Lípínguapua dopo Pêpê

Question

Língua do Pê, or P Language, is a language game spoken in Brazil and Portugal with Portuguese. It is also known in other languages, such as Dutch and Afrikaans. (Wikipedia)

There are some dialects in this language game. The different languages the game is played with even have their own unique dialects. Some people are fluent in speaking P Language and the best can even translate any text to their preferred dialect on the spot!

P Language

In this challenge, we will use the Double Talk dialect.

To translate text into P Language, any sequence of vowels in the text is appended with a single p character followed by a copy of the sequence of vowels.

Challenge

Write a function or program that accepts a string as input and outputs its translation in P Language.

The input consists only of printable ASCII characters.
The output consists only of the translated input and optionally a trailing newline.
Vowels are any of the following characters aeiouyAEIOUY.
A sequence of vowels is delimited by any other character. The string "Aa aa-aa" has three vowel sequences.
Leading and trailing whitespace may optionally be omitted from the translated output string.

Examples

""                              =>   ""
"Lingua do Pe"                  =>   "Lipinguapua dopo Pepe"
"Hello world!"                  =>   "Hepellopo woporld!"
"Aa aa-aa"                      =>   "AapAa aapaa-aapaa"
"This should be easy, right?"   =>   "Thipis shoupould bepe eapeasypy, ripight?"
"WHAT ABOUT CAPS?"              =>   "WHApAT ApABOUpOUT CApAPS?"
"   Hi "                        =>   "   Hipi " or "Hipi"

The double quotes character " is used to delimit the input and output strings in the examples but obviously this character may also appear in any valid input string.

I'm not sure if anyone has welcomed you to the stack yet, so: welcome to Code Golf Stack Exchange! Nicely specified first challenge. :-) — Giuseppe, Aug 14 '19 at 15:05

Community · Accepted Answer · 2020-06-17 09:04:33Z

9

JavaScript (ES6), 35 bytes

s=>s.replace(/[aeiouy]+/gi,'$&p$&')

Try it online!

Where the special replacement pattern $& means matched substring.

edited Jun 17 '20 at 9:04

Community♦

1

answered Aug 13 '19 at 21:41

Arnauld

138k12 gold badges145 silver badges508 bronze badges

8

\$\begingroup\$ I did not know about $&. All these years, I've been wrapping the whole regex in a capture group. Who said code golf is impractical?! \$\endgroup\$ – recursive Aug 13 '19 at 22:50
2

\$\begingroup\$ Is $& the more common way? In Java it's $0 a.f.a.i.k., and Retina allows both. Didn't knew $& came from JavaScript. Or is it a .NET flavored regex, and JavaScript also uses it? \$\endgroup\$ – Kevin Cruijssen Aug 14 '19 at 6:37

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Zachary Cotton · Accepted Answer · 2019-08-14 05:21:20Z

9

Sed, 30, 25, 21, 19 Bytes

-5 Bytes Thanks to Arnauld!
-4 Bytes Thanks to Shaggy!
-2 Bytes Thanks to Leo Tenenbaum!

s/[aeiouy]\+/&p&/gi

Try it Online!

edited Aug 14 '19 at 5:21

answered Aug 13 '19 at 21:49

Zachary Cotton

6893 silver badges6 bronze badges

\$\begingroup\$ 19 bytes: tio.run/##BcGxCsIwEAbg/Z7id@mikkco0SWDYMEUB3U4yZEcBCONofTp4/… \$\endgroup\$ – pommicket Aug 14 '19 at 1:11

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

Japt, 10 bytes

r"%y+"_+ip

Try it

r"%y+"_+ip     :Implicit input of string
r              :Replace
 "%y+"         :RegEx /[aeiouy]+/gi
      _        :Pass each match through a function
       +       :  Append a copy of the match
        ip     :  Prepended with "p"

edited Jun 17 '20 at 9:04

Community♦

1

answered Aug 13 '19 at 21:36

Shaggy

30k3 gold badges25 silver badges83 bronze badges

\$\begingroup\$ Darn it, 2 minutes too late! I had the exact same answer, except it used Japt 2.0 short regexes "%v" -> \v \$\endgroup\$ – Gymhgy Aug 13 '19 at 21:39
\$\begingroup\$ @EmbodimentofIgnorance, you'd need \y instead of \v. \$\endgroup\$ – Shaggy Aug 13 '19 at 22:44

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Java 8, 40 bytes

s->s.replaceAll("(?i)[aeiouy]+","$0p$0")

Try it online.

