How do I manipulate matrix elements conditionally without loops?

I have a practise problem that I'm really stuck with.

There is an input variable A that is a random matrix with a random number of rows and columns. The element values of the matrix are random double precision numbers that fall in the range from to . The matrix has between 5 and rows and between 5 and columns. The operations must be solved with only vectorised code.

%given function parameters
function [B, C, D] = matrixFun(A)

%1) change matrix elements<0 to 999:
A(A<0)=999
%assign values to B

%2) replace positive elements with square root of element:
A(A>0)=sqrt(A) 
%Assign values to C

%3) column vector from elements between -2 and 2:
     a= A(-2 <= A <= 2)
     D=sort(a)
end 

How do I assign the changed values to the B and C matrices? And is there an alternative method than using A(A<0)? I can't find any relavent documentation.

[Walter Roberson]

The problem with

A(A>0)=sqrt(A) 

is that the left side selects only some of the locations in A, but the right side selects all of the locations in A. You should be selecting the same locations in A on both the right and the left.

By the way, after you change all the negative elements to 999, every element will be positive except for the entries that are 0, and the square root of 0 is 0 so you might as well just take the square root of everything.

Likewise, after you set all of the negatives to 999, none of the entries can be between -2 and 0.

Perhaps you are not intended to do the changes in that order?

a= A(-2 <= A <= 2)

As far as MATLAB is concerned, that means the same as

a= A(((-2 <= A) <= 2))

The first part, -2 <= A, results in an array of 0 (false) and 1 (true) values the same size as A. Then you compare that array of 0 and 1 entries <= 2, which would be true for all of them, because all 0 and 1 are less than 2.

[Emily Stewart]

This makes sense, thanks

Answer 1

[Stephen23]

Just assign A to a new variable first, and then use indexing afterwards. It will be clearer if you define the index as its own variable, e