Eh, codegolf shmodegolf

Question

This challenge is about implementing Shm-reduplication, originating in Yiddish, where one takes a word, duplicates it, and replaces the first syllable in the second word with "Shm" in order to indicate that one does not care. Some examples include:

• "Isn't the baby cute?", "Eh, baby shmaby"
• "Come buy from my store, we have a sale!", "Pff, sale shmale"
• "I got this smartphone yesterday, isn't it cool?", "Meh, smartphone shmartphone"

Rules

In the real world there are a bunch of special cases, but for this challenge we will use rules that describe a simplified version of the full linguistic phenomenon that is more suitable for programming:

1. Consonants up to first vowel are replaced by "shm".
2. Prepend shm to words beginning with vowels.
3. If a word contains no vowels it is returned unchanged. No vowel, no shm.
4. The vowels are: a, e, i, o, u.
5. The input will be restricted to lowercase letters.
6. The output should be a lowercase string containing the reduplicated version of the input. If the input is "string" the output should be "shming".

This is code golf, shortest code wins!

Example solution (ungolfed python)

This is an example of code that would solve the challenge:

def function_shmunction(string):
    vowels = "aeiou"
    for letter in string:
        if letter in vowels:
            index = string.index(letter)
            shming = "shm" + string[index:]
            return shming
    return string

Test cases

• function -> shmunction
• stellar -> shmellar
• atypical -> shmatypical
• wwwhhhat -> shmat
• aaaaaaaaaa -> shmaaaaaaaaaa
• lrr -> lrr
• airplane -> shmairplane
• stout -> shmout
• why -> why

An answer has been accepted since there haven't been any new responses in a while, but feel free to add more!

Answer 1

sed, 29 bytes

s/[^aeiou]*\([aeiou]\)/shm\1/

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Answer 2

Jelly, 16 12 bytes

œlØḄṭ⁸¹?“shm

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-4 thanks to ovs remembering there's a strip builtin

œlØḄ            Strip all leading consonants.
    ṭ   “shm    Prepend "shm"
      ¹?        if there's anything left, otherwise
     ⁸          give the original word.

Answer 3

Haskell, 50 bytes

snd.break(`elem`"aeiou")>>=(?)
""?s=s
r?_="shm"++r

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Thanks to Roman Czyborra for -2 bytes

Answer 4

sed, 25 bytes

/[aeiou]/s/[^aeiou]*/shm/

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/[aeiou]/ is a conditional that only runs the substitution s/[^aeiou]*/shm/ if there is a vowel in the line. This prevents needing to capture the first vowel and then using a backreference as in the more obvious s/[^aeiou]*\([aeiou]\)/shm\1/.

Answer 5

Stax, 11 bytes

ÄFï¥░$º{═┘ç

Run and debug it

1. Left-trim consonants.
2. If non-empty, prepend "shm".
3. Else restore original input.

Answer 6

Zsh, 36 bytes

<<<${(S)1/#(#b)*([aeiou])/shm$match}

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• <<<: print
• ${1}: the input...
  • //: replace the first match
    • #: matching only at the beginning of the string
    • (#b): activate backreferences, so the () group gets stored in the $match variable
    • (S): use the shortest possible match, rather than the longest
    • *: anything
    • (): store this match in the variable:
      • [aeiou]: followed by a vowel
  • with shm$match: the string shm, plus the matched vowel

$match is actually an array which contains all the backreference groups, but since there's only one group, we don't need to access any specific element with [1]

Answer 7

Python 3.8 (pre-release), 62 bytes

lambda s:(q:=s.lstrip('bcdfghjklmnpqrstvwxyz'))and'shm'+q or s

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Answer 8

05AB1E, 28 13 bytes

žNSÛ©ai®…shmì

Try it online or verify all test cases.

Explanation:

žNSÛ           # Trim all leading consonants of the (implicit) input-string
    ©          # Store this new string in variable `®` (without popping)
     ai        # Pop, and if any letters are left (thus the input-string contains
               # a vowel):
       ®       #  Push string `®` again
        …shmì  #  Prepend "shm"
               #  (implicitly print it as result)
               # (implicit else)
               #  (implicitly output the implicit input-string)

Answer 9

Powershell, 36 bytes

$args-replace('^.*?([aeiou])','shm$1')

Explanation

Simple replace with a named group reference.

