')),6,1,5)==x,3)+1)


Try it online!

• Better than my attempts :-) still longer than I would have thought before I tried it though... You had a very different approach to it! Apr 18, 2018 at 6:01

Jelly, 19 bytes

Oị“ÆẠḃbṂƬɠF#ṁ²’ḃ⁵¤S


A monadic link accepting a list of upper-case characters which returns an integer

Try it online! Or see the test-suite.

How?

Oị“ÆẠḃbṂƬɠF#ṁ²’ḃ⁵¤S - Link: list of characters
O                   - ordinals ('A'->65, B->66...)
“ÆẠḃbṂƬɠF#ṁ²’     -   literal 14011114485013321424185131
⁵   -   literal 10
ḃ    -   bijective-base = [1,3,10,1,1,1,1,4,4,8,4,10,1,3,3,2,1,4,2,4,1,8,5,1,3,1]
ị                  - index into (1-based & modular) (vectorises)
-  i.e. mapping from: O P  Q R S T U V W X Y  Z A B C D E F G H I J K L M N)
S - sum


R, 90 63 bytes

function(W,u=utf8ToInt)sum(u('

')[u(W)-64])


Try it online!

Takes input as an uppercase string. R handles unprintables and multiline strings without issues, so that's nice. Now we're almost twice the external package!

And because CRAN has so many random goodies:

R + ScrabbleScore 31 bytes

ScrabbleScore::sws(scan(,""),F)


Try it online!

Sadly, sws checks for validity by default.

• Had a play with the score list and trimmed a couple Apr 18, 2018 at 20:44
• @MickyT nice! I played around with unprintables and re-using utf8ToInt instead of match and managed to get a few more down! Apr 18, 2018 at 22:04

Emojicode, 358 bytes

🐖🔥➡️🔡🍇🍮s 0🔂l🍡🐕🍇🍮s➕s🍺🐽🍯🔤a🔤1🔤e🔤1🔤i🔤1🔤l🔤1🔤n🔤1🔤o🔤1🔤r🔤1🔤s🔤1🔤t🔤1🔤u🔤1🔤d🔤2🔤g🔤2🔤b🔤3🔤c🔤3🔤m🔤3🔤p🔤3🔤f🔤4🔤h🔤4🔤v🔤4🔤w🔤4🔤y🔤4🔤k🔤5🔤j🔤8🔤x🔤8🔤q🔤10🔤z🔤10🍆🔡l🍉🍎🔡s 10🍉


Try it online!

Explanation:

I changed the variable names from single letters to more meaningful words, and expanded some parts of my code to hopefully make it more readable for people unfamiliar with the language. You can test the expanded program here.

🐋🔡🍇      👴 define a class that takes a string
🐖🔥➡️🔡🍇    👴 define a method that returns a string
🍦values🍯    👴 create int dictionary
🔤a🔤1 🔤e🔤1 🔤i🔤1 🔤l🔤1 🔤n🔤1 🔤o🔤1 🔤r🔤1 🔤s🔤1 🔤t🔤1 🔤u🔤1 🔤d🔤2 🔤g🔤2
🔤b🔤3 🔤c🔤3 🔤m🔤3 🔤p🔤3 🔤f🔤4 🔤h🔤4 🔤v🔤4 🔤w🔤4 🔤y🔤4 🔤k🔤5 🔤j🔤8 🔤x🔤8
🔤q🔤10 🔤z🔤10
🍆        👴 ^ dictionary contains letters(keys) and their numerical values

🍮score 0                         👴 declare 'score' variable and set to 0
🍦iterator🍡🐕                     👴 transform input string to iterator
🔂letter iterator🍇                👴 iterate over each byte in input string
🍮score➕score 🍺🐽values 🔡letter   👴 add value of each letter to score
🍉
🍎🔡score 10    👴 return the score as a string
🍉
🍉

🏁🍇          👴 begin the program here
😀🔥🔤abc🔤    👴 call scoring method and print the score
😀🔥🔤ppcg🔤    👴 repeat with other test cases
😀🔥🔤stewie🔤
😀🔥🔤fizzbuzz🔤
😀🔥🔤abcdefghijklmnopqrstuvwxyz🔤
🍉

• ouch... my eyes... is there an option on golf.se to hide some specific langages? ^^ Apr 19, 2018 at 12:15
• @OlivierDulac There's probably a way to prevent the browser from rendering emoji specially. They do each have standard Unicode black and white characters associated with them. Apr 20, 2018 at 15:19

sum.map(\c->1+read["02210313074020029000033739"!!(fromEnum c-65)])


Try it online! Same approach as DanielIndie's JavaScript answer.

$s  Try it online! 05AB1E, 21 bytes Takes input as a lowercase list of characters. •_JÊ¿ùã$Ƶ½œM•11вAIkèO


Try it online! or as a Test suite

Octave, 73 bytes

@(x)sum('09977433333222211'(([~,y]=ismember(x,'QZJXKFHVWYBCMPDG'))+1)-47)


Try it online!

