Spoonerise words

Question

Given two words, spoonerise them by switching their initial consonant clusters. Fewest bytes wins.

plaster man -> master plan
blushing crow -> crushing blow
litigating more -> mitigating lore
strong wrangler -> wrong strangler
def ghi -> ghef di
few years -> yew fears

The consonants appearing before the first vowel (aeiou) are switched.

Input: Two lowercase strings that start with different consonants and each contain a vowel aeiou.

Output: The two spoonerised strings, in the right order.

For input and/or output, the two strings may also be in a list or the like, or as a single string with separator.

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**Leaderboard:**

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This is a near-dupe of this old question, but I'm hoping to write a spec that is clearer and more stable so that the old one can be closed as a dupe.

Answer 1

Vim, 18 keystrokes

/[aeiou]<CR>dbwvnhp0P

The crucial trick is that pasting over something in Visual mode actually puts the old contents of the visual selection in the paste buffer.

Answer 2

brainfuck, 238 207 bytes

,[[<+<+>>-]++++[<-------->-]-[<<-->>-----]<<+++++[----[----[------[------>>]]]]>[>>]>,]<<++++++++[<[++++<<]<++++<[++++<<]>>[>>]>>[>>]<-]<[<<]-[<+>---]>[>.>]<[<<]<<[<<]>[>>]>[.>>]<.<[<<]>[>.>]>[>>]>[>>]>[.>>]

Requires , to return 0 on EOF, 8-bit wrapping cells and the ability to move left from cell 0. In my experience, these are the most common default settings.

I'm pretty happy with this one. My first try was 314 bytes and this is certainly an improvement. :)

It works by storing every byte from input in two cells; one with the actual value of the byte and the other with the output of the following code when given (the value of the byte - 97):

[
    ----
    [
        ----
        [
            ------
            [
                ------>>
            ]
        ]
    ]
]

If the character is a consonant, it comes out of that with a non-zero value. If it's a vowel, it becomes 0. From there it's just a matter of finding where the second word starts and printing everything in order.

The 238 byte version of this program found the space character after collecting all the input. It was a headache because doing so involved creating a cell that contained 0 right where I was trying to loop over it. The way I solved that had the effect of subtracting 30 from every character in the first word and 32 from every character after it. A fairly large portion of the code was dedicated to handling that nonsense.

Now, 32 is subtracted from every character in the input loop, which is shorter and has a similar side effect that's easier to deal with. As a bonus, doing it this way allowed me to create my own space character in a shorter way: Instead of subtracting 139 from 171 (171 is what you get when you run a space through the vowel detector above), the loop that adds 32 to every character goes out of its way to also add 32 to the 171 cell. This costs four bytes there, but means I can subtract 171 from it later (instead of subtracting 139) for a net total of 3 bytes saved.

With comments:

For every input character
,[

  Make a copy
  [<+<+>>-]

  Subtract 32 from one
  ++++[<-------->-]

  Subtract 97 from the other
  -[<<-->>-----]<<+++++

  If nonzero:
  [

    Subtract 4
    ----

    If nonzero:
    [

      Subtract 4
      ----

      If nonzero:
      [

        Subtract 6
        ------

        If nonzero:
        [

          Subtract 6
          ------>>

        ]

      ]

    ]

  ]

>[>>]>,]

Add 32 to every character and the 171 that the space left behind
<<++++++++[<[++++<<]<++++<[++++<<]>>[>>]>>[>>]<-]

Go to the space
<[<<]

Subtract 171 from (32 plus 171)
-[<+>---]

~~~~~~~~~~~~~~~

Ready to print!

~~~~~~~~~~~~~~~

Print letters from the second word until the value to the left is zero
>[>.>]

Go to the beginning of the first word
<[<<]<<[<<]

Look at cells to the left of letters in the first word until reaching a zero
>[>>]

Print the rest of the letters in the first word
>[.>>]

Print a space
<.

