Niedawno masz nowy telefon, ale nie podoba ci się sposób, w jaki wibruje, zdecydowałeś, że chcesz tworzyć własne wzorce wibracji. Napisałeś więc program, w którym użyłeś słów kluczowych long, shorti pauseżeby twój telefon wibrował zgodnie z tymi słowami kluczowymi.

Zadanie

Stworzyć mały program, który akceptuje ciąg long, shortoraz pausei wyprowadza kolejny ciąg reprezentujący fonetyczną dźwięk telefonu wibrującym;Rrrr - Rr

longDźwięki to Rrrr
shortDźwięki to Rr
(Sprawa obudowy)
pauseto myślnik, -
wszystkie dźwięki są rozdzielane myślnikiem z otaczającymi spacjami' - '

Przypadki testowe

wejście:    long long short long short
wyjście:Rrrr - Rrrr - Rr - Rrrr - Rr

wejście:   long long long short short short
wyjście:Rrrr - Rrrr - Rrrr - Rr - Rr - Rr

wejście:   short short short pause short short short
wyjście:Rr - Rr - Rr - - - Rr - Rr - Rr

wejście:   long short short long long pause short short
wyjście:Rrrr - Rr - Rr - Rrrr - Rrrr - - - Rr - Rr

To jest pytanie w golfa kodu, więc odpowiedzi będą oceniane w bajtach, przy najmniejszej liczbie wygranych bajtów.

Pyke , 22 20 bajtów

cFh.o6.&\R*\-|l4)J" - 

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c                      -  split(input, " ")
 Fh.o6.&\R*\-|l4)      -  for i in ^:
  h                    -        ^[0]
   .o                  -       ord(^)
     6.&               -      ^ & 6
        \R*            -     ^
           \-|         -    ^ or "-"
              l4       -   ^.title()
                 J" -  - " - ".join(^)

Sednem tej odpowiedzi jest przekształcenie ["long", "short", "pause"]w [4, 2, 0]. Pobiera kod pierwszej litery każdego słowa i ANDma 6. 6. Na szczęście przypadek przekształca się w wartości, których szukamy. (Przeszukałem kilka innych dłuższych rozwiązań, zanim znalazłem to). Gdy to zrobisz, możemy dalej przekształcić tę listę liczb całkowitych ["RRRR", "RR", ""], mnożąc naszą liczbę wewnętrzną, przez "R"którą następnie zamienia się ["RRRR", "RR", "-"]i na końcu tytuł, aby ją zdobyć ["Rrrr", "Rr", "-"]. Następnie dołączamy do wynikowej listy przez" - "

JavaScript, 70 63 bajtów

2 bajty zapisane dzięki Łukaszowi

a=>a.replace(/./g,a=>[['Rr','rr','-',' - ']['onp '.search(a)]])

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Haskell , 71 66 59 bajtów

g 'o'="Rr"
g 'n'="rr"
g 'p'="-"
g ' '=" - "
g _=""
f=(g=<<)

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No tak, =<<jest concatMap.

Wykorzystuje fakt, że "long"i "short"oba mają list o.

JavaScript (ES6), 65 59 bajtów

s=>s.split` `.map(x=>x<'m'?'Rrrr':x<'q'?'-':'Rr').join` - `

let f =

s=>s.split` `.map(x=>x<'m'?'Rrrr':x<'q'?'-':'Rr').join` - `

console.log(f("long long short long short")); // => Rrrr - Rrrr - Rr - Rrrr - Rr
console.log(f("long long long short short short")); // => Rrrr - Rrrr - Rrrr - Rr - Rr - Rr
console.log(f("short short short pause short short short")); // => Rr - Rr - Rr - - - Rr - Rr - Rr
console.log(f("long short short long long pause short short")); // => Rrrr - Rr - Rr - Rrrr - Rrrr - - - Rr - Rr

05AB1E , 33 27 25 21 bajtów

#εÇн6&'m×™'-)éθ}… - ý

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Wyjaśnienie

#                       # split input on spaces
 ε             }        # apply to each
  Çн                    # get the character code of the head
    6&                  # AND with 6
      'm×               # repeat "m" this many times
         ™              # title case
          '-)           # wrap in a list with "-"
             éθ         # get the longest string       
                … - ý   # join to string using " - " as separator

