Maximum average ord

Question

Your task

Take a list of strings as the input, and output the maximum average ord.

Example

Given the list ['hello', 'world', 'bye']:

• The average ord of 'hello' is:
  • (ord(h) + ord(e) + ord(l) + ord(l) + ord(o)) / len('hello')
  • = 106.4
• The average ord of 'world' = 110.4
• The average ord of 'bye' = 106.7

The maximum average ord is 110.4. This is your output.

Test cases

Note: For the last two test cases, I have given the roundings for 1, 2, and 3 decimal places. As mentioned in the rules below, you can round to any number of decimal places.

Input                                        Output
['hello', 'world', 'bye']                    110.4
['code', 'golf', 'stack', 'exchange']        106.8
['!@#', '$%^', '&*(']                        55.7 / 55.67 / 55.667 / etc.
['qwertyuiop[', 'asdfghjkl;', 'zxcvbnm,']    110.9 / 110.91 / 110.909 / etc.

Rules/clarifications

• The output must be only the average ord. You may not output anything else.
• It may be rounded (up or down) to any number (\$\ge\$ 1) of decimal places. If your language does not support floating points, you may output the average multiplied by 10.
• Floating point innacuracies are ok.
• You may assume that the input list will always have a length of 2 or more
• You may assume that the strings will never be empty
• You may assume that the strings will not contain whitespace or non-ASCII characters
• This is code-golf, so shortest code in bytes wins!

Answer 1

Pyth, 7 bytes

eSm.OCM

Test suite

Explanation:

eSm.OCM   | Full program
eSm.OCMdQ | with implicit variables
----------+-------------------------------
  m     Q | For each word in the input,
     CMd  |  Get the ord of each character
   .O     |  Get the average
eS        | Get the max

Answer 2

Python, 46 bytes

lambda l:max(sum(map(ord,s))/len(s)for s in l)

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Unfortunately, the only built-in mean function Python has is in the statistics module, and that's too expensive to import.

Answer 3

K (ngn/k), 18 11 bytes

|/{+/x%#x}'

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Massive -7 bytes golf thanks to Steffan!

Damn everyone is answering quick. Anyways here's my solution. Takes input as a list of strings.

Explanation:

|/{+/x%#x}'  Main function. Takes implicit input
          '  For each string in the input...
  {      }   Execute a function that...
   +/x       Takes sum of every character in the string
      %      And divide it by
       #x    The length of the string
|/           Get the maximum value

Answer 4

Wren, 68 bytes

Fn.new{|x|x.reduce(0){|y,z|y.max(z.bytes.reduce{|a,b|a+b}/z.count)}}

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Answer 5

APL(Dyalog Unicode), 14 13 bytes ^SBCS

-1 bytes thanks to Adám

⌈/+⌿∘⎕UCS¨÷≢¨

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A tacit function which finds the maximum ord. Originally wrote a dfn, but I tried random stuff to convert it to tacit and it somehow worked out.

⌈/+⌿∘⎕UCS¨÷≢¨
          ¨              ⍝  for each element in the input...
     ⎕UCS               ⍝  convert the characters to their ascii values...
  +⌿∘                   ⍝  and sum the resulting ascii values...
           ÷             ⍝  divided by...
            ≢¨           ⍝  the length of each element...
⌈/                       ⍝  and take the maximum of that

Answer 6

><>, 80 51 48 bytes

0>0001.
$i:"!"(?v+$1+
(@:{:,$&/
v?(0&~$?/.02
\n;

Animated Version

Requires a extra space at the end of the input. Also assumes no 2 spaces in a row. (no 0 length words). Anything with ORD of 32 or lower is considered white-space.

Explanation

Top row: Push 3 0s on to the stack. The first is the current best ORD, the second is the length of the current word, and the last is the sum of the ord values of the current word. 01. is a jump to the next row.

The second row sums a word. If a character is less than "!" (char value 33) go down. Otherwise sum the word and increment the length. `$1+

In the third row, we fist push the last value to the register & . We use this to later check if this was the end of the string. Then we divide by the length to get the new ORD value. Then we use the :{:@ trick to copy the stack if it has only 2 elements (which is always the case) from the stack and compare them. If the new one is better we swap the stack with $. In either case we delete the top item.