Go to the beginning of the first word
<[<<]

Print letters from the first word until the value to the left is zero
>[>.>]

Go to the beginning of the second word
>[>>]

Look at cells to the left of letters in the second word until reaching a zero
>[>>]

Print the rest of the letters in the second word
>[.>>]

Answer 3

vim, 23

/[aeiou]<cr>"xd0wndb0Pw"xP

I wonder if vim is actually competitive in this challenge. Probably not with the golfing languages, but perhaps with Ruby/Python/Perl/etc.

/[aeiou]<cr>  Find the first vowel
"xd0          "Delete" (cut), to register "x, everything from here to BOL
w             Go to next word
n             Find next vowel
db            "Delete back" - delete from here to the beginning of the word
0             Go to BOL
P             Paste what we just "deleted" (cut) behind the cursor
w             Next word
"xP           Paste backwards from register "x

Answer 4

Python, 68 63 60 bytes

import re
lambda s:re.sub('([^aeiou]+|.+ )'*3,r'\3\2\1',s,1)

Try it online on Ideone.

How it works

The pattern string is repeated thrice (*3), resulting in the pattern

([^aeiou]+|.+ )([^aeiou]+|.+ )([^aeiou]+|.+ )

All instances will either match a non-empty string of non-vowels, or a non-empty string followed by a single space character.

The first instance will match the consonants at the beginning of the first word. Since + is greedy, it will attempt to match as many as possible, i.e., all up to the first vowel.

Because of this, the second instance will start at a vowel, so it will match the remainder of the first word, up to and including the space that separates the words.

Similarly to the first instance, the third will match all consonants at the beginning of the second word, resulting in a successful match for the entire pattern.

The raw string \3\2\1 (r prevents Python from replacing \3 etc. with control characters) reverses the order of the matches of the parenthesized patterns, i.e., it replaces them with the consonants at the beginning of the second word, then all characters from the first vowel of the first word up to the space, and finally the consonants at the beginning of the first word.

The final argument to sub (1) makes it return immediately after the first successful replacement, to avoid nonsensical replacements in the remainder of the second word. This is required since the pattern can match any string of three or more consecutive consonants.

Answer 5

JavaScript (ES6), 54 bytes

s=>s.replace(r=/\b[^aeiou ]+/g,(_,i)=>s.match(r)[+!i])

Explanation

s=>
  s.replace(
    r=/\b[^aeiou ]+/g,     // r = regex to match consonants
    (_,i)=>s.match(r)[+!i] // replace consonants with the consonants from the other word
  )

Test

var solution = s=>s.replace(r=/\b[^aeiou ]+/g,(_,i)=>s.match(r)[+!i])

<input type="text" id="input" value="plaster man" />
<button onclick="result.textContent=solution(input.value)">Go</button>
<pre id="result"></pre>

Answer 6

Python 3, 108 101 99 bytes

(No use of regex)

This function expects input via 2 arguments, e.g. f('blushing','crow'). Returns the new words in a tuple.

S=lambda s,p="":s[0]in"aeiou"and(p,s)or S(s[1:],p+s[0])
def f(x,y):a,b=S(x);c,d=S(y);return c+b,a+d

There are a lot of regex solutions, so I wanted to write something that didn't use Python's re library.

How it works

The only complicated part is the lambda expression S (the abbreviation means "Split before the first vowel"). It recursively "iterates" over the given word, moving one character at a time from the beginning of s (which starts with the whole word) to the end of p (which starts empty). At the first vowel encountered it returns (p,s), i.e. (prefix, suffix). Notice that that is the wrong order compared to the parameters!

I thought it made more sense for the returned order to be prefix, then suffix (because generally a prefix goes before a suffix). This order might make the a,b=S(x) code slightly easier to read.

