for loop that changes specific letters to numbers

5 Ansichten (letzte 30 Tage)
Seaturtle
Seaturtle am 2 Okt. 2019
Kommentiert: Adam Danz am 4 Okt. 2019
I want to replace the vowels AEIOU with the number 0 and all other letters with the number 1. For example the output should be ans = [0 1 1 1 0] if the user inputs apple. I know I must be misunderstanding how to get the loop to go through my whole string. This is what I have managed so far.
party = input('What is your answer? ', 's');
n = length(party)
for i = 1:n
if i == 'A'
disp(0)
elseif i == 'B'
disp(1)
elseif i == 'C'
disp(1)
end
end

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 2 Okt. 2019
party = input('What is your answer? ', 's');
n = length(party)
for i = 1:n
if party(i) == 'A'
party(i) = 0;
elseif i == 'B'
party(i) = 1;
elseif i == 'C'
party(i) = 1;
end
end
disp( double(party) )
  7 Kommentare
Guillaume
Guillaume am 3 Okt. 2019
It would be nice if 'vowel' were a category option.
The problem with that is what is a vowel or not depends on the language. Some letters such as 'y' in english can also qualify as a vowel or consonant depending on the word.
Adam Danz
Adam Danz am 3 Okt. 2019
True. It would be neat to play around with one of the vowel classification algorithms that are not specific to any language such as this one (link below) that is based on character co-occurrences. It wouldn't necessarily solve the fuzzy set problem but it would be a semi-objective, language-independent classification.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Adam Danz
Adam Danz am 2 Okt. 2019
Bearbeitet: Adam Danz am 2 Okt. 2019
No loop needed.
str = 'apple';
isConsonant = ~ismember(lower(str),'aeiou') %lower() makes it not case sensitive
If you really wanted to do that in a loop,
n = numel(party);
isConsonant = true(1,n);
for i = 1:n
if ismember(party(i),'aeiou')
isConsonant(i) = false;
end
end
In both cases, isConsonant is a logical vector. If you want a double vector of 0/1 instead of false/true,
isConsonant = double(isConsonant);
  8 Kommentare
Seaturtle
Seaturtle am 3 Okt. 2019
Bearbeitet: Seaturtle am 3 Okt. 2019
Thanks for this answer! This really helped with my understanding of not using a loop to do something with shorter code.
Adam Danz
Adam Danz am 3 Okt. 2019
Glad I could chip in! :)

Melden Sie sich an, um zu kommentieren.


Jos (10584)
Jos (10584) am 3 Okt. 2019
Another option:
str = 'apple';
TF1 = any(lower(str) ~= 'aeiou'.')
  6 Kommentare
Jos (10584)
Jos (10584) am 4 Okt. 2019
Thanks for the corrections :-)
Adam Danz
Adam Danz am 4 Okt. 2019
+1
This implicit expansion solution is faster and neater than my ismember() solution.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by