Lastly, we take the value of the register and check if it is negative, &0(?. If so, we print the maximum ord and exit. If not, we jump back to 00. To get the correct direction we use the 00!.| trick.

Answer 7

><>, 65 45 44 40 bytes

000i:"!"(?v+$1+$20.
$~$0(?n00.>@$,:{:@(?

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Takes input as a space separated string.

Answer 8

Factor, 24 bytes

[ 0 [ mean max ] foldl ]

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Just fold with the max of the mean.

Answer 9

R, 38 bytes

\(x)max(sapply(Map(utf8ToInt,x),mean))

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               Map(utf8ToInt,x)        # for every string in the input, store the vector of its codepoints in a list
        sapply(                ,mean)  # apply mean to each of the list elements and return a vector
\(x)max(                             ) # take max

Answer 10

Jelly, 4 bytes

OÆmṀ

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Answer 11

05AB1E, 10 9 4 bytes

ÇÅAà

-5 thanks to Sʨɠɠan

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Explained:

ÇÅAà  # Implicit input as a list
      # Implicit map over the input:
Ç     #   Get the ord of each character
 ÅA   #   Get the average of the ords
   à  # Get the maximum value

Answer 12

Pip, 12 bytes

M:$+A*_/#_Mg

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Answer 13

Knight (v2), 56 bytes

;=mF;W=pP;=s!=i~1;W>Lp=i+1i=s+*10A Gp iTs&<m=t/sLp=m tOm

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Because Knight doesn't support floating point values, it outputs the maximum average ord multiplied by \$10\$, as allowed by the rules:

It may be rounded to any number (\$\ge1\$) of decimal places. If your language does not support floating points, you may output the average multiplied by 10.

Answer 14

Julia, 34 bytes

!a=max(sum.(Int,a)./length.(a)...)

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Answer 15

Excel (ms365), 70 bytes

Formula in F1:

=MAX(MAP(A1:C1,LAMBDA(a,SUM(CODE(MID(a,SEQUENCE(LEN(a)),1)))/LEN(a))))

Answer 16

MathGolf, 4 bytes

$m▓╙

Input as a list of lists of characters.

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Explanation:

$     # Get the codepoint of each inner-most character of the (implicit) input-list
 m    # Map over each list of integers:
  ▓   #  Pop and push the average of the list
   ╙  # After the map: pop and push the maximum
      # (after which the entire stack is output implicitly as result)

Answer 17

J, 18 bytes

[:>./(1#.3&u:%#)&>

Accepts list of boxed strings

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[:>./(1#.3&u:%#)&>
[:                  NB. cap, [: f g y -> f (g y)
                &>  NB. for each input item, result will be unboxed
     (         )    NB. monadic fork
              #     NB. length
         3&u:       NB. convert input to list of char codes
             %      NB. vectorized division
      1#.           NB. sum the result to compute average
  >./               NB. max item

Answer 18

PHP 8.x, 75 73 bytes

Creates an anonymous function that returns the expect values, hopefully.

fn($x)=>max(array_map(fn($z)=>array_sum(unpack('C*',$z))/strlen($z),$x));

---

-2 bytes thanks to Sʨɠɠan.

How does it work?

This uses the unpack('C*', ...) function to convert all characters into an unsigned char (0-255).
This assumes all characters are part of the ASCII table.

Then it calculates the average using array_sum([...])/strlen([...]).
The function array_sum() takes an array and returns the sum of all elements, and strlen() returns the length of the string.

The function max() takes the array with the averages and returns the highest value from it.

Anonymous functions implicitly return values.

