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Context

As a conlanger, I am interested in creating a uniform, naturalistic language. One of the tricks is to create vocabulary according to certain structures of words. An example from English: In English, we have the word “tap” structured consonant-vowel-consonant. Usually, this means that there are many other words of this structure: “cat”, “dog”, “rock”, “fog”, “good”, etc.

Task

As input, you have:

  • an array C containing strings: consonants (C is the first letter in the word consonants). Identical consonants cannot be repeated on this list. For example, this list cannot contain ['b', 'b'].
  • an array V containing strings: vowels (V is the first letter of the word vowels). Identical vowels cannot be repeated on this list.
  • string S, which contains something like this "CVCCV" (any combination of "C" and "V")

Your task is to replace “C” in the string with a randomly taken string from array C and replace “V” in the string with a randomly taken string from array V and return (or display) this string.

"randomly" is defined as all possibilities having an equal chance of being selected. This is a kind of simplification: in real languages as well as in conlangs, ​​there are very frequent (for example 'r' in English) and not very frequent sounds but this is just a code-golf.

Rules

This is code-golf so the lowest byte count wins.

Examples

Input:
C = ['p', 'b', 't', 'd', 'k', 'g', 'm', 'n', 'w']
V = ['a', 'o', 'u', 'i', 'e', 'ä']
S = 'CVCCV'

Output:
pakto

Input:
C = ['p', 'b', 't', 'd', 'k', 'g', 'v', 's', 'r']
V = ['a', 'o', 'u', 'i', 'e', 'ä', 'ᵫ']
S = 'CVVCCVCCV'

Output:
koebrᵫvtä
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0

12 Answers 12

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5
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05AB1E, 4 bytes

Takes input as S, [V,C]

Çè€Ω

Try it online!

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0
4
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Jelly,  5  4 bytes

-1 thanks to Neil!

OịX€

A dyadic Link accepting a list of characters, S, on the left and a list of two lists of characters, [C, V], on the right which yields a list of characters.

Try it online!

If we could take S as a list of 1s and 0s we'd have the three byte solution ịX€.

How?

OịX€ - Link: S, [C,V]
O    - ordinals (of S)          i.e. 'C':67 'V':86
 ị   - index into [C,V] (vectorises)      C      V   [Jelly indexing is 1-indexed and modular]
   € - for each:
  X  -   random choice
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3
  • \$\begingroup\$ I like both solutions! The 5 and the 3 byte ones +1 \$\endgroup\$
    – RGS
    Feb 22, 2020 at 16:54
  • \$\begingroup\$ What does Jelly do if an index is out of range? (Charcoal automatically reduces it modulo the range, so I don't need to extract the bit manually.) \$\endgroup\$
    – Neil
    Feb 22, 2020 at 17:42
  • \$\begingroup\$ @Neil you just saved a byte - thanks! I even noted in my code break-down that "Jelly indexing is 1-indexed and modular". \$\endgroup\$
    – Jonathan Allan
    Feb 22, 2020 at 17:47
3
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Python 3, 92 68 61 bytes

Takes input as S, C, V:

lambda s,*w:[random.choice(w[a>"u"])for a in s]
import random

You can try it online! How it works:

The *w is used so that the lists of consonants and vowels are packed in a list w.

When we do w[a > "u"] we are essentially checking if a (a character of s) is "v" or not. If it is, then a > "u" returns True, which indexes as 1 into w. If a is "c", then a > "u" returns False, which indexes as 0.

We use those indices to retrieve the correct list of letters from w and then choose randomly from that list, picking the list of choices.

Thanks @Arnauld for saving one byte and to @Jonathan Allan for citing this meta, saving me 7 bytes.

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3
  • \$\begingroup\$ This is what I would've done in my C solution had it not been for that the lists of consonants and vowels are not of equal length. \$\endgroup\$
    – S.S. Anne
    Feb 22, 2020 at 17:57
  • \$\begingroup\$ I believe that you should be able to take s as a list of characters and return the same (see this meta) and as such can replace the "".join(...) with [...]. \$\endgroup\$
    – Jonathan Allan
    Feb 22, 2020 at 19:48
  • \$\begingroup\$ @JonathanAllan thanks for this reference! Shaved 7 bytes :) \$\endgroup\$
    – RGS
    Feb 22, 2020 at 20:51
3
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APL+WIN, 35 34 bytes

One byte saved thanks to Adam

Prompts for consonants and vowels as continuous strings and the desired output as a comma separated string

c←⍞⋄v←⍞⋄C←',c[?⍴c]'⋄V←',v[?⍴v]'⋄⍎⎕

Try it online! Courtesy of Dyalog Classic

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10
  • \$\begingroup\$ If you liked this golf question, please vote to reopen it. \$\endgroup\$
    – USERNAME GOES HERE
    Feb 23, 2020 at 14:01
  • \$\begingroup\$ If you don't like this golf challenge, please retain its closed state. \$\endgroup\$
    – user92069
    Feb 23, 2020 at 14:23
  • \$\begingroup\$ This is the first time I have been lobbied from both sides of a question closure. Personally I had no problem understanding the question and as I got two up votes I assume at least two people agreed with my interpretation and my answer \$\endgroup\$
    – Graham
    Feb 23, 2020 at 15:04
  • \$\begingroup\$ By using ⍞ instead of ⎕ you can specify that input shouldn't have quotes, and then use ⎕ instead of ⍎⎕ to save a byte: Try it online! \$\endgroup\$
    – Adám
    Feb 25, 2020 at 13:48
  • \$\begingroup\$ @Adám Thanks for the suggestion but whilst it works in Dyalog Classic I cannot get it to work in APL+WIN. If I run the code as you suggest it simply outputs the concatenated code string which needs the second ⍎ to execute. \$\endgroup\$
    – Graham
    Feb 25, 2020 at 16:28
2
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Ruby, 45 40 bytes

->s,*l{s.bytes.map{|x|l[x%2].sample}*''}

Try it online!