Uses ismember to map each character in the input stream x onto it's index in the lookup string 'QZJXKFHVWYBCMPDG'. Any element not found will be mapped to an index of 0 (this will include the 1-point characters).

Next we add 1 to the index to make the 0's become valid 1-index references, and lookup into the string '09977433333222211'. This is one element longer than the first lookup string. The digits represent the point value of each element in the original string, minus 1, with the extra element being a '0' at the beginning .

Finally the resultant string is converted to integers by subtracting 47 ('0'-1), yielding the point value for each letter, and all point values are then summed.

• Very clever! :) Apr 18, 2018 at 7:44

C++, 95 bytes

char*m="02210313074020029000033739";
int f(char*p){int n=0;while(*p)n+=m[*p++-65]-47;return n;}


Try it online (not a TIO link sorry)

Explanation:

• Declares m, an array of the values of each letter in order, minus 1. The minus 1 is because of Q and Z: I couldn't have a two digit number in there
• Iterates through through the string p until we get to null character, and adds the score of the number (*p gives us the letter, and -65 so we can properly index the array). Since m is a char* it converts to a char so we minus 48 so bring it back to 0, but add 1 since m is declared as one score less for each character.

I'm not an avid poster here so I hope I've done this correctly. I believe they returning n counts as printing the value, and that declaring a function is fine.

• Very nice! The only byte you can save, is the newline: Try it online! Apr 30, 2019 at 21:39
• @movatica Why you need m?
– l4m2
May 21, 2023 at 4:38
• Nice find. But it's not my post, so you should ask @Tas ;) Jun 4, 2023 at 5:14

K (oK), 60 38 bytes

Solution:

+/1+.:'"02210313074020029000033739"65!


Try it online!

Explanation:

Index into the scores, sum up result.

+/1+.:'"02210313074020029000033739"65! / the solution
65! / input modulo 65 to get position in A-Z
"02210313074020029000033739"    / index into the scores (1 point lower)
.:'                                / value (.:) each (') to convert to ints
1+                                   / increase by 1
+/                                     / sum up


PowerShell, 92 bytes

$args|% T*y|%{$i+=switch -r($_){[DG]{2}[BCMP]{3}[FHVWY]{4}K{5}[JX]{8}[QZ]{10}default{1}}};$i


Try it online!

Didn't notice @Add-TheNewOne's answer before writing this, but I think it's fairly different in any case.

05AB1E, 18 bytes

•Q<ß÷λv¸Ïàœ¶•IÇè>O


Try it online!

Explanation:

•Q<ß÷λv¸Ïàœ¶•           # compressed integer 30740200290000337390221031
IÇ         # codepoints of the input
è        # index into the digits of the large integer
>       # increment each
O      # sum


The large integer encodes the point value of each letter. Since encoding numbers in the range 1-10 would be inconvenient, we subtract 1 so that each one fits on a single digit (this is why the > is needed).

Note that indexing in 05AB1E wraps around, so the first digit of the number actually corresponds to "h" ("h"'s codepoint is 104, which is 0 mod 26). This is convenient: since numbers can't start with 0, we don't want the first digit to correspond to "a".

Alternative 18-byter:

A•Θ¡₆öηÖ&'/α°•R‡>O


Try it online!

A pushes "abcdefghijklmnopqrstuvwxyz". ‡ transliterates, replacing each letter in the input with the corresponding digit in the compressed integer. A‡ saves one byte over IÇè; unfortunately, this encoding puts "a" first, so we have to use R (reverse) to get around the issue that numbers can't start with 0.

Yet another 18-byter:

žW•Δÿ¦ÝZ=áí4_ø•‡>O


Try it online!

Same idea as above, but this one uses the obscure žW built-in to push "qwertyuiopasdfghjklzxcvbnm". žW is one byte longer than A, but it saves the R since "q" isn't a 1-point letter.

Beatnik, 379 bytes

K ZZZZKF ZD ZD ZD ZD ZD ZD ZD
K ZZZZZZZZA J Z ZD ZB B
K XA KD ZD ZF F ZA ZD KF KF ZD ZB B
K D Z ZD ZB B
K ZF Z ZD ZF KA ZA ZD ZD KF KF KF ZD ZB B
K A KD ZD ZF D ZA KF ZD ZB B
K ZF KD ZD ZB B
K D Z ZD ZB B
K A Z ZD ZB B
K ZF Z ZD ZB B
K D Z ZD ZF D ZA KF ZD ZB B
K Z KD ZD ZB B
K B Z ZD ZB B
K Z Z ZD ZB B
K A Z ZD ZF F ZA KF ZB Z
K K KD ZD ZB B
K B Z ZF A KF
KF
XX ZZZZZZZZZZZZZX


Try it online!