+4

Thanks caird coinheringaahing

-5

Thanks mazy

Answer 10

Python, 53 bytes

lambda s:re.sub("^.*?([aeiou])","shm\\1",s)
import re

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Another regex answer.

---

Alternative:

Python, 53 bytes

lambda s:re.sub(".*?(?=[aeiou])","shm",s,1)
import re

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Answer 11

Haskell, 48 bytes

g.break(`elem`"aeiou")
g(s,"")=s
g(_,r)="shm"++r

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Answer 12

Japt, 15 bytes

r"^.*?%v"ÈÌi`¢m

Try it (Includes all test cases)

r"^.*?%v"ÈÌi`¢m     :Implicit input of string
r                   :Replace
 "^.*?%v"           :  RegEx /^.*?[aeiou]/gi
         È          :  Pass each match through a function
          Ì         :    Last character
           i        :    Prepend
            `¢m     :      Compressed string "shm"

Answer 13

Retina, 19 bytes

Just a regex replacement, there might be some language feature which allows for a shorter solution.

^.*?(?=[aeiou])
shm

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Answer 14

SNOBOL4 (CSNOBOL4), 49 bytes

	I =INPUT
	I BREAK('aeiou') ='shm'
	OUTPUT =I
END

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BREAK "matches zero or more characters provided they are not in the set of characters in the argument string. That is, it matches up to, but not including, a character from the argument string."

Answer 15

C (gcc), 81 bytes

i;f(char*s){for(i=0;*s-"aeiou"[++i%6];i%6||s++);printf("shm%s"+3*!*s,s-i*!*s/6);}

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Answer 16

Vyxal, 13 bytes

k⁰øl:[«88«p|_

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Port of Jelly.

How?

k⁰øl:[«88«p|_
k⁰øl           # Strip leading consonants
    :[         # If there's anything left:
      «88«p    # Prepend "shm"
           |_  # Else, pop, returning the implicit input

Answer 17

JavaScript, 36 bytes

s=>s.replace(/.*?([aeiou])/,"shm$1")

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Answer 18

Excel, 149 bytes

=IFERROR(REPLACE(A1,1,FIND("$",SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"a","$"),"e","$"),"i","$"),"o","$"),"u","$"))-1,"shm"),A1)

102 bytes just to find the vowels. If anybody knows a better way to do this. please share.

From inside out:

• SUBSTITUTE vowels with $
• Find the index of the first $
• Replace chars before this index with shm
• If no vowels, return input.

Answer 19

Perl 5 + -p, 20 bytes

Uses the ; separator (as ; is implicitly added via -p).

s;.*?(?=[aeiou]);shm

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---

Perl 5 + -lF/^[^aeiou]+/ -M5.10.0, 17 bytes

Slightly more cheaty with the regex in the flags, but saves a few bytes

say@F?(shm,@F):$_

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Answer 20

C (gcc), 125 113 bytes

-8 bytes thanks to @ceilingcat

#define T&&*s-
main(s,v)char**v,*s;{for(s=*++v;*s T'a'T'e'T'i'T'o'T'u'||!(*v=memcpy(s-3,"smh",3));++s);puts(*v);}

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Answer 21

Charcoal, 25 bytes

≔⌕Ｅθ№aeiouι¹η¿⊕η«shm✂θη»θ

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

≔⌕Ｅθ№aeiouι¹η

Count the number of vowels in each character and find the first 1 i.e. the index of the first vowel in the input.

¿⊕η

If there was in fact a vowel, then...

«shm✂θη»

... output the literal string shm followed by the input sliced starting at that index.

θ

Otherwise just output the input string.

Answer 22

Java 8, 40 bytes

s->s.replaceAll("^.*?([aeiou])","shm$1")

Port of the JavaScript and Retina answers.

