How can I make a code with if/ else function?

 for i=1:sz(1) %sz(1) is 21
     k = find(xi == XX(i) & yi == YY(i));
     disp(zi(k))
     w(i,1)=zi(k)
 end 
 

When I run above code, it shows error with this statement

"Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side

is 0-by-1.

Error in sur3 (line 151)

w(i,1)=zi(k)"

so I made another code to run the code anyway, and set the data that cannot be found in my data(xi, yi) as o(zero)

 for i=1:siz1(1)
   if k == find(xi == XX(i) & yi == YY(i));
         disp(zi(k))
         w(i,1)=zi(k)
   else k==[]
        w(i,1)=0
   end
     w(i,1)=zi(k)
 end 

 for i=1:siz1(1)
    if (XX(i)==xi) & (YY(i) == yi )
         k = find(xi == XX(i) & yi == YY(i));
        disp(zi(k))
        w(i,1)=zi(k)
        else (XX(i) ~= xi)  | (YY(i) ~= yi);
        w(i,1)=0
    end
 end 
 

but it didnt work as well..

I want to find data which satisfy xi == XX(i) & yi == YY(i))

If there is no data for that, I just wanna set the w(i,1) as zero n jump to the next step to find data..

How can i fix this problem..?

Answer 1

[John D'Errico]

Part of your problem is in this test:

else k==[]

Why is that an issue? Because you can't do it. For example, suppose I create an explicitly empty array, then do that same test.

E = [];
E == []
ans =
  0×0 empty logical array

Hmm. It did not recognize E as empty, at least not by that test. Instead, you need to use isempty.

isempty(E)
ans =
  logical
   1

Or, I suppose you could have tested the number of elements in E.

numel(E) == 0
ans =
  logical
   1

Another issue is in the else itself. An if statement will be in this form:

help if
 if Conditionally execute statements.
    The general form of the if statement is
 
       if expression
         statements
       ELSEIF expression
         statements
       ELSE
         statements
       END

So if you wanted to do a test, then you need to use elseif.

elseif isempty(k)

[Jiyeon Kim]

 for i=1:siz1(1)
 
    if k == find(xi == XX(i) & yi == YY(i));
        disp(zi(k))
        w(i,1)=zi(k)
     elseif isempty(k) 
        w(i,1)=0
    end
 
 end

I tried this code using elseif that u recommend ,, but it doesnt work becase I got w matrix consist of only zeros..

[Steven Lord]

Define k to be the result of that find call then use if and isempty to test if it is empty. If it is, set the corresponding element of w to 0. If it is not, set the corresponding element of w to the associated element in zi.