Alphabet completion rate

Question

Introduction

How much of the English alphabet does a given string use? The previous sentence uses 77%. It has 20 unique letters (howmucftenglisapbdvr), and 20/26 ≃ 0.77.

Challenge

For an input string, return the percentage of letters of the English alphabet present in the string.

• The answer can be in percentage or in decimal form.

• The input string can have upper and lower case, as well as punctuation. However you can assume they have no diacritics or accentuated characters.

Test cases

Input

"Did you put your name in the Goblet of Fire, Harry?" he asked calmly.

Some valid outputs

77%, 76.9, 0.7692

Input:

The quick brown fox jumps over the lazy dog

All valid outputs:

100%, 100, 1

The expected output for "@#$%^&*?!" and "" is 0.

Answer 1

Python 3, 42 bytes

lambda s:len({*s.upper()}-{*s.lower()})/26

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We filter all the non-alphabetic characters out of the string by taking the (set) difference of the uppercase and lowercase representations. Then, we take the length and divide by 26.

Python 3, 46 bytes

lambda s:sum(map(str.isalpha,{*s.lower()}))/26

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Count the unique alphabetic (lowercase) characters, and divide by 26. In Python 2 it would require 3 more characters; two for changing {*...} to set(...), and one for making 26 a float: 26., to avoid floor division.

Python 3, 46 bytes

lambda s:sum('`'<c<'{'for c in{*s.lower()})/26

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Same length, essentially the same as the previous one, but without "built-in" string method.

Answer 2

MATL, 8 bytes

2Y2jkmYm

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Explanation

2Y2    % Predefined literal for 'abcdefghijklmnopqrstuvwxyz'
j      % Explicitly grab input as a string
k      % Convert to lower-case
m      % Check for membership of the alphabet characters in the string. 
       % Results in a 26-element array with a 1 where a given character in 
       % the alphabet string was present in the input and a 0 otherwise
Ym     % Compute the mean of this array to yield the percentage as a decimal
       % Implicitly display the result

Answer 3

05AB1E, 8 7 6 bytes

lASåÅA

-1 byte thanks to @LuisMendo.

Try it online or verify a few more test cases.

6 bytes alternative provided by @Grimy:

láÙg₂/

Try it online or verify a few more test cases.

Both programs output as decimal.

Explanation:

l       # Convert the (implicit) input-string to lowercase
 AS     # Push the lowercase alphabet as character-list
   å    # Check for each if it's in the lowercase input-string
        # (1 if truthy; 0 if falsey)
    ÅA  # Get the arithmetic mean of this list
        # (and output the result implicitly)

l       # Convert the (implicit) input-string to lowercase
 á      # Only leave the letters in this lowercase string
  Ù     # Uniquify it
   g    # Get the amount of the unique lowercase letters by taking the length
    ₂/  # Divide this by 26
        # (and output the result implicitly)

Answer 4

Octave / MATLAB, 33 bytes

@(s)mean(any(65:90==upper(s)',1))

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Explanation

@(s)                               % Anonymous function with input s: row vector of chars
             65:90                 % Row vector with ASCII codes of uppercase letters
                    upper(s)       % Input converted to uppercase
                            '      % Transform into column vector
                  ==               % Equality test, element-wise with broadcast. Gives a
                                   % matrix containing true and false
         any(                ,1)   % Row vector containing true for columns that have at
                                   % least one entry with value true
    mean(                       )  % Mean

Answer 5

C# (Visual C# Interactive Compiler), 56 49 bytes

a=>a.ToUpper().Distinct().Count(x=>x>64&x<91)/26f

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-6 bytes thanks to innat3

Answer 6

K (oK), 19 15 bytes

Solution:

1%26%+/26>?97!_

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Explanation:

Convert input to lowercase, modulo 97 ("a-z" is 97-122 in ASCII, modulo 97 gives 0-25), take unique, sum up results that are lower than 26, and convert to the percentage of 26.

1%26%+/26>?97!_ / the solution
              _ / lowercase
           97!  / modulo (!) 97
          ?     / distinct
       26>      / is 26 greater than this?
     +/         / sum (+) over (/)
  26%           / 26 divided by ...
1%              / 1 divided by ...

Notes:

• -1 bytes thanks to ngn, 1-%[;26] => 1-1%26%
• -3 bytes inspired by ngn #(!26)^ => +/26>?

Answer 7

PowerShell, 55 52 bytes

($args|% *per|% t*y|sort|gu|?{$_-in65..90}).count/26

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First attempt, still trying random ideas

EDIT: @Veskah pointed out ToUpper saves a byte due to the number range, also removed extra () and a space

Expansion:
($args|% ToUpper|% ToCharArray|sort|get-unique|where{$_-in 65..90}).count/26

Changes string to all loweruppercase, expands to an array, sorts the elements and selects the unique letters (gu needs sorted input), keep only characters of ascii value 97 to 122 (a to z) 65 to 90 (A to Z), count the total and divide by 26 for the decimal output

Answer 8

APL (Dyalog Extended), 10 bytes^SBCS

Anonymous tacit prefix function. Returns decimal fraction.