But I had no choice of order in the lambda's parameters, so I couldn't define p before s. The first parameter, s, had to take the entire word because p had a default value, and default parameters go last. Doing that, I didn't need to call the function S with an empty string twice, and a few bytes could be saved. However, perhaps it was simply a bad decision to return prefix/suffix in the opposite order as it was used within the lambda expression.

As for the choice of lambda expression over function, it takes more bytes to say def S(s,p=""):return than S=lambda s,p="":. I can make that choice because Python has short-circuit evaluation, and the ternary operator. However, I cannot adequately explain how I used short-circuits; it's hard to reason about.

---

This is my first answer. I hope I did this correctly, and that there's value to posting a solution that cannot win.

Edits: Thank-you commenters: Reduced the byte count a bit, twice, and removed unnecessary info. Attempted to improve writing. Hopefully made no mistakes.

Answer 7

C#6, 115 bytes

string S(string a)=>System.Text.RegularExpressions.Regex.Replace(a,"([^aeiou]*)(.*) ([^aeiou]*)(.*)","$3$2 $1$4");

It's just a pain the namespace for regex is so long.

Answer 8

CJam, 27 24 22 bytes

2{l_{"aeiou"&}#/(N\}*o

I/O is one word per line. Try it online!

How it works

2{                 }*   Do the following twice:
  l                       Read a line from STDIN.
   _                      Push a copy.
    {"aeiou"&}#           Find the index of the first character that has a non-
                          empty intersection with the string "aeiou".
               /          Split the string into chunks of that size.
                (         Shift out the first chunk, i.e., all leading consonants.
                 N\       Swap the shifted out chunk with a linefeed.
                     o  Print the topmost stack item.

Answer 9

Perl, 33 bytes

s/([^aeiou]+)(.+ )((?1))/\3\2\1/

The above code is 32 bytes long and requires the -p switch (+1 byte).

The substitution is very similar to the one @DigitalTrauma's Retina answer, but this answer takes advantage of PCRE's recursive subpatterns.

Try it online on Ideone.

Answer 10

Retina, 34

([^aeiou]+)(.+ )([^aeiou]+)
$3$2$1

Try it online.

Answer 11

JavaScript ES6, 93 58 52 bytes

Saved 6 bytes thanks to ETHProductions!

x=>x.replace(/([^aeiou]+)(.+ )([^aeiou]+)/,"$3$2$1")

Test it! (ES6 only)

F=x=>x.replace(/([^aeiou]+)(.+ )([^aeiou]+)/,"$3$2$1");

console.log=_=>u.innerHTML+=_;

console.log(`plaster man -> master plan
blushing crow -> crushing blow
litigating more -> mitigating lore
strong wrangler -> wrong strangler
def ghi -> ghef di`.split`\n`.map(e=>(e=e.split` -> `,`${e[0]} => ${e[1]} ||> `+(F(e[0])==e[1]))).join`<br>`);

html { font-family: Consolas, monospace; }

<div id=u></div>

Answer 12

C, 255 201 199 bytes

I don't see a lot of C answers around here, so enjoy; Also, first time golfing, suggestions and critique are welcome.

#define S(w,p)p=strpbrk(w,"aeiou")-w
#define c ;strcpy(r
#define l c+strlen(v[1])+b+1
main(int q,char**v){char r[64],S(v[1],a),S(v[2],b)c,v[2])c+b,v[1]+a);strcat(r," ")l-a,v[1])l,v[2]+b);puts(r);}

If main() is not required we can save 24 bytes, getting to 179 bytes

#define S(w,p)p=strpbrk(w,"aeiou")-w
#define c ;strcpy(r
#define l c+strlen(x)+b+1
s(char*x,char*y){char r[64],S(x,a),S(y,b)c,y)c+b, x+a);strcat(r," ")l-a,x)l,y+b);puts(r);}

Ungolfed:

void spoonerise(char* w1, char* w2)
{
    char rt[64];

    int p1 = strpbrk(w1, "aeiou")-w1;
    int p2 = strpbrk(w2, "aeiou")-w2;

    strcpy(rt, w2);
    strcpy(rt+p2, w1+p1);

    strcat(rt, " ");

    strcpy(rt+strlen(w1)+p2+1-p1, w1);
    strcpy(rt+strlen(w1)+p2+1, w2+p2);

    puts(rt);
}

EDIT: Thanks to feersum's suggestion I saved 54 bytes. =D

Answer 13

sed, 38 characters

s/([^aeiou]+)(.+ )([^aeiou]+)/\3\2\1/

Using extended regular expression from the Retina solution.