Zapisane 3 bajty wykorzystujące AND 6sztuczki z muddyfish za PYKE odpowiedź

Python 2, 76 69 64 bytes

lambda s:' - '.join('Rrrr'[:ord(b[0])&6]or'-'for b in s.split())

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Alternates:

Python 2, 76 69 bytes

lambda s:' - '.join(['Rrrr','-','Rr'][ord(b[1])%3]for b in s.split())

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Python 2, 69 bytes

lambda s:' - '.join('-Rrrr'['o'in b:8-ord(b[1])%9]for b in s.split())

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Mathematica, 81 bytes

StringReplace[#,{"long"->"Bzzz -","short"->"Bz -","pause"->"- -"}]~StringDrop~-2&

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Retina, 31 29 bytes

1 byte saved thanks to Dom Hastings
1 byte saved thanks to Neil

[^lhp ]

l
hrr
h
Rr
p
-
 
 - 

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This is different approach than @DomHastings's answer.

Perl 5, 39 bytes

37 bytes code + 2 for -pa.

$_=join" - ",map/p/?"-":Bz.zz x/l/,@F

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R, 77 bytes

cat(c('Rrrr','Rr','-')[match(scan(,''),c('long','short','pause'))],sep=' - ')

Takes input through STDIN, checks whether the input matches long, short or pause and swaps the matches for Rrrr, Rr or - respectively.

This is then printed with - padded with spaces as separator, matching the desired output.

Röda, 73 57 47 46 40 44 bytes

f&a{a~=*`s\w+|l;Rr;ong;rr;p\w+;-; ; - `/";"}

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+4 bytes due to rule change (must use Rrrr instead of any 4 letter variant).

Previous code:

{[[split()|["Bzzz"]if[_="long"]else["Bz"]if[_1="short"]else["-"]]&" - "]}

C (gcc), 93 77 76 bytes

-2 bytes thanks to Scepheo!
-1 byte thanks to Cyoce!

Takes a NULL terminated **char or equivalent as an input.

f(char**a){for(;*a;*++a&&printf(" - "))printf(**a&6?**a&1?"Rr":"Rrrr":"-");}

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Explanations:

f(char**a){
  // While the string at the current position is not NULL
  for(;*a;
    // Advances the pointer to the next string
    // Then if the current string is not NULL, prints a delimiter
    *++a&&printf(" - ")
  )
    /* 
      If the 1st char of the string is not a 'p'
        If the 1st char is not a 'l'
          Prints "Rr"
        Else
          Prints "Rrrr"
      Else:
        Prints "-"
     */
    printf(**a&6?**a&1?"Rr":"Rrrr":"-");
}

R, 72 bytes

Takes input from stdin, prints to stdout.

cat(sapply(scan(,''),switch,long="vvvv",short="vv",pause="-"),sep=" - ")

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Batch, 88 bytes

@set/ps=
@set s=%s: = - %
@set s=%s:long=Rrrr%
@set s=%s:short=Rr%
@echo %s:pause=-%

Takes input on STDIN. Unfortunately loop overhead costs 26 bytes so this is just boring replacements.

Retina, 31 bytes

short|l
Bz
ong
zz
pause
-
 
 - 

-1 byte thanks to @fergusq!

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PHP, 113 bytes

<?$s=[];for($i=1;$i<$argc;$i++){$c=$argv[$i][0];$s[]=($c<'m')?'Rrrr':(($c<'q')?'-':'Rr');}echo implode(' - ',$s);

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First attempt at code golf, so probably a lot of optimisations available!

Vim (52 bytes)

:s/long/Rrrr/ge|s/short/Rr/ge|s/pause/-/ge|s/ / - /genter

Can probably be made shorter...

Excel, 100 bytes

=REPLACE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"long","- Bzzz"),"short","- Bz"),"pause","- -"),1,2,"")

Per examples, Input is SPACE separated string, as is output.