Example usage

You need to assign this to a variable, or call with call_user_func() or call_user_func_array():

<?php
$fn = fn($x)=>max(array_map(fn($z)=>array_sum(unpack('C*',$z))/strlen($z),$x));

// Should output: float(110.4)
var_dump($fn(['hello', 'world', 'bye']));

You can try this here (with test examples): https://onlinephp.io/c/2a424

Answer 19

MATLAB, 39 31 bytes

@(s)max(cellfun(@(x)mean(x),s))

Matlab unfortunately doesn't convert strings ("") directly to ASCII values using the double function, and if converted to char type then a string array will be padded with spaces to make a proper matrix. As a result, the input is a cellstr type, which needs to be looped over using the cellfun function with an anonymous function that takes the average of the unicode values of the strings (''). The output is then the maximum of those values.

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edit: -8 bytes thanks to Guiseppe

Answer 20

T-SQL, 125 bytes

WITH C as(SELECT len(x)m,0b,*FROM @
UNION ALL SELECT m-1,b+ascii(right(x,m)),x
FROM C WHERE
m>0)SELECT max(b*1./len(x))FROM c

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Answer 21

Raku, 27 bytes

*».&{.ords.sum/.chars}.max

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Edit: as per advice from @Steffan

Answer 22

Vyxal G, 3 bytes

Cvṁ

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Answer 23

Ruby, 48 bytes

->l{l.map{_1.chars.sum(&:ord)/_1.size.to_f}.max}

As the division of two integers in ruby gives an integer, and what was expected was a float, it's long :/
And .fdiv is one byte longer

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Answer 24

Sequences, \$9 \log_{256}(96) \approx 7.41\$ bytes

[$v$aH]gM

Explanation

[$v$aH]gM  // Implicit list of string input
[     ]    // Loop through the input list:
 $v$       //   Get a list of ords of each character
    aH     //   Get the average and append to `g`
       g   // Push the list `g`
        M  // Get the maximum value
           // Implicit output

Answer 25

Mathematica, 29 bytes

Max[Mean/@ToCharacterCode@#]&

View it on Wolfram Cloud!

Answer 26

Fig, \$6\log_{256}(96)\approx\$ 4.939 bytes

]KemMC

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]KemMC
     C - Charcodes
  emM  - Mean of each
 K     - Sort
]      - Last

Answer 27

Java 8, 81 79 56 bytes

a->a.mapToDouble(s->s.chars().average().orElse(0)).max()

Input as a Stream<String> and output as an OptionalDouble.

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Explanation:

a->                  // Method with String-Stream parameter & OptionalDouble return-type
   a.map             //  Map over the input-Stream
        ToDouble(s-> //  and convert them to a Stream of Doubles
      s.chars()      //   Convert the String to an IntStream of its codepoint integers
       .average()    //   Then take the average of that IntStream (as OptionalDouble)
       .orElse(0))   //   And convert the OptionalDouble to double
                     //   (`.orElse(0)` is shorter than `.getAsDouble()`)
    .max()           //  After the map: leave the maximum (again as OptionalDouble)

Answer 28

BQN, 13 12 bytes

-1 byte inspired by Aiden Chow's APL answer

⌈´+´∘-⟜@¨÷≠¨

Anonymous tacit function; takes a list of strings and returns a float. Try it at BQN online!

Explanation

⌈´+´∘-⟜@¨÷≠¨
          ≠¨  Length of each string
         ÷    Divide the following by the above:
  +´∘           Sum of
     -⟜@       Subtract null character from (each character in)
        ¨       Each string
⌈             Get the maximum of the resulting list of averages

Answer 29

Go, 132 108 bytes

func f(S[]string)(m int){for a,s:=range S{a=0
for _,r:=range s{a+=int(r)}
if a=a*10/len(s);a>m{m=a}}
return}

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• -24 bytes by @Sʨɠɠan: return as an int, multiplied by 10

Answer 30

Japt -h, 9 bytes

Takes input as an array of character arrays.

®xc /ZlÃñ

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®xc /ZlÃñ     :Implicit input of 2D array of characters
®             :Map each sub-array Z
 x            :  Reduce by addition of
  c           :    Codepoints
    /         :  Divide by
     Zl       :    Length of Z
       Ã      :End map
        ñ     :Sort
              :Implicit output of last element