26÷⍨∘≢⎕A∩⌈

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⌈ uppercase

⎕A∩ intersection with the uppercase Alphabet

≢ tally length

∘ then

26÷⍨ divide by twenty-six

Answer 9

Perl 6, 27 24 bytes

-3 bytes thanks to nwellnhof

*.uc.comb(/<:L>/).Set/26

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Answer 10

Bash and Gnu utils (81 78 68 60 42 bytes)

bc -l<<<`grep -io [a-z]|sort -fu|wc -l`/26

-8 bytes thanks to @wastl

-18 bytes thanks to Nahuel using some tricks I didn't know:

• sort -f and grep -i ignore case
• sort -u is a replacement for | uniq

Answer 11

R, 47 bytes

function(x)mean(65:90%in%utf8ToInt(toupper(x)))

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Converts to upper case then to ASCII code-points, and checks for values 65:90 corresponding to A:Z.

Answer 12

Retina 0.8.2, 45 bytes

T`Llp`ll_
+`(.)(.*\1)
$2
.
100$*
^
13$*
.{26}

Try it online! Link includes test cases. Explanation:

T`Llp`ll_

Lowercase letters and delete punctuation.

+`(.)(.*\1)
$2

Deduplicate.

.
100$*

Multiply by 100.

^
13$*

Add 13.

.{26}

Integer divide by 26 and convert to decimal.

Answer 13

APL (Dyalog Extended), 8 bytes

⌹∘≤⍨⎕A∊⌈

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loosely based on Adám's answer

⌈ uppercase

⎕A∊ boolean (0 or 1) vector of length 26 indicating which letters of the English Alphabet are in the string

⌹∘≤⍨ arithmetic mean, i.e. matrix division of the argument and an all-1 vector of the same length

Answer 14

Java 8, 62 59 bytes

s->s.map(c->c&95).distinct().filter(c->c%91>64).count()/26.

-3 bytes thanks to @OlivierGrégoire.

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Explanation:

s->                     // Method with IntStream as parameter and double return-type
  s.map(c->c&95)        //  Convert all letters to uppercase
   .distinct()          //  Uniquify it
   .filter(c->c%91>64)  //  Only leave letters (unicode value range [65,90])
   .count()             //  Count the amount of unique letters left
    /26.                //  Divide it by 26.0

Answer 15

Charcoal, 11 bytes

Ｉ∕ＬΦβ№↧θι²⁶

Try it online! Link is to verbose version of code. Output is as a decimal (or 1 for pangrams). Explanation:

  Ｌ         Length of
    β       Lowercase alphabet
   Φ        Filtered on
     №      Count of
        ι   Current letter in
      ↧     Lowercased
       θ    Input
 ∕          Divided by
         ²⁶ Literal 26
Ｉ           Cast to string
            Implicitly printed

Answer 16

Batch, 197 bytes

@set/ps=
@set s=%s:"=%
@set n=13
@for %%c in (A B C D E F G H I J K L M N O P Q R S T U V W X Y Z)do @call set t="%%s:%%c=%%"&call:c
@cmd/cset/an/26
@exit/b
:c
@if not "%s%"==%t% set/an+=100

Takes input on STDIN and outputs a rounded percentage. Explanation:

@set/ps=

Input the string.

@set s=%s:"=%

Strip quotes, because they're a headache to deal with in Batch.

@set n=13

Start with half a letter for rounding purposes.

@for %%c in (A B C D E F G H I J K L M N O P Q R S T U V W X Y Z)do @call set t="%%s:%%c=%%"&call:c

Delete each letter in turn from the string. Invoke the subroutine to check whether anything changed, because of the way Batch parses variables.

@cmd/cset/an/26

Calculate the result as a percentage.

@exit/b
:c

Start of subroutine.

@if not "%s%"=="%t%" set/an+=100

If deleting a letter changed the string then increment the letter count.

Answer 17

Pepe, 155 138 bytes

rEeEeeeeeEREeEeEEeEeREERrEEEEErEEEeReeReRrEeeEeeeeerEEEEREEeRERrErEErerREEEEEeREEeeRrEreerererEEEEeeerERrEeeeREEEERREeeeEEeEerRrEEEEeereEE

Try it online! Output is in decimal form.

