Can you Remove Double Letters?

[2DG Blog]

In the blocks, when you press button, the Label text is replaced with whatever the user typed in TextBox. Also all the Vowels are removed.

Is it possible to also remove any DOUBLE letters? Example: if user typed "APPLE", I would want it to return as "APLE". With Vowels gone it should only return, "PL"

Make sense?

[SteveJG]

segment PP

replacement P      will take one of the Ps away

[2DG Blog]

Hey thanks for the reply. That only seems to work if the Ps are next to each other. I am thinkin more like, Take away any of the same letters in a sentence. I should also say, Apple was just an example. Ideally I want ANY double of a letter to be removed from a sentence. No set letter, just any double of a letter needs to go. lol

[Abraham Getzler]

A different way to attack this would be to build up a second (output) string from empty,

walking through the input string letter by letter, and for each letter,

skip it if it's a vowel or if it's already in your output string,

otherwise JOIN it onto the end of your output string.

ABG

[2DG Blog]

Heya Abraham thank you for your input. I'm not exactly sure what you mean, Today is my first day using App Inventor. Is it possible you could screenshot a very simple example of what you mean? All I need is a starting point, not sure how to set it uo tho : )

[2DG Blog]

I'm not really sure what you mean by "Skip" thats basically my issue, I dont know what would cause it to skip or remove letters.

[SteveJG]

See chapter 19 here http://www.appinventor.org/book2  it may help...also, the entire Inventor's Manual can be read in about an hour.  This  book is one of the best introductions to AI Blocks anywhere.

Regards,

Steve

[2DG Blog]

I tried this, but it only gathers one letter instead of both. 

[Abraham Getzler]

Here's the algorithm in a C-like pseudocode...

set global vowels = 'aeiou'

set global another = empty string

for i = 1 to length(input_string) by 1

   set local variable L = letter at input_string position i

   if AND(not(contains(vowels,L)),not(contains(another,L)) // not a vowel or already in another

      then set another = JOIN(another,L)

end for i

// another contains all  consonants of input_string

ABG

P.S.  This can be made even shorter using DeMorgan's Law and a JOIN.