write a program that creates two vectors from x—one (call it P) that contains the positive elements of x, and a second (call it N) that contains negative. I cant create the arrays

clear all
clc
for x=[-3.5 -5 6.2 11 0 8.1 -9 0 3 -1 3 2.5]
    if x>0
        P=x
    elseif x<0
        N=x
    end
end

[Stephen23]

@Muhammad Waseem: what do you expect that code to do?

Writing code by randomly guessing what to write is not productive or efficient, and it is a very bad way to learn how to use MATLAB. It is quite clear that you did not read about either indexing or if statements.

If you really want to learn how to use MATLAB then you need to start reading the documentation. For a start, you need to do the introductory tutorials, which are highly recommended for all beginners:

https://www.mathworks.com/help/matlab/getting-started-with-matlab.html

And also read this:

https://www.mathworks.com/matlabcentral/answers/228557-experts-of-matlab-how-did-you-learn-any-advice-for-beginner-intermediate-users

PS: The best solution to your problem is not to use loops and ifs, but to use logical indexing. You can read about it in the MATLAB documentation.

[Muhammad Waseem]

I cant use logical operators in this. the Question has restrictions

[Stephen23]

@Muhammad Waseem: okay, then read about for and if and indexing in the MATLAB documentation: these are covered in the Introductory Tutorials which I gave you a link to. Did you do them yet?

Try the examples. Practice. Don't give up just because the first thing you thought of does not work.

The more you read the documentation the more you will learn and the easier it will be to find information in future.

Answer 1

[Image Analyst]

Hint: use the sign() function and you can do it without logical operations or logical indexing.

Answer 2

[OCDER]

Then try growing P and N per iteration (don't use in real codes - it's slow due to matrix creation and copying). Currently, P and N are being replaced by a single scalar value since there's no change in matrix index.

Also, if there is no positive or negative value in x, you'll never create variable P or N - this causes issues when a function needs to return P and N that were never created.

Lastly, avoid using x as the loop counter directly - this makes it hard for the function accept any x and will lead to headaches in the future.

Here's something to get you started, and use the tutorial Stephen sent you to figure out how to manipulate matrices P and N.

function [P, N] = fun(x) 

P = [];
N = [];
for j = 1:length(x)
   %grow P if x(j) > 0
   %grow N if x(j) < 0
end

When done, try comparing the speed of this method versus using logical indices (way faster).

Good luck!

[Image Analyst]

That uses logical operators, which he says he can't use.

[OCDER]

Hm, I think he means no logical index? Looks like he can use > and < operators in an if statement, but not to create a logical index. Please clarify the restrictions, Muhammad. Also, the issue was P and N weren't arrays.

[Stephen23]

"Then try growing P and N per iteration. "

Please do not do that. Always preallocate!

"if there is no positive or negative value in x, you'll never create variable P or N - this causes issues when a function needs to return P and N that were never created."

There are both positive and negative values in x, yet neither P nor N are created. This is because of the vector condition provided to if, which you do not mention at all...

[OCDER]

Yes, I know it's slow as growing matrices creates and copies matrices. Hence I'm urging him to try logical indexing later. But it looks like he's restricted from using the simple code...

P = x(x>0);
N = x(x<0);

I'm assuming x that he gave was an example, and not for all x. If x = [1 2 3 0 0 1], N is never created.

Surprisingly, his for loop DOES work to create P and N (scalar value). I never use this, but apparently you can use vector x as the for loop counter.

for x = [-1 0 2]
  disp(x)
end

is the same as

x = [-1 0 2]
for j = 1:length(x)
   disp(x(j))
end

Preallocate is ideal, but then he'd have to either use a logical index to count how many positive / negative he has. OR, he can create P and N of length(x), and then delete out unused ones.

P = zeros(1, sum(x > 0)); %Preallocated P
N = zeros(1, sum(x < 0)); %Preallocated N
for j = 1:length(x)
  %fill in P and N
end

[Stephen23]

@Donald Lee: you are right, I did not consider the strange restriction "I cant use logical operators". It does turn the problem into something much more awkward than what it seemed to me originally.

Your comment would have made a great answer!

[OCDER]

Thanks! And yes, it indeed is a strange restriction...