can someone check my code?

Hello!

I have a problem in my code, can anyone help me please?

i have a matrix A(1826*3600) randomly generated contains such values.

i will fill another matrix B based on elements of A, but not all elements(such specific columns)

the first step i did is to generate a vector contains number of specific columns i want to fill.

this is my case :

i created a vector from 1 to 720 element(number of specific columns i want to fill).

S=linspace(1,720,720)

i will check the value in matrix A (in this case for value =50), then i will fill P with my formula.

for i=1:1826
    for j=1:3600
        for k=1:720
            if A(i,j)==50
                P(i,j)=(50*S(k))/3600;
            else P(i,j)=0;
            end
        end
    end
end

this code doesn't work, also i need to elements can't exceed the bounds of my matrix.

i will be grateful if you could help me!

Please HELP!!!!

[Safia]

@Walter Roberson did you write somthing?

[David Goodmanson]

Hi Safia,

you can probably do this with fewer do loops, but the first thing is to get the code to do what you intend. Inside the innermost for loop, k = 1:720, you are doing the same if check every time through. That's because the if check depends on i and j, but not k. Then if the if check passes, with the line

P(i,j)=(50*S(k))/3600

you are writing into the P(i,j) element 720 times, so P(i,j) will end up having only the last value involving S(720). The other 719 values are overwritten. So you need to determine what is supposed to happen.

Also, unless the A(i,j) data is sure to be integers, the if check with A(i,j)==50 may fail, for example if A(i.j) = 50+1e-15 or some such.

[Walter Roberson]

Safia: no, I reformatted your code

[Safia]

@David Goodmanson Hi! thank you for your reply! i want to fill 720 columns in matrix B , that's why i put the vector S. i'm trying to find easier solution than fill each column with loop if.

[Walter Roberson]

        for k=1:720
            if A(i,j)==50

Does A(i, j) change inside the for k loop? The next k value, would you not be testing the same A(i, j)? So the end result is going to be the same as it would be if you only did for k=720 and no other iterations.

Answer 1

[Sulaymon Eshkabilov]

Here is the corrected code:

A=randi([0, 111], 1826, 3600);
S=1:720;
P= zeros(1826, 3600);
for i=1:1826
    for j=1:3600
        for k=1:720
            if A(i,j)==50
                P(i,j)=(50*S(k))/3600;
            end
        end
    end
end

[Safia]

@Sulaymon Eshkabilov thank you for you reply! but it displays the same result as the code i posted above