Explanation:

rEeEeeeeeE REeEeEEeEe # Push 65 -> (r), 90 -> (R)
REE # Create loop labeled 90 // creates [65,66,...,89,90]
  RrEEEEE # Increment (R flag: preserve the number) in (r)
  rEEEe # ...then move the pointer to the last
Ree # Do this while (r) != 90

Re # Pop 90 -> (R)
RrEeeEeeeee rEEEE # Push 32 and go to first item -> (r)
REEe # Push input -> (R)
RE RrE # Push 0 on both stacks, (r) prepend 0
rEE # Create loop labeled 0 // makes input minus 32, so the
    # lowercase can be accepted, since of rEEEEeee (below)
  re # Pop 0 -> (r)
  rREEEEEe REEee # Push item of (R) minus 32, then go to next item 
  RrE # Push 0 -> (R)
ree # Do while (R) != 0

rere # Pop 0 & 32 -> (r)
rEEEEeee # Remove items from (r) that don't occur in (R)
         # Remove everything from (r) except the unique letters
rE # Push 0 -> (r)
RrEeee # Push reverse pointer pos -> (r)
REEEE # Move pointer to first position -> (R)
RREeeeEEeEe # Push 26 -> (R)
rRrEEEEee reEE # Divide it and output it

Answer 18

K (oK), 19 bytes

1%26%26-#(!26)^97!_

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J, 30 bytes

26%~26-u:@(97+i.26)#@-.tolower

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Answer 19

Retina, 57 46 35 bytes

.
$L
[^a-z]

D`.
.
100*
^
13*
_{26}

-11 bytes taking inspiration from @Neil's trick of adding unary 13 before dividing.
Another -11 bytes thanks to @Neil directly.
Rounds (correctly) to a whole integer.

Try it online.

57 46 40 bytes version which works with decimal output:

.
$L
[^a-z]

D`.
.
1000*
C`_{26}
-1`\B
.

Same -11 bytes as well as an additional -6 bytes thanks to @Neil.

Outputs with one truncated^† decimal after the comma (^† i.e. \$0.1538\$ (\$\frac{4}{26}\$) is output as 15.3 instead of 15.4). This is done by calculating \$\lfloor{\frac{1000 × \text{unique_letters}}{26}\rfloor}\$ and then inserting the decimal dot manually.

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Explanation:

Convert all letters to lowercase:

.
$L

Remove all non-letters:

[^a-z]

Uniquify all letters:

D`.

Replace every unique letter with 1000 underscores:

.
1000*

Count the amount of times 26 adjacent underscores fit into it:

C`_{26}

Insert a dot at the correct place:

-1`\B
.

Answer 20

Julia 1.0, 34 bytes

s->sum('a':'z'.∈lowercase(s))/26

Uses the vectorized version of the ∈ operator, checking containment in the string for all characters in the range from a to z. Then sums over the resulting BitArray and divides by total number of possible letters.

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Answer 21

C, 96 bytes

float f(char*s){int i=66,l[256]={};for(;*s;)l[1+*s++&~32]=1;for(;i<92;*l+=l[i++]);return*l/26.;}

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Answer 22

Perl 5 -MList::Util=uniq -p, 24 bytes

$_=uniq(lc=~/[a-z]/g)/26

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Answer 23

Stax, 9 bytes

░║üy$}╙+C

Run and debug it

Answer 24

Jelly, 8 bytes

ŒuØAe€Æm

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Explanation

Œu       | Convert to upper case
  ØAe€   | Check whether each capital letter is present, returning a list of 26 0s and 1s
      Æm | Mean

Answer 25

JavaScript (Node.js), 45 bytes

s=>~-s.match(/$|([a-z])(?!.*\1)/ig).length/26

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---

JavaScript (Node.js), 47 bytes

s=>(s.match(/([a-z])(?!.*\1)/ig)||[]).length/26

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Answer 26

Pyth, 9 bytes

cl@rw0G26

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c      26 # Float division by 26 (alternative: replace 26 by lG)
 l        #   Length of
  @       #     intersection of
   r 0    #       lowercase
    w     #         input
      G   #       and the lowercase alphabet

Answer 27

Swift 5.9, 105 100 89 bytes

let p={Float((65...90).filter(($0+"").uppercased().map(\.asciiValue).contains).count)/26}

Using Float instead of Double saved a byte.

Answer 28

Japt, 9 bytes

;CoU Ê/26

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;CoU Ê/26     :Implicit input of string U
;C            :Lowercase alphabet
  oU          :Remove the characters not included in U, case insensitive
     Ê        :Length
      /26     :Divide by 26

Answer 29

Python 2, 57 bytes

lambda i:len(set(o for o in i.lower()if o.isalpha()))/26.

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A bit longer than the Python 3 answer from ArBo but posted as a different approach in Python 2 anyway.

Answer 30

Ruby -n, 38 34 bytes

-4 bytes from @historcrat!

p (?A..?Z).count{|c|~/#{c}/i}/26.0

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