Japt, 36 bytes

¬£2+dg
bcmp
fhvwy
k

jx

qz·bøX
x


Takes input as a lowercase string, returns a number.
Short explanation:

¬£2+dg
¬       // Split the input into chars,
£      // then map over each char, returning
2+dg // 2 plus

qz·bøX
bøX // the char's index in
qz·    // the hardcoded string split by newlines.
x       // And finally sum the whole thing.


Try it online!

Ruby, 60 bytes

->s{s.sum{|c|"BDDCBECEBIFBDBBDKBBBBEEIEK"[c.ord-65].ord-65}}


Try it online!

A lambda, accepting input as an array of (uppercase) characters and returning an integer.

Perl 5-pF, 50 bytes

_1DG2BCMP3FHVWY4K7JX9QZ=~/\d\D*$_/,$\+=1+$&for@F}{  Try it online! Gforth, 109 Bytes : V s" 1332142418513113:11114484:" ; : C 0 NAME 0 DO DUP C@ 65 - V DROP + C@ 48 - ROT + SWAP 1+ LOOP DROP . ;  Input must be uppercase: C PPCG 11 OK Readable \ String used as table with values for each letter in the alphabet \ : follows 9 in the ASCII-table : V s" 1332142418513113:11114484:" ; : C 0 \ Initialize sum ( sum ) NAME \ Get the string ( sum c-addr count ) 0 DO \ Start of loop ( sum c-addr ) DUP C@ \ Get letter ( sum c-addr char ) 65 - \ Calculate table index ( sum c-addr index ) V DROP + C@ \ Get table entry ( sum c-addr entry ) 48 - \ Calculate entry value ( sum c-addr value ) ROT + SWAP \ Update sum ( sum' c-addr ) 1+ \ Next character ( sum' c-addr' ) LOOP DROP . \ Drop c-addr and print result ;  Try it online! Perl 6, 52 bytes {TR/A..Z/02210313074020029000033739/.comb.sum+.ords}  Try it online! Maps every character to a digit, and sums them. And adds 1 for each character because there isn't a digit 10 without incurring unicode bytes. Retina 0.8.2, 41 bytes TBCDGJKMPQXZF\HVWY221174229793 .$*..
.


Try it online! Link includes test cases. Explanation: Like the Haskell answer, nontrivial letters are translated to 1 less than their score, and 1 is added later when the characters are converted to unary. Putting FHVWY last allows them all to map to a score of 3 + 1.

C (gcc), 78 72 bytes

i;f(char*s){for(i=0;*s;)i+="\n\n"[*s++-65];s=i;}


There are actually 26 characters in that string. See the code rendered properly and run it here.

Thanks to gastropner for golfing 6 bytes.

Ungolfed version:

i; // declare a variable to store the score; it is implicitly of type int
f(char* s) { // function taking a string as argument and implicitly returning an int
for(i = 0; // initialize the score to 0
*s; ) // iterate over the string until we hit terminating NUL byte
i += "\n\n"[*s++ - 65]; // this is a 26-char string containing the ASCII equivalent of each numeric scrabble value; 65 is ASCII code for 'A', mapping the alphabet onto the string
s = i; // implicitly return the score
}


Excel, 91 bytes

{=LEN(A1)+SUM(0+("0"&MID("02210313074020029000033739",CODE(MID(A1,ROW(A:A),1)&"z")-64,1)))}


Explanation:

• Input is in cell A1
• The formula must be entered as an array formula with Ctrl+Shift+Enter, which adds the curly brackets { } to both ends.
• MID(A1,ROW(A:A),1) pulls out each character in turn (and a lot of empty values, too, since it's going to return as many values as there are rows in the sheet)
• CODE(MID(~)&"z") pulls out the ASCII value for the each character. The &"z" appends a z to the end of the MID() result because CODE() doesn't like empty inputs. The ASCII value for z is higher than every capital letter, though, so it's effectively ignored later.
• MID("02210313074020029000033739",CODE(~)-64,1) pulls out a letter from the score string based on its ASCII value adjusted down by 64 so the letters run 1-26 instead of 65-90.
• "0"&MID(~) prepends a zero to the MID() result because Excel won't let you do math with empty strings, of which there will be several.
• 0+("0"&MID(~)) turns all those strings into numbers.
• SUM(0+("0"&MID(~))) adds up all those strings that are now numbers.
• LEN(A1)+SUM(~) adds the length of the input to the sum because all the values in the score string (02210313074020029000033739) were adjusted down by one so they would all be one digit long.

There's a very similar solution in Google Sheets but it comes in at 97 bytes because ArrayFromula() is longer than {} (but at least it can handle 0 + "" = 0).

=Len(A1)+ArrayFormula(Sum(0+Mid("02210313074020029000033739",Code(Mid(A1,Row(A:A),1)&"z")-64,1)))

• Well done. I have a Excel solution using 26 SUBSTITUTE(), coming in at a hefty 527 bytes. Apr 20, 2018 at 12:35