If-statements within a for loop does not respond to my counter appropriate?

Hello, my if statements within my for loop does not work appropiate. When i'm printing the x-vector to the screen, the maximum value is sadly 25000 in x(i)

The counter only reach 1, but i'll like to count further if the condition name(i) & name1(i) holds. Name and name1 are also vectors with dimensions 100x1. Another concern is for instance when printing name(1) and name1(1) to the screen I've seen that the counter, op, runs despite that name1(1) is <10. Is the && command wrong?

x = zeros(100,1); %Preallocation
name1 = randi(30,100,1) %vector
name = randi(([0 1], 100,1) %vector

op=0; %counter index

for i=1:length(x)
if name(i)==1 && name1(i)>=10

    op=op+1; %counter

else
    if op~=0;

            if 0<op<=1;
            x(i)=x(i)+25000;

            elseif 1<op<=2;
            x(i)=x(i)+75000;

            elseif 2<op;
            x(i)=x(i)+200000;

            end 
op=0;
        end

    end
end

Answer 1

[Guillaume]

a<x<=b is not doing what you thing it's doing and is not the way to check that a number is between two numbers. What it does is check that a<x. This results in either false (0) or true (1). It then compares that 0 or 1 to b. When b is 1, this is always true, thus

0<op<=1

will always be true. The proper syntax is

0 < op && op <= 1

However, since you've already checked that op is not 0 and since it is integer you could just have

if op == 1
    %...
 elseif op == 2
    %...
 else
    %...
 end

Other comments

• name1 and name seem like poor names for your variables. They certainly do not describe at all what these variables are used for.
• ; are not needed at the end of an if statement.
• considering that x(i) is always 0 why not assign directly the values to x(i) rather than adding them?

[Guillaume]

And here is the way to do the same without loops:

x = zeros(100,1); %Preallocation
name1 = randi(30,100,1) %vector
name = randi(([0 1], 100,1) %vector
increments = [25000; 75000; 200000];  %needs to be column vector to match x being a column vector

condition = name & name1 >= 10;  
startruns = strfind([0, condition.'], [0 1]);  %transpose required because strfind only works on row vector
endruns = strfind(condition.', [1 0]);   %same
if numel(startruns) > numel(endruns)   %could be 1 more if condition ends with a run of 1s
    startruns(end) = [];
end
x(endruns + 1) = x(endruns + 1) + increments(min(3, endruns - startruns + 1));

Answer 2

[Andrei Bobrov]

val = [25000; 75000; 200000];
op0 = name1 >= 10 & name == 1;
x = zeros(size(name)+[1 0]);
a = diff([0;op0(:);0]);
n = accumarray(cumsum(a == 1).*[op0;false] + 1,1);
x(a == -1) = val(n(2:end));