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

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.

[Steven Lord]

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.

Answer 1

[OCDER]

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

[John Lutz]

Thank you very much! This worked, I just ignored the use of non word count for this and it worked perfectly. Is there any way you can explain this out a little more? Thank you so much!

[OCDER]

You're Welcome! You can see what each line does by removing the semicolons and running the code line-by-line.

strcell is just str, but splitting by space storing words into a cell array.

wordloc uses regexp to figure out if there is ANY non-letter character in the word. cellfun applies this search to every cell in strcell. It returns a logical array showing where the words (1) and non-words are (0).

[John Lutz]

Is there any way I can get it so it only displays the words with a letter in the first part of the word. Ex: if you have lion2 and 2lions , lion2 is a word, but 2lions isn't. I've tried to restrict it but i can't seemingly find a way to. Thanks again.

[OCDER]

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

[Cedric]

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

Answer 2

[Walter Roberson]

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;