Uiua,₁₈ 15 bytes

/+♭⊞="ij.::;!?"

Try it:Uiua pad

Explanation: Sum of the flattened equality table of the input with"ij.::;!?".

• \$\begingroup\$/+♭ is equivalent to⧻⊚ for booleans, as demonstrated inmy answer.\$\endgroup\$

  janMakoso

  – janMakoso

  2024-10-15 00:04:31 +00:00

  CommentedOct 15, 2024 at 0:04
• \$\begingroup\$@janMakoso Clever, upvoted.\$\endgroup\$

  noodle person

  – noodle person

  2024-10-15 01:27:04 +00:00

  CommentedOct 15, 2024 at 1:27

Google Sheets, 45 bytes

=len(regexreplace(A1,"(:)|[^ij.;!?]","$1$1"))

Put the text string in cellA1 and the formula in cellB1.

_{(-1 thanks to emanresu A)}

• \$\begingroup\$.split`:`.join`::` saves a few bytes on the latter (edit: actually, replacing'::' with10 is better)\$\endgroup\$

  emanresu A

  – emanresu A

  2024-10-10 20:04:03 +00:00

  CommentedOct 10, 2024 at 20:04
• \$\begingroup\$@emanresuA thanks, but using abetter regex saves many more.\$\endgroup\$

  doubleunary

  – doubleunary

  2024-10-10 20:23:12 +00:00

  CommentedOct 10, 2024 at 20:23

Python 3,42 37 bytes

lambda a:sum(map(a.count,'ij.;!?::'))

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• \$\begingroup\$You can save some bytes withmap\$\endgroup\$

  xnor

  – xnor

  2024-10-11 02:16:26 +00:00

  CommentedOct 11, 2024 at 2:16

JavaScript (Node.js), 50 bytes

x=>[...'ij.;!?::'].flatMap(c=>x.split(c)).length-8

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Perl 5-p0, 20 bytes

$_=y/ij.;!?://+y/://

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• \$\begingroup\$This doesn't seem to work for inputs with newlines (per TIO at least). Do we care?\$\endgroup\$

  Greg Martin

  – Greg Martin

  2024-10-11 19:22:00 +00:00

  CommentedOct 11, 2024 at 19:22
• 1
  \$\begingroup\$@GregMartin It does if you add a-0 on the command line. There's no change needed to the code itself to support it.\$\endgroup\$

  Xcali

  – Xcali

  2024-10-11 19:39:34 +00:00

  CommentedOct 11, 2024 at 19:39
• 2
  \$\begingroup\$@GregMartin Byconsensus, printable ASCII doesn’t include newlines.\$\endgroup\$

  Jordan

  – Jordan

  2024-10-11 19:51:46 +00:00

  CommentedOct 11, 2024 at 19:51

JavaScript (V8), 44 bytes

t=>t.replace(/(:)|[^ij.;!?]/g,'$1$1').length

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Port of myGoogle Sheets answer.

_{(-1 thanks to emanresu A)}

• \$\begingroup\$44\$\endgroup\$

  emanresu A

  – emanresu A

  2024-10-10 20:55:50 +00:00

  CommentedOct 10, 2024 at 20:55

Vyxal, 14 bytes

`ij.::;!?`fvO∑

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            O  # Count in input          fv   # Each of`ij.::;!?`     # String "ij.::;!?"             ∑ # sum counts

R, 43 bytes

\(s)nchar(gsub("(:)|[^ij.;!?]","\\1\\1",s))

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Port of@doubleunary's Google Sheets answer.

R,48 44 bytes

Edit: -4 bytes porting@Luis Mendo's MATL answer.

\(s)sum(outer(utf8ToInt("ij.;!?::"),s,`==`))

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Takes input as a vector of character codes.

Zsh, 28 bytes

<<<${#${1//:/..}//[^ij;.!?]}

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Replace: with.., then remove all non-dot characters and print the string length

Jelly, 14  13bytes

ċⱮ“::ij.;!?”S

A monadic Link that accepts a list of characters and yields the number of dots.