The above is 37 characters long and requires the -r switch (+1 character).

Example:

$ cat input | tee /dev/tty | sed -r 's/([^aeiou]+)(.+ )([^aeiou]+)/\3\2\1/'
plaster man
blushing crow
litigating more
strong wrangler
def ghi
master plan
crushing blow
mitigating lore
wrong strangler
ghef di

Answer 14

Python 2, 364 352 269 251 bytes

EDIT: Thanks so much to @Cyoce for helping me golf 83 bytes!

Please help me golf this! Anyway, at least I am the first Python answerer. And I hope I can beat Java, if there ever is a Java answer!

i=input()
f=l=''
A='aeiou'
for j in A:
 I=i[0];J=I.find(j)
 if ~J:
  if f:
   if f>I[0][:J]:f=I[:J];break
  else:f=I[:J]
for j in A:
 Y=i[1];J=Y.find(j)
 if ~J:
  if l:
   if l>Y[:J]:l=Y[:J];break
  else:l=Y[:J]
print I.replace(f,l,1),Y.replace(l,f,1)

Try it here

Answer 15

Japt, 26 25 bytes

Fortunately, I added a vowel class to Japt's regex features a few days ago. Unfortunately, I didn't add a non-vowel class or a way around the double backslashes in regex strings.

#¿s4 £Uf"[^\\v]+|.+
?" gX

The ¿ should be the raw char U+0093. Input is a multi-line string, one word/line. Try it online!

EDIT: I've now added the non-vowel class \V and a way around \\ (with %), so this code now works for 21 bytes: (Try it online)

#¿s4 £Uf"%V+|.+
?" gX

How it works

                     // Implicit: U = input string
#¿s4                 // Convert the char code (147) into a base-4 string "2103".
     £               // Map each char X in this string to:
      Uf"[^\\v]+|.+  //  Match each group in U of non-vowels or non-newlines,
?"                   //  with an optional trailing newline.
   gX                //  Get the item at position X in the resulting array.
                     // Implicit: output last expression

Old version (26 bytes):

UrA="\\b[^\\v ]+"@UfA gÂ!Y

Try it online!

How it works

UrA="\\b[^\\v ]+"@UfA gÂ!Y  // Implicit: U = input string
  A="\\b[^\\v ]+"           // Set variable A to this regex string, which matches
                            // a word boundary followed by one or more non-vowels.
UrA              @          // Take U and replace each match X of A and its index Y with:
                  UfA       //  Take U.match(A).
                      gÂ!Y  //  Get the item at index ~~!Y (converts 0 to 1, anything else to 0).
                            // Implicit: output last expression

Answer 16

Python 3, 100 (or 99) bytes

import re
def f(*g):a,b=[re.split("[aeiou]",r)[0] for r in g];return b+g[0][len(a):],a+g[1][len(b):]

Played with a few versions but can't seem to get it below. Can get it down 99 bytes by using def f(g) instead so it takes a list of strings rather than two seperate strings, but I prefer the two arg roue.

The alternative is equal length:

import re
s=re.split
def f(x,y):v="[aeiou]";a,b=s(v,x)[0],s(v,y)[0];return b+x[len(a):],a+y[len(b):]

I tried replace, like @TanMath uses, but I couldn't get it any shorter. Also, TanMath can get his answer shorter by a byte by also using "[aeiou]" instead of "[aeiou]+" because we only need to match the single instances. Finally, the implementation of input() seems to have changed between py2 and py3 - it automatically evaluates stdin as a string.

