MATLAB: Separating strings one by one

countstrings

   N=length(str);
   count=1;
   for k=1:N
       x=str2double(str(k));
       if isnan(x)==0
           str(count)=x(k);
           count=count+1;
       disp(x);
       end
   end

This is currently the code I have right now but it keeps giving me an error because the "index exceeds matrix dimensions on the line: str(count)=x(k); What am I doing wrong?

Best Answer

your ‘x’ variable is a scalar, by definition a (1x1) ‘array’. So ‘x(k)’ is only valid for x=1.

This will likely eliminate that error:

str(count)=x;

Related Solutions

MATLAB: Help me differentiate between two codes?

Ah, yes. This curious bit of syntax. What you ended up with was sum(x)! Why? Well, MATLAB evaluates the first condition a <= x(k). This returns either 1 (true) or 0 (false). Then there's a second condition (previous result) <= b. Given that b = 3, and the previous result is 0 or 1, this is always true! Hence all the elements of x are added to your sum.

IOW, a <= x(k) <= b is parsed as ((a <= x(k)) <= b), and T/F in MATLAB is equivalent to 1/0.

[Note: it's possible that the order of operations is the other

As an aside, I hope they will teach you this, but the best way to do this operation in MATLAB is

idx = (a <= x) & (x <= b);
mysum = sum(x(idx))

Logical indexing is sweet!

MATLAB: Switch case on an array

Yes, it's a lot clearer. So, the transition to a char array is a complete red herring. It's totally unnecessary.

Using sequentially named variables is a bad idea. It will force you to write complicated code. If you have 10 different numbers that you want to count would you add another 7 cases to your switch? How about 100 numbers, 10000 numbers. Will you write all these cases?

The answer is to use array indexing, T(1) the count of your first, T(2) the count of your 2nd number, etc. Then you can use a loop. For example,one (inneficient) way to then implement your code would be:

%input
x = [12 24 64 24 12 12]
lookup = [12, 24, 64];  %values to compute the histogram of
%code
count = zeros(size(lookup))
for i = 1:numel(x)
    [found, where] = ismember(x(i), lookup);
    if found
        count(where) = count(where) + 1;
    end
end

But the whole thing can be achieve without loops, if or case. You want to compute an histogram, use the histcounts function:

x = [12 24 64 12 12];
values = unique(x)';   %get unique values of
count = histcounts(x, [values; Inf])';  %inf is right edge of last bin
table(values, count)  %for pretty display

returns

ans =
  3×2 table
    values    count
    ______    _____
      12        3  
      24        1  
      64        1

Best Answer

Related Solutions

MATLAB: Help me differentiate between two codes?

MATLAB: Switch case on an array

Related Question