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How?

ċⱮ“::ij.;!?”S - Link: list of characters, T  “::ij.;!?”  - list of characters -> "::ij.;!?" Ɱ            - map with:ċ             -   {T} count {char}            S - sum

Arturo, 41 bytes

$=>[replace&":"{..}|match"[ij.;!?]"|size]

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Vyxal 3.3.0d, 11 bytes

ᵒ="ij.::;!?

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I downgraded to Vyxal 3.3.0 because that's the most recent version which still supportsNilad moving, a feature which was removed because it makes debugging super annoying but makes this program a byte shorter because it lets me move the string to the end of the line instead of having it at the front, saving a byte on the closing quotation mark.

The logic is super simple: Take the outer product of equality with the stringij.::;!?, andd deeply sums the result.

• 1
  \$\begingroup\$Who removed their upvote when I made my solution shorter lol\$\endgroup\$

  noodle person

  – noodle person

  2024-10-11 20:36:54 +00:00

  CommentedOct 11, 2024 at 20:36

Japt-mx,171615 14bytes

This could be12 bytes in v2 but there seems to be a bug withS.è(x) that only counts a maximum of 1 occurrence ofx.

Takes input as an array of characters.

"!.::;?ij"oU Ê

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"..."oU Ê     :Implicit map of each U in the input array"..."         :The necessary characters, with : duplicated     oU       :Keep only those characters that appear in U        Ê     :Length              :Implicit output of sum of resulting array

Uiua, 14 bytes

⧻⊚⊞="ij.;!?::"

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Explanation

• ⊞=: The outer product of equality between the literal"ij.;!?::", and the input
• ⧻⊚: count where elements equal one (true)

K (ngn/k), 19 bytes

+//"ij.;!?::"=\:

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iogii, 13 bytes

rq__"ij.;!?::

run it here!!

Repeat the input so that the following eQual check is essentially a cartesian product instead of element by element, then sum and sum again.

The string of dot characters is placed at the end so it can be unterminated, it becomes the left arg to equal using iogii's rotation trick.

Racket, 60 bytes

(λ(s'(curryr regexp-match* s)`length)(+`'"[ij.;!?:]"`'":"))

Takes a string and runs a regexp twice, nothing fancy, but I'm pleased with the garble of symbols. Heavily relies on default arguments, whichwrite and some formatting elucidates:

(λ (s [quote (curryr regexp-match* s)] [quasiquote length])  (+ (quasiquote (quote "[ij.;!?:]")) (quasiquote (quote ":"))))

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MATL, 15 bytes

0h'ij.;!?::'!=z

Try it online! Orverify all test cases.

How it works

0h         % Implicit input. Attach character 0. This is needed to avoid empty string'ij.;!?::' % Push this string. Note that ':' appears twice!          % Transpose into a column vector=          % Matrix of all pair-wise comparisonsz          % Number of non-zeros. Implicit display

JavaScript (ES6), 49 bytes

Expects an array of characters.

a=>a.map(c=>t+=c==":"?2:/[ij.;!?]/.test(c),t=0)|t

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APL+WIN, 29 bytes

Prompts for string:

(+/s∊'ij.;!?.')+2×+/(s←⎕)∊':'

Try it online! Thanks to Dyalog Classic

05AB1E, 13bytes

"!.::;?ij"S¢O

Try it online orverify all test cases.

Explanation:

"!.::;?ij"    # Push this string (note the "::")          S   # Convert it to a list of characters           ¢  # Count each of those characters in the (implicit) input-string            O # Sum the counts together              # (after which the result is output implicitly)

Unfortunately, the shortest compressed version of this character-list I could find is 1 byte longer:•м€2Q¸Ü•žQÅв¢O. Also, there are some minor variations possible for the same byte-count:":!.;?ij"Ć€¢O or"!.;?ij:"¤ª¢O.

