Shortened words

Question

My cell phone carrier charges a cent per character in a text! I need a program to shorten the words for me, while still being understandable. I will be texting this program to my friends, so it needs to be as short as possible.

Your program will take a String (a word is separated by a non-alphanumeric character) of words from STDIN (or closest alternative) reduce any similar (a and A are similar, b and a are not, 1 is similar only to itself) consecutive alphanumeric characters down to one character, and remove all the vowels (aeiou), and print it to STDOUT (or closest alternative), only letters, numbers, and spaces are kept: everything else is removed at the end. The first character of a word must remain the same, and only shorten the word if the resulting shortened word is 4 or more characters long. Examples:

homework -> hmwrk
llamas -> llamas (lms<4 characters)
botany -> btny
y3333llow -> y3lw
shorten -> shrtn
aabcddde -> abcd
abb'bbc -> abbbbc (abb -> abb and bbc -> bbc = abb'bbc --(remove non-alphanumeric-space characters)-> abbbbc)
SHARRRKnado -> shrknd (or with capitalization bonus: SHrKnd)
abracadabra -> abrcdbr
everything is awesome! -> evrythng is awsm
you can't break what is already broken -> you cant break what is alrdy brkn
the person's cat's bed's blanket's poka-dots are orangy-yellow -> the prsns cats beds blnkts pokadots are orngyyellow

-5% bonus if kept a i o or u at end of word

sharknado -> shrkndo
abracadabra -> abrcdbra

-10% bonus if kept capitalization of non-consecutive characters

aAbCDdDE -> abCd

This is code-golf so the shortest program in bytes wins.

Answer 1

Japt, 57 bytes 58 80 88 90 95 98

Saved 6 bytes thanks to @ETHproductions

Japt is a shortened version of JavaScript. Interpreter

Uv qS m@(A=Xg0 +Xs1 r"(.)\\1+",@Y r"[aeiou]",P)g3 ?A:X)qS

Uses RegExp to do most of the work.

Try it online

Explanation

This explanation is old but the method used is still generally the same.

UqS       // Splits input 
   m@     // Loops through words
     Xr"(.)\\1+","$1"     // Removes consecutive sequences
      r"[aeiouAEIOU]",P   // Removes vowels
   m@     // Loops again
     Xl >3?               // If length of word is greater than 3
           X[0]==UqS [Y][0]?  // If first letter is right, keep it
              X
           :UqS [Y][0]+X      // Otherwise, use the correct first letter
     :UqS [Y]  qS v       // Otherwise return the original word

Answer 2

TeaScript, 60 bytes 64 58 63 70

xL¡l»(b=l0+lS(1)g(/(.)\1+/,"$1")g`[aeiou]`)n>3?b:l,ø)O`a-z `

I should be able to use the unique function to get rid of repeated characters but for now I'm using a regex.

I believe this should work fine. I'll be heading off to sleep so notify me if anything is wrong and I'll address it next morning.

Ungolfed

xL()l(#(b=l0+lS(1)g(/(.)\1+/,"$1")g`[aeiou]`)n>3?b:l,` `)O`a-z `

Answer 3

CJam, 54 bytes

qelS/" ,"*__eu&S-A,s-NerN%{_(\"aeiou"-+e`1f=_S-Z>\@?}/

Try this fiddle or this test suite in the CJam interpreter.

How it works

qel        e# Read all input and cast to lowercase.
S/" ,"*    e# Replace spaces with " ,". This way, we don't have to consider
           e# spaces word delimiters, which will make them easier to preserve.
__eu       e# Push two copies and convert the second one to uppercase.
&          e# Intersect the copies. This will keep only non-letters.
S-A,s-     e# Remove spaces and digits from the non-letters.
Ner        e# Replace their occurrences in the input with linefeeds.
N%         e# Split at linefeeds.
{          e# For each resulting chunk:
  _(\      e#   Push a copy and shift out the first character.
  "aeiou"- e#   Remove all vowels from the tail.
  +        e#   Concatenate the result with the first character.
  e`       e#   Perform run-length encoding.
  1f=      e#   Select all characters.
  _S-      e#   Push a copy and remove spaces.
           e#   This is for the length check; spaces don't actually form part
           e#   of the word.
  Z>       e#   Discard the first three characters.
  \@?      e#   If there are characters left, pick the modified word; else
           e#   pick the unmodified one.
}/         e#

Answer 4

C# - 404 bytes

using System;using System.Text;using System.Text.RegularExpressions;class j{static bool v(char l){return l=='e'||l=='a'||l=='o'||l=='i'||l=='u';}static void Main(){string a=Console.ReadLine();string z=Regex.Replace(a,"[^a-zA-Z ]","");string[]b=z.Split(' ');string c="";foreach(string d in b)    {if(d.Length>4){foreach(char e in d.ToCharArray()){if(!v(e))c+=e;}}else{c+=d;}c+=' ';}Console.WriteLine(c);}}

It takes input from Console.ReadLine(). I used a regex to find the non alphanumeric characters and spaces and I calculated if the size is less than 4. I tested using the final example.

C# - 279 bytes (from LegionMammal798)

using C=System.Console;class j{static void Main(){var c="";foreach(var d in System.Text.RegularExpressions.Regex.Replace(C.ReadLine(),"[^a-zA-Z0-9 ]",c).Split(' ')){if(d.Length>4){foreach(var e in d)if(!"aeiou".Contains(e.ToString()))c+=e;}else c+=d;c+=' ';}C.Write(c);}}

Thank you for the golfing tips!

Answer 5

𝔼𝕊𝕄𝕚𝕟, 57 chars / 97 bytes (noncompetitive)

Ɱ(ï,↪(a=$ù⬮ċ/⊙\1+⌿,`⑴”ċ/[ᶌ]⌿,⬯⸩Ꝉ>3?a⎖0≔$⎖0?a:$⎖0+a:$,⬭,⬭)

Try it here (Firefox only).

Uses regex aliasing and other syntax made after the challenge was posted.

Answer 6

Haskell, 187 176 bytes

Adhering to written description, not given examples:

import Data.List
import Data.Char
f=filter
d w@(_:_:_:_)=map head$groupBy(\a b->toLower a==toLower b)w
d w=w
s=unwords.map(f isAlphaNum.d.(\(h:t)->h:f(`notElem`"aeiou")t)).words

Call s.

Improved late after vote reminded me of its existence.