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2
  • \$\begingroup\$ I didn't think that ruby could outgolf python. Great! \$\endgroup\$
    – USERNAME GOES HERE
    Feb 22, 2020 at 16:59
  • 1
    \$\begingroup\$ Ruby code is usually shorter than the equivalent python code. \$\endgroup\$
    – G B
    Feb 23, 2020 at 7:14
2
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Charcoal, 7 bytes

⭆η‽§θ℅ι

Try it online! Link is to verbose version of code. Takes the input in the form [V, C], S where each value can be either a string or an array of characters as desired. Explanation:

 η      `S`
⭆       Map over elements/characters
    θ   `[V, C]`
   §    Cyclically indexed by
     ℅  Ordinal of
      ι Current element/character
  ‽     Random element/character
        Implicitly print
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0
2
\$\begingroup\$

C# (Visual C# Interactive Compiler), 66 bytes

(C,V,s)=>s.Select(x=>(x<68?C:V).OrderBy(l=>Guid.NewGuid()).Last())

Try it online!

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2
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Japt, 6 bytes

ËcgV ö

Takes input as S, [V,C]

Try it

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6
  • \$\begingroup\$ @a'_' Ok.6 more to go... \$\endgroup\$
    – USERNAME GOES HERE
    Feb 23, 2020 at 14:28
  • \$\begingroup\$ @a'_' By the way, did you like this code-golf question? \$\endgroup\$
    – USERNAME GOES HERE
    Feb 23, 2020 at 14:29
  • \$\begingroup\$ @VictorVosMottorthanksMonica I loved it. But it was blatant that I can't re-open it, because I don't have enough reputation to do so. \$\endgroup\$
    – user92069
    Feb 23, 2020 at 14:30
  • 2
    \$\begingroup\$ @VictorVosMottorthanksMonica, liking a challenge is not a valid reason to VTR it. Address the issues that led to it being closed and, once done so to the community's satisfaction, it will be reopened. \$\endgroup\$
    – Shaggy
    Feb 23, 2020 at 19:21
  • 3
    \$\begingroup\$ @a'_', similarly, not liking a challenge is not grounds for closing it or leaving it closed. There have been many challenges here over the years I haven't personally liked, as there have been for everyone - if you don't like a challenge, just skip it and move on to the next one. \$\endgroup\$
    – Shaggy
    Feb 23, 2020 at 19:26
2
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W f j, 4 bytes

Hmm, let's do a pure-ASCII port of that.

C[gr

Explanation

C    % Convert the string to its codepoints
 [   % Index into the other list
   r % For every indexed item:
  g  % "g"et a random item in this list

Flag:f % Flatten the output list
Flag:j % Join the flattened output list
```
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1
  • 1
    \$\begingroup\$ @KevinCruijssen I absolutely can't understand the challenge, so feel free to post your answer so that I can get a better understanding of this. (I guess I'd keep it deleted because I don't know how to fix it.) \$\endgroup\$
    – user92069
    Feb 27, 2020 at 11:35
1
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PHP, 68 bytes

function($C,$V,$s){for(;$r=$s[$i++];)echo$$r[rand(0,count($$r)-1)];}

Try it online!

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1
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C (gcc), 92 bytes

f(s,c,v,m,n)char*s,**c,**v;{for(srand(&s);*s;)printf("%s",*s++/86?v[rand()%n]:c[rand()%m]);}

Takes as input a string (s), an array of consonant strings (c), an array of vowel strings v, the length of c (m), and the length of v (n).

Could use a temporary variable for the result of rand but it wouldn't save me anything.

Try it online!

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0
1
\$\begingroup\$

JavaScript (ES6),  65  64 bytes

(C,V,s)=>s.replace(/./g,x=>eval(x).sort(_=>Math.random()-.5)[0])

Try it online!

Or 55 bytes if we can use an array of characters as I/O.

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3
  • \$\begingroup\$ The eval trick is really nice! Sadly enough I don't think I can use it in my answer to save some bytes :( \$\endgroup\$
    – RGS
    Feb 22, 2020 at 16:49
  • \$\begingroup\$ 55 seems fine unless OP has explicitly said not since I==O and if it quacks like a duck, it is a duck (i.e. strings are lists of characters) \$\endgroup\$
    – Jonathan Allan
    Feb 22, 2020 at 19:46
  • \$\begingroup\$ @a'_' I think I'm privileged enough to hammer this question, which I think would be inappropriate either way. \$\endgroup\$
    – Neil
    Feb 24, 2020 at 10:19

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