Question itself does not mention a SPACE requirement, allowing for a slightly shorter 97 byte solution:

=REPLACE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"long","-Bzzz"),"short","-Bz"),"pause","--"),1,1,"")

AutoIt, 145 bytes

EXECUTE(STRINGREPLACE('MSGBOX(0,0,STRINGSTRIPWS(====INPUTBOX(0,0),"PAUSE",""),"LONG","Rrrr"),"SHORT","Rr")," "," - "),4))',"=","STRINGREPLACE("))

(AutoIt is really bad choice for code golf, tried my best to make it small as possible)

Alice, 37 bytes

/ lRnrhR
\""orgrp-""!yi'."?-e"ySNo?@/

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Explanation

This program makes the following substitutions:

• l, h → R
• o, n, g → r
• p → -
• Space → Space
• Everything else → Nothing

"longhp "!i.?eyN?"RrrrR- "y' " - "So@

"longhp "    Push this string
!            Immediately move to tape
i            Take input string
.            Duplicate
?ey          Remove all instances of the characters "longhp " from copy
N            Remove the remaining characters from the original, leaving only "longhp "
?"RrrrR- "y  Replace the characters in "longhp " with the corresponding characters in "RrrrR- "
' " - "S     Replace all spaces with " - "
o            Output
@            Terminate

Sed, 50 bytes

Takes input from stdin, prints to stdout

s/l\w*/Rrrr -/g
s/s\w*/Rr -/g
s/p\w*/- -/g
s/ -$//

Edit - saved 2 bytes

Sed, 40 bytes

Copying idea from Nitrodon's answer

s/[srtaue]//g
y/lhongp/RRrrr-/
s/ / - /g

Edit: saved another 2 bytes

Python 2, 69 bytes

lambda s:" - ".join((ord(i[0])&6)*"R"or"-"for i in s.split()).title()

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Port of my Pyke answer

Paradoc (v0.2.10), 21 bytes (CP-1252)

Wμ‹6&'r\°"-":Ãu}« rTc

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Takes a string on the stack and results in a string on the stack. Prepend i to turn into a full program that reads from STDIN.

Uses &6 like the Pyke answer and everybody else, but joins the tokens together slightly differently, by adding a "-" token after each noise, deleting the last one, and then joining these tokens by spaces. Seems to save a byte over joining by " - ".

Explanation:

W                     .. Break into words
 μ             }      .. Map over this block:
  ‹                   .. Take the first character
   6&                 .. Binary AND with 6, to get 4, 2, or 0
     'r               .. Character "r"
       \              .. Swap top two of stack
        °             .. Replicate, to get "rrrr", "rr", or ""
         "-"          .. Push string "-"
            :         .. Duplicate on stack
             Ã        .. Compute the max...
              u       .. ... underneath the top of the stack (so, of the
                      .. second and third elements on the stack, i.e. the
                      .. string of "r"s and "-")
                      .. The mappped block ends here; we now have
                      .. something like ["rrrr", "-", "-", "-", "rr", "-"]
                «     .. Take all but the last
                  r   .. Join with spaces (this built-in's name is two
                      .. characters, the first of which is a space)
                   Tc .. Title-case

v0.2.11 will support shaving two more bytes by replacing \° with x and "-" with '-.

SOGL V0.12, 28 bytes

θ{K;⁄5κ« r*; p=?X┌}}¹" - ”∑ū

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fun fact: if it was allowed to delimit with - or - it'd be a byte shorter

Ruby, 67 bytes

p ARGV[0].split(' ').map{|w|w<'m'?'Rrrr':w<'q'?'-':'Rr'}.join ' - '

This is Johan Karlsson's JavaScript solution ported to Ruby. If you like this answer, you should upvote Johan's answer.

The key idea is to compare the word strings 'short', etc. to a single character in order to distinguish between words.

| Word  | < 'm' | < 'q' | Output |
|-------|-------|-------|--------|
| short | false | false | 'Rr'   |
| long  | true  | N/A   | 'Rrrr' |
| pause | false | true  | '-'    |

Or, in alphabetical order:

• long
• m
• pause
• q
• short

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Ruby, 78 bytes

p ARGV[0].chars.map{|c|{p:'-',o:'Rr',g:'rr',' '.to_sym=>' - '}[c.to_sym]}.join

The only important parts of the input are p, o, g, and space... ignore the rest.

• short becomes o
• long becomes og
• pause becomes p

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V, 32 bytes

Í /ò
çl/CRrrr
ço/CRr
çp/-
HòJa- 

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Python 3, 66 bytes

' - '.join(['Rrrr','-','Rr'][ord(s[1])%3]for s in input().split())

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