Red, 69 bytes

func[s][n: 0 foreach c s[n: n + case[find"ij.;!?"c 1 c =#":"2 1 0]]n]

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Bash+coreutils, 33 bytes

tr -cd ij.\;!?<<<${1//:/..}|wc -c

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Is also optimal for zsh, I think.

explanation:

                 ${1//:/..} #replace all : with .. in first argument              <<<           #and pass it to standard input of...tr -cd ij.\;!?              # from stdin, output only characters specified to stdout                           |wc -c # pass stdout to stdin of: count chars in stdin#implicit output stdout to terminal

Retina 0.8.2,14 13 bytes

:..[!.;?ij]

Try it online! Link includes test cases. Explanation::s are "rotated", then the relevant characters are counted which produces the desired result. Edit: Saved 1 byte thanks to @m90.

• \$\begingroup\$If: is replaced with.. instead, then: can be omitted from the last line.\$\endgroup\$

  m90

  – m90

  2024-10-11 15:29:15 +00:00

  CommentedOct 11, 2024 at 15:29

Haskell, 34 bytes

f x=sum[1|c<-x,v<-"ij.;!?::",c==v]

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J, 19 bytes

1#.1#.'ij.::;!?'=/]

Works the same as most.

Attempt This Online!

1#.1#.'ij.::;!?'=/]                  ]   NB. input string      'ij.::;!?'      NB. string literal                =/    NB. equality table   1#.                NB. sum rows1#.                   NB. sum resulting list

Pyth, 12 bytes

sm/"ij.::;!?

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Javascript -290242 240 bytes

Counting the dots on a canvas.

s=>[...s].map(a=>{for(x=document.createElement`canvas`.getContext`2d`,x.font="40Q A",x.fillText(a,10,50),d=x.getImageData(i=0,0,99,99).data;++i<1e4;)t+=d[j=i*4+3]>9&&g(j,p=0)<25},t=0,g=i=>d[i]?g(i+4,p++,d[i]=0,g(i-4),g(i+396),g(i-396)):p)|t

• -45 -46 bytes thanks to @Shaggy!

JSFiddle

• 1
  \$\begingroup\$You're always allowed to do an intentionally suboptimal approach and still golf that approach.\$\endgroup\$

  Unrelated String

  – Unrelated String

  2024-10-13 15:31:59 +00:00

  CommentedOct 13, 2024 at 15:31
• 1
  \$\begingroup\$Some very quick savings to get you down to 245, but there's a lot more room for improvement.\$\endgroup\$

  Shaggy

  – Shaggy

  2024-10-14 10:07:26 +00:00

  CommentedOct 14, 2024 at 10:07
• 2
  \$\begingroup\$Forgot to include this above:40px can be40Q\$\endgroup\$

  Shaggy

  – Shaggy

  2024-10-14 16:16:38 +00:00

  CommentedOct 14, 2024 at 16:16
• 1
  \$\begingroup\$Doesn't work fine here, get 0 onij.\$\endgroup\$

  l4m2

  – l4m2

  2024-10-15 00:31:05 +00:00

  CommentedOct 15, 2024 at 0:31
• 1
  \$\begingroup\$<30 works fine\$\endgroup\$

  l4m2

  – l4m2

  2024-10-15 01:04:24 +00:00

  CommentedOct 15, 2024 at 1:04

Charcoal, 14 bytes

ＩΣＥ!.::;?ij№θι

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

   !.::;?ij     Literal string `!.::;?ij`  Ｅ             Map over characters           №    Count of             ι  Current character            θ   In input string Σ              Take the sumＩ               Cast to string                Implicitly print

Dart, 120 bytes

main(List<String>y,{i=0,j=0,x}){x=y[0];for(;i<x.length;i++){if('ij.;!?'.contains(x[i]))j++;if(x[i]==':')j+=2;}print(j);}

Try it online!

Only takes one string at a time.

• \$\begingroup\$85 bytes\$\endgroup\$

  jdt

  – jdt

  2024-10-14 18:09:27 +00:00

  CommentedOct 14, 2024 at 18:09

• code-golf
• string
• printable-ascii