Answer 17

C#6, 165 bytes

string S(string a,string b){int i=V(a),j=V(b);return b.Remove(j)+a.Substring(i)+" "+a.Remove(i)+b.Substring(j);}int V(string s)=>s.IndexOfAny("aeiou".ToCharArray());

Expanded:

string S(string a,string b){
    int i=V(a),
        j=V(b);
    return b.Remove(j)+a.Substring(i)+" "+a.Remove(i)+b.Substring(j);
}
int V(string s)=>s.IndexOfAny("aeiou".ToCharArray());

Answer 18

𝔼𝕊𝕄𝕚𝕟, 24 chars / 42 bytes

ïē/⟮(⁅ᶌ]+)⟯(.+ )Ⅰ/,`⑶⑵⑴`

Try it here (Firefox only).

If you need help comprehending this, this translates to ES6 as

input.replace(/([^aeiou]+)(.+ )([^aeiou]+)/g,'$3$2$1')

Answer 19

PowerShell, 52 bytes

"$args "-replace('(.*?)([aeiou]\S+) '*2),'$3$2 $1$4'

e.g. 
PS C:\scripts\> .\Spoonerise.ps1 'blushing crow'
crushing blow

It's a regex replace with four capture groups; with:

• String multiplication to expand:
  • ('(.*?)([aeiou]\S+) '*2) to '(.*?)([aeiou]\S+) (.*?)([aeiou]\S+) '
• The "$args " forces the args array into a string, and adds a trailing space so the trailing space in the regex won't break it.

Answer 20

JavaScript (ES6), 120 107 102 101 99 92 bytes

• Thanks downgoat!

This takes into account if the parameters were an object like this, and backward:
var a = {x: "man", y:"plaster"]}

a.b=(e=>e.slice(e.search`[aeiou]`));with(a){c=b(x),d=b(y),x=x.replace(c,d);y=y.replace(d,c)}

Answer 21

Python, 129 108 105 109 bytes

This program takes in a list of words like this ["master","plan"]

EDIT: Thanks @Volatility

EDIT: Now using re.split

import re
a=re.split
c='[aeiou]+'
i,j=input()
f=a(c,i)[0]
l=a(c,j)[0]
print i.replace(f,l,1),j.replace(l,f,1)

This answer uses regex as most of them do.

Try it here

Answer 22

Java 8, 343 Bytes

Here you have your first Java-Answer. Not that experienced with golfing, so every suggestion is appreciated!

java.util.function.Function;public class C{public static void main(String[] a){Function<String, Integer>f=s->{for(int i=0;i<s.length();++i)if("aeiou".contains(s.substring(i,i+1)))return i;return 0;};int i1=f.apply(a[0]),i2=f.apply(a[1]);System.out.println(a[1].substring(0,i2)+a[0].substring(i1)+" "+a[0].substring(0,i1)+a[1].substring(i2));}}

Ungolfed:

public class SpooneriseWords {
    public static void main(String[] args)
    {       
        Function<String, Integer> f = s -> {
            for(int i = 0; i < s.length(); ++i)
                if("aeiou".contains(s.substring(i, i + 1))) 
                    return i;
            return 0;
        };

        int i1 = f.apply(args[0]);
        int i2 = f.apply(args[1]);
        System.out.println(args[1].substring(0, i2) + args[0].substring(i1));
        System.out.println(args[0].substring(0, i1) + args[1].substring(i2));
    }
}

Answer 23

Octave, 96 bytes

Thanks to Octave's inline assignments and 'index-anywhere' functionality, this entire thing is just the definition of a single, anonymous function. Basically, we reconstruct the spoonerized string while storing the cut-off points in a and b. I'm especially happy with the inline f function, which finds the cut-off point, and prevented me from having to use the whole 'find(ismember(a,b),1)' thing twice. Also, no regex :).

