how to use for loop?

if we have an array a= [1,2,3,1,2,3,4,1,2] and second array b=[1,2]

we want to know what is the number of occurrences that b are in a

which means the answer should be 3.

then how to write that using for loop?

Answer 1

[Star Strider]

Unless you absolutely must use a loop, use the strfind (link) function:

a = [1,2,3,1,2,3,4,1,2];
b = [1,2];
c = strfind(a,b);
Result = numel(c)

producing:

Result =
     3

[aya qassim]

if I want to use the "for" is it possible?

because when I tried to do so but matlab warned that the matrix deminisions are not true

I did not understand why

[Star Strider]

It is. You must compare 2 consecutive elements of ‘a’ to ‘b’ in each iteration of the loop.

Try this:

a = [1,2,3,1,2,3,4,1,2];
b = [1,2];
for k = 1:numel(a)-1
    z(k,:) = a(k:k+1) == b;
end
Result = nnz(z(:,1))                    % Using A ‘for’ Loop

producing:

Result =
     3 

Note that you can replace the nnz call with a sum call.

[aya qassim]

thank you fis of all;

for i=1:numel(x)-1;
    z(i:1)=x(i:i+1)==y;
   
end
    r=sum(z(:,1));
    

I entered x=[1,2,1,2,1,2,1,2,1,2,1] and y=[1,2,1,2,1]

the answer should be 4

but matlab said:

Matrix dimensions must agree

I don't understand

[Star Strider]

You must change my code to accommodate the length of the vector you want to compare, here ‘y’.

This works:

x=[1,2,1,2,1,2,1,2,1,2,1];
y=[1,2,1,2,1];
for i=1:numel(x)-numel(y)+1             % Limits Must Accommodate The Length Of ‘y’
    z(i,:)=x(i:i+numel(y)-1)==y;        % The Comparison Must Consider Segments Of ‘x’ Equal To The Length Of ‘y’
end
r=sum(z(:,1))

producing:

r =
     4

[aya qassim]

why did you put (i+4) what the 4 mean I don't understand

[Star Strider]

I edited my previous Comment to make it more general. The current code should address your concerns.

[aya qassim]

thank you, but the code did not work it gives me an error answer.

[Star Strider]

As always, my pleasure.

What error did it throw?

It worked when I ran it. Please post the code you used. Also, what MATLAB version (release) are you using?

[aya qassim]

I have the version of matlab R2017b

for i=1:numel (x)-numel(y)+1
    z(i:1)=(x(i:i+numel(y)-1)==y);
end
a=sum(z(:,1));

x=[1,2,1,2,1,2,1,2,1,2,1]

y=[1,2,1,2,1]

matlab answer:

error, subscripted assignment dimensions mismatch.

thank you,

[Star Strider]

Your ‘z’ assignment subscrpting is incorrect.

Your code:

z(i:1)=(x(i:i+numel(y)-1)==y);

Your code will either assign the right-hand-side of the equation (a logical vector) to a single scalar (if i=1) or to an empty array (if i>1), neither of which will work.

My code:

z(i,:)=x(i:i+numel(y)-1)==y;

My code assigns the right-hand-side to a matrix row. It works. Use my code.

Please see the documentation section on Matrix Indexing (link) to understand the difference between my code (that works) and your code (that errors), and how to index a matrix correctly.

[aya qassim]

Thank you, I understood my mistake.

[Star Strider]

As always, my pleasure.