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Or a minor alternative:

s->s.replaceAll("^.*?(?=[aeiou])","shm")

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Explanation:

s->             // Method with String as both parameter and return-type:
  s.replaceAll( //  Modify and return the String: Regex-replace all
    "...",      //  these matches
    "...")      //  with these replacements

Regex-explanation 1:

^.*?([aeiou])    # Match:
^                #  At the start of the string
 .*              #  Match zero or more characters
   ?             #  Which are optional, to give other matches precedence
     [aeiou]     #  Followed by a vowel
    (       )    #  captured in capture group 1

shm$1            # Replacement:
shm              #  Literal "shm"
   $1            #  And the vowel of capture group 1

Regex-explanation 2:

^.*?([aeiou])    # Match:
^                #  At the start of the string
 .*              #  Match zero or more characters
   ?             #  Which are optional, to give other matches precedence
    (?=       )  #  Followed by a positive (non-matching) look-ahead to:
       [aeiou]   #   A vowel

shm              # Replacement:
shm              #  Literal "shm"

Minor note: the replaceAll with ^ is basically the same as a replaceFirst without the ^, but 1 byte shorter: try it online.

Answer 23

Pip, 15 bytes

a~XVaH$(:"shm"a

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Explanation

a~XVaH$(:"shm"a
a                Command-line argument
 ~               Find first match of
  XV             regex `[aeiou]`
      $(         Index of that match
    aH           Prefix of cmdline arg containing that many characters
        :        Set to
         "shm"   that string
                 If a vowel was not found, $( is nil, which means no assignment is done
                 and a's value remains unchanged
              a  Autoprint the (possibly changed) value of a

Answer 24

APL (Dyalog Extended), 31 bytes

{⍵≢⍛≡n←⌊/⍵⍳'aeoiu':⍵⋄'shm',n↓⍵}

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Answer 25

Haskell, 58 57 bytes

f s|g s>""="shm"++g s|1>0=s
g=dropWhile(`notElem`"aeiou")

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-1 byte thanks to Wheat Wizard ♦

Answer 26

V (Eh, vim shmim), 23 bytes

:%s/.\{-}\ze[aeiou]/shm

vim non-greedy regex with look-ahead.

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Same in “magic very very” mode:

23 bytes

:%s/\v.{-}[aeiou]@=/shm

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Further 3 bytes can be saved by replacing :%s/ with V’s specific í

Answer 27

Go, 103 bytes

import."regexp"
func f(s string)string{return MustCompile(`^.*?([aeiou])`).ReplaceAllString(s,"shm$1")}

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Answer 28

tinylisp 2, 64 61 bytes

(\(S)(: T(drop-while(. !(p contains?"aeiou"))S)(? T(,"shm"T)S

Try it at Replit. Example session:

tl2> (\(S)(: T(drop-while(. !(p contains?"aeiou"))S)(? T(,"shm"T)S
(() (S) (: T (drop-while (. ! (p contains? "aeiou")) S) (? T (, "shm" T) S)))
tl2> (def shm _)
shm
tl2> (map shm (list "codegolf" "stellar" "airplane" "lrr"))
("shmodegolf" "shmellar" "shmairplane" "lrr")

Explanation

Port of Unrelated String's Jelly answer, among others.

First, we need a function that tests whether a character is not a vowel:

(. !(p contains?"aeiou"))
(.                      ) ; Compose
   !                      ; Logical negation
    (p                 )  ; with the partial application of
       contains?          ; the contains? function
                "aeiou"   ; to "aeiou"

Then our main function is as follows (where (...) represents the above function):

(\(S)(: T(drop-while(...)S)(? T(,"shm"T)S)))
(\                                         ) ; Lambda function
  (S)                                        ; that takes a string S:
     (: T                                 )  ;  In what follows, let T be
         (drop-while     S)                  ;   Drop characters from S while
                    (...)                    ;   they are not vowels
                           (? T          )   ;  Is T truthy (nonempty)?
                               (,"shm"T)     ;  If so, concatenate "shm" with T
                                        S    ;  If not, return the original string S

Answer 29

Nibbles, 15.5 bytes

?,;>~$-~?"aeiou"$:"shm"@@

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Nibbles has no built-in for vowels, which makes it slightly worse than other golfing languages for this.

Explanation

?         If
,          non-empty
;           x :=
>~           drop-while
$             from input
-~            not
? "aeiou"      contained in "aeiou"
$              it
          then
: "shm"    prepend "shm"
@          to x
          else
@          input

Answer 30

Factor, 70 61 45 bytes

[ R/ ^[^aeiou]*(?=[aeiou])/ "shm"re-replace ]

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