@(p,q)[q(1:(a=(f=@(m)find(ismember(m,'aeoui'),1))(q))-1),p(b=f(q):end),32,p(1:b-1),q(a:end)];

Answer 24

Python (no regexes), 85 bytes

t=lambda s:s[1]in'aeiou'or-~t(s[1:])
lambda a,b:(b[:t(b)]+a[t(a):],a[:t(a)]+b[t(b):])

Sample run:

>>> t=lambda s:s[1]in'aeiou'or-~t(s[1:])
>>> lambda a,b:(b[:t(b)]+a[t(a):],a[:t(a)]+b[t(b):])
<function <lambda> at 0x7f495e534398>
>>> _('plaster', 'man')
('master', 'plan')

t is a recursive function that computes the index of the earliest vowel after the first character in its argument s. If the second character s[1] is a vowel, it evaluates to True, which has int value 1. Otherwise it makes a recursive call with the first character removed, and adds 1 to the resulting index using -~ (two’s complement of one’s complement). Finally, the results of t are used as indices for string slicing to compute the spoonerism.

Answer 25

TeaScript, 27 bytes

xg/(\w+)(.+ )(\w+)/,"$3$2$1

\w is actually [^aeiou].

Answer 26

PowerShell, 61 bytes

"$args"-replace'([^aeiou]+)(.*)\s([^aeiou]+)(.*)','$3$2 $1$4'

Uses regex to swap the first non-vowel characters of each word

Answer 27

Elixir, 143 117 Bytes

s=fn w->String.split_at(w,elem(hd(Regex.run(~r/[aeiou]/,w,return: :index)),0))end
{v,w}=s.(a)
{x,y}=s.(b)
[x<>w,v<>y]

Split up the two strings (a, b) on the first vowel and construct new strings to return.

EDIT: Got a few bytes off by using pattern matching instead of clunky elem calls to get the values out of the tuples.

Answer 28

Java, 147 Bytes

I'm assuming just a function is fine too.

String s(String[]i){String r="[aeiou]",f=i[0].split(r)[0],s=i[1].split(r)[0];return s+i[0].substring(f.length())+" "+f+i[1].substring(s.length());}

split(regex) unfortunately consumes the delimiter, which means that I have to use substring to get the suffixes.

Answer 29

Pyth, 30 28 bytes

Takes input and gives output as a list of the two words.

ACm,Kh:d"[aeiou]"3.-dKQ+V_GH

Try it online.

Answer 30

Brainfuck, 141 bytes

,[[[<+>>+<-]----[>-----<--]>+[++++++[++++++[++++[++++[,>]]]]]<]<[>+<-]>>>+[,[[<+
>>+<-]++++[>--------<-]]>]>,]<<<<[[<]<[<]>[.,+>]+<]>[-[+.,]>]

Formatted:

,
[
  [
    [<+> >+<-]
    ----[>-----<--]>+
    [
      not u
      ++++++
      [
        not o
        ++++++
        [
          not i
          ++++
          [
            not e
            ++++
            [
              not a
              ,>
            ]
          ]
        ]
      ]
    ]
    <
  ]
  <[>+<-]>>>+
  [
    ,
    [
      [<+> >+<-]
      ++++[>--------<-]
    ]
    >
  ]
  >,
]
<<<<
[
  [<]<[<]
  >[.,+>]
  +<
]
>[-[+.,]>]

Expects the two words separated by a space, with or without a trailing newline.

Try it online.

This takes the input and stores it on the tape with three zero cells added: one before the first vowel of each word, and one after the space between the two words. For example, plaster man becomes:

pl\x00aster \x00m\x00an

The three fragments m, aster, and pl are then printed by looping backwards through the tape while creating dummy \x01 cells, and the final an is printed by going forwards and ignoring the dummy cells.