can someone check my code?

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Safia
Safia 2022-9-30
评论： Walter Roberson 2022-10-1
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!!!!
  5 个评论
显示 3更早的评论
Safia
Safia 2022-10-1
@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
Walter Roberson 2022-10-1
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.

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回答（1 个）

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
  1 个评论
Safia
Safia 2022-10-1
@Sulaymon Eshkabilov thank you for you reply! but it displays the same result as the code i posted above

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