Explanation:

s->                              // Method with String as both parameter and return-type
  s.replaceAll("(?i)[aeiouy]+",  //  Replace the regex matches
               "$0p$0")          //  With this replacement

Regex explanation:

(?i)[aeiouy]+                    // MATCH:
(?i)                             //  Enable case insensitivity
            +                    //  Match one or more
    [aeiouy]                     //  Adjacent vowel characters

$0p$0                            // REPLACEMENT:
$0                               //  The entire match (the vowel 'sequence')
  p                              //  Appended with a literal "p"
   $0                            //  Appended with the entire match again

wastl · Accepted Answer · 2019-08-13 22:52:33Z

4

Perl 5 `-p`, 20 bytes

s/[aeiouy]+/$&p$&/gi

Try it online!

answered Aug 13 '19 at 22:52

wastl

3,5539 silver badges33 bronze badges

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Jitse · Accepted Answer · 2019-08-14 12:03:15Z

Python 3, 55 bytes

lambda s:re.sub('([aeiouy]+)',r'\1p\1',s,0,2)
import re

Try it online!

Sans regex:

Python 3, 101 bytes

def f(s,q=''):i=s[:1];t=i in{*'aeiouyAEIOUY'};return(q+(q!='')*'p'+q+i)*0**t+(s and f(s[1:],(q+i)*t))

Try it online!

Python 3.8 (pre-release): 99 bytes

Explanation

Recursive function, accepting a string s and an optional argument q. If the first character of s (i) is a vowel, it is stored in the queue q. If not, a string is returned which is composed of q, the letter 'p', q again, the character i and the result of the recursive function with the first character of the string stripped off. Recursion stops when the function encounters an empty string s.

\$\begingroup\$ Love the solution without regex! \$\endgroup\$ – Maarten Bamelis Aug 14 '19 at 10:34 — Maarten Bamelis, Aug 14 '19 at 10:34

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

05AB1E, 22 20 bytes

.γžÁyå}vyžÁyнåi'py}J

05AB1E doesn't have any regexes unfortunately.
I don't really like the duplicated žÁyнå, but I'm currently a bit too busy to look for alternatives..

-2 bytes thanks to @Grimy for showing me a constant I didn't even knew existed (and was missing from the Wiki page.. >.>)

Try it online or verify all test cases.

Explanation:

.γ               # Group the characters in the (implicit) input-string by:
  žÁ             #  Push vowels builtin: "aeiouyAEIOUY"
    yå           #  And check if the current character is in this string
 }v              # After grouping: loop over each group `y`:
   y             #  Push group `y`
    žÁyнåi   }   #  If the first character of the group is a vowel:
          'p    '#   Push a "p"
            y    #   And push group `y` again
              J  #  Join everything on the stack together to a single string
                 # (after the loop, implicitly output the result)

edited Jun 17 '20 at 9:04

Community♦

1

answered Aug 14 '19 at 11:40

Kevin Cruijssen

85.4k7 gold badges109 silver badges303 bronze badges

\$\begingroup\$ žÁ instead of žO lets you drop the l. \$\endgroup\$ – Grimmy Aug 14 '19 at 12:09
\$\begingroup\$ @Grimy Maybe I should start using the info.txt file instead of the Wiki page.. -_-' \$\endgroup\$ – Kevin Cruijssen Aug 14 '19 at 12:29
\$\begingroup\$ They're both missing lots of commands, though not the same ones (hopefully. Sometimes I wonder if there are secret commands missing from both). \$\endgroup\$ – Grimmy Aug 14 '19 at 12:31
1

\$\begingroup\$ @Grimy Yeah, maybe we should use the source code instead of the info.txt or Wiki pages. ;) \$\endgroup\$ – Kevin Cruijssen Aug 14 '19 at 12:32

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Neil · Accepted Answer · 2019-08-13 22:41:32Z

2

Retina 0.8.2, 17 bytes

i`[aeiouy]+
$&p$&

Try it online! Link includes test cases. Explanation: Trivial regexp approach; the i flag turns on case insensitivity (Retina already defaults to a global match).

