for loop that changes specific letters to numbers
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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
采纳的回答
Walter Roberson
2019-10-2
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 个评论
Adam Danz
2019-10-3
The conditions following the initial one all contain the index value i in the antecedent.
Instead, it should be
for i = 1:n
if party(i) == 'A'
party(i) = 0;
elseif party(i) == 'B'
party(i) = 1;
elseif party(i) == 'C'
party(i) = 1;
end
end
Is the idea to create a conditional statement for each individual character in both upper and lower case, presumably? That would result in over 100 conditions. An improvement would be to use lower() or upper() which would eliminate 26 of those conditions. Why not just use the non-loop, 1-liner?
Walter Roberson
2019-10-3
~isstrprop(party, 'vowel')
Adam Danz
2019-10-3
It would be nice if 'vowel' were a category option.
party = 'apple';
~isstrprop(party, 'vowel')
Error using isstrprop
vowel is an invalid qualifier.
r2019b
Walter Roberson
2019-10-3
Atch, I was getting questions confused, as someone recently asked about upper vs lower case.
Seaturtle
2019-10-3
Guillaume
2019-10-3
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
2019-10-3
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.
更多回答（2 个）
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 个评论
Walter Roberson
2019-10-2
The test is not for consonants: the test is for non-vowels. For example a space or period should have 1 in that location, not 0
disp() of one digit at a time does not satisfy ans = [0 1 1 1 0]
Adam Danz
2019-10-2
The no-loop version does exactly that though the variable name is misleading. The question specifies the encoding of "letters" as opposed to any character and since all non-vowel letters are consonants, the variable name would be OK if the only input were letters.
But my for-loop method indeed had issues that were fixed.
Stephen23
2019-10-3
+1 very neat solution
Walter Roberson
2019-10-3
In English, y can act like a vowel (and often does). w as well, but that is rare. There is a third letter as well like this but I keep forgetting which one.
Adam Danz
2019-10-3
J, maybe?
This solution below categorizes j, w, and y as a vowel based on a binomial random draw with a probability of being a vowel at 50% although that probability should be lowered to reflect the true frequency of those letters representing vowels :D
str = 'jabberwocky';
vowels = 'aeiou';
semiVowels = 'jwy';
isConsonant = double(~ismember(lower(str),vowels))
isSemiVowel = ismember(str,semiVowels);
isConsonant(isSemiVowel) = binornd(1, .5, [1, sum(isSemiVowel)]); %Just for fun :)
Walter Roberson
2019-10-3
semi-vowel does sound like the correct term.
In practice, y probably acts like a vowel more than it acts like a consontant .
Adam Danz
2019-10-3
Glad I could chip in! :)
Jos (10584)
2019-10-3
Another option:
str = 'apple';
TF1 = any(lower(str) ~= 'aeiou'.')
6 个评论
Walter Roberson
2019-10-3
Cute.
(Needs R2016b or newer.)
(I wonder if I can stop saying that now that R2019b is out? I think there are still a fair number of people on older releases though.)
Adam Danz
2019-10-3
TF1 = any(lower('apple') ~= 'aeiou'.')
TF1 =
1×5 logical array
1 1 1 1 1
I think what you meant was,
TF = ~any(lower('apple') == 'aeiou'.')
Walter Roberson
2019-10-3
TF1 = all(lower(str) ~= 'aeiou'.')
Jos (10584)
2019-10-4
Thanks for the corrections :-)
Adam Danz
2019-10-4
+1
This implicit expansion solution is faster and neater than my ismember() solution.
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