Adding continuously to get a sum of how many words there are

Question

John Lutz 2017-9-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/358945-adding-continuously-to-get-a-sum-of-how-many-words-there-are

评论： Steven Lord 2017-9-29

function getWords
str = char('jeff john 2td');
n = size(str,1);
for t=0
    for i = 1:n
        if str(i,1) == isletter(str(i,1))
            t+1;
        end
        if str(i,1) == isnumeric(str(i,1))
            disp('For', str(i,1), 'it isnt a word')
        end
    end
end
disp(t);
end

I want to continuously add +1 to my t value of originally 0 when I get another word into my string, but as of right now it keeps displaying "t" as 0 for some reason. Trying to get the function to be capable of displaying how many words there are, but if the one of the words starts with a number the function WILL NOT count it as a word. For example: if the string is 'my 2sons had lunch' it would display it as the t variable having 3 words.

Thank you for your assistance.

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Steven Lord 2017-9-29

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How many words would you say this sentence contains?

Jean-Luc counted to 3 then asked "Q, what are you doing here?"

Do you consider Jean-Luc to be zero, one, or two words? Does the hyphen disqualify it from being a word, split those two collections of characters into separate words, or join the two collections of characters together into one word?

Do you consider 3 to be a word?

Do you consider "Q, to be a word? In this context Q is the name of the character to whom Jean-Luc asked the question.

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Answer 1

OCDER 2017-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/358945-adding-continuously-to-get-a-sum-of-how-many-words-there-are#answer_283614

编辑：OCDER 2017-9-29

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NEW ANSWER: treats a word as one that starts with a letter

function wordcount = getWords
str = 'jeff john 2lion lion2 this2that dog.';
strcell = regexpi(str, '\w*', 'match');
%Defining word as one that starts with a letter (this2that would still be a word)
wordloc = cellfun(@(x) ~isempty(regexpi(x(1), '[a-z]')), strcell);
wordcount = sum(wordloc); %number of words
nonwordcount = sum(~wordloc); %number of non-words
%Displaying words /  nonwords
fprintf('Is a word: %s\n', strcell{wordloc})
fprintf('Not a word: %s\n', strcell{~wordloc})

Outputs:

Is a word: jeff
Is a word: john
Is a word: lion2
Is a word: this2that
Is a word: dog
Not a word: 2lion

OLD ANSWER: One way to do this without loops:

function getWords
str = 'jeff john 2td tomorrow 3e4 a23 dog.';
strcell = regexp(str, '\s+', 'split');
wordloc = cellfun(@(x) isempty(regexp(x, '[^a-zA-Z\.]', 'once')), strcell); %gets location of string with only letter a-zA-Z
wordcount = sum(wordloc); %number of words, 4
nonwordcount = sum(~wordloc); %number of non-words

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OCDER 2017-9-29

I've updated the answer above that should answer this question. Hope this helps!

Cedric 2017-9-29

编辑：Cedric 2017-9-29

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As you are using regular expressions, note that the following would work too (here applied to your test str):

 >> numel( regexp( str, '(^|\s)[a-zA-Z]' ))
 ans =
     5

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Answer 2

Walter Roberson 2017-9-29

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https://ww2.mathworks.cn/matlabcentral/answers/358945-adding-continuously-to-get-a-sum-of-how-many-words-there-are#answer_283615

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t+1;

retrieves the value of t, adds 1 to the value, then throws away the value because the phrase ends in a semi-colon. You are not changing t. Perhaps you are thinking of C's t++ or t++1 which are not valid MATLAB syntax.

t = t + 1;