answered Aug 13 '19 at 22:41

Neil

114k9 gold badges54 silver badges227 bronze badges

add a comment |

Erik the Outgolfer · Accepted Answer · 2019-08-13 22:48:35Z

1

Jelly, 20 bytes

Żṁe€ØyŒg$$,j”pƊ€ÐeẎḊ

Try it online!

answered Aug 13 '19 at 22:48

Erik the Outgolfer

39k4 gold badges33 silver badges116 bronze badges

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Galen Ivanov · Accepted Answer · 2019-08-14 09:00:17Z

1

Red, 92 bytes

func[s][v: charset"aeiouyAEIOUY"parse s[any[to v copy t any v insert(rejoin["p"t])| skip]]s]

Try it online!

Of course Red's Parse is much more verbose than regex.

edited Aug 14 '19 at 9:00

answered Aug 14 '19 at 7:40

Galen Ivanov

18.2k3 gold badges17 silver badges56 bronze badges

1

\$\begingroup\$ More verbose than regex for sure, but perhaps that makes it more interesting as well. I like it! \$\endgroup\$ – Maarten Bamelis Aug 14 '19 at 9:07
\$\begingroup\$ @MaartenBamelis Thank you! Yes, Red (like Rebol) is an interesting language. \$\endgroup\$ – Galen Ivanov Aug 14 '19 at 9:16

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Giuseppe · Accepted Answer · 2019-08-14 14:48:54Z

1

SNOBOL4 (CSNOBOL4), 97 bytes

	I =INPUT
S	I ARB . L SPAN('AEIOUYaeiouy') . V REM . I	:F(O)
	O =O L V 'p' V	:(S)
O	OUTPUT =O
END

Try it online!

Vowel clusters match SPAN('AEIOUYaeiouy').

answered Aug 14 '19 at 14:48

Giuseppe

23.4k3 gold badges25 silver badges89 bronze badges

add a comment |

Khuldraeseth na'Barya · Accepted Answer · 2019-08-16 02:00:33Z

Stax, 17 bytes

àº`≈Zö=q╦ⁿ↔èblTï÷

Run and debug it at staxlang.xyz!

Unpacked (20 bytes) and explanation:

Vv'y+c^+:}'++{'pnL}R
                   R    Regex replace using
                        Pattern:
Vv'y+                     Push "aeiou", push "y", and concatenate
     c^+                  Copy, convert copy to all caps, and concatenate
        :}                Enclose in []
          '++             Push "+" and concatenate
                        And replacement:
             {    }       Block:
              'p            Push "p"
                n           Copy second item (matching substring) to top
                 L          Listify
                          Implicit concatenate
                        Implicit print

No case-insensitive regular expressions in Stax, and the vowel builtins don't include Y. The documentation won't tell you about using a block as a replacement, but it's a working feature nevertheless.

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Lípínguapua dopo Pêpê

P Language

Challenge

Examples

12 Answers 12

JavaScript (ES6), 35 bytes

Sed, 30, 25, 21, 19 Bytes

Japt, 10 bytes

Java 8, 40 bytes

Perl 5 `-p`, 20 bytes

Python 3, 55 bytes

Python 3, 101 bytes

Explanation

05AB1E, 22 20 bytes

Retina 0.8.2, 17 bytes

Jelly, 20 bytes

Red, 92 bytes

SNOBOL4 (CSNOBOL4), 97 bytes

Stax, 17 bytes

Unpacked (20 bytes) and explanation:

Your Answer

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Lípínguapua dopo Pêpê

P Language

Challenge

Examples

12 Answers 12

JavaScript (ES6), 35 bytes

Sed, 30, 25, 21, 19 Bytes

Japt, 10 bytes

Java 8, 40 bytes

Perl 5 -p, 20 bytes

Python 3, 55 bytes

Python 3, 101 bytes

Explanation

05AB1E, 22 20 bytes

Retina 0.8.2, 17 bytes

Jelly, 20 bytes

Red, 92 bytes

SNOBOL4 (CSNOBOL4), 97 bytes

Stax, 17 bytes

Unpacked (20 bytes) and explanation:

Your Answer

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Perl 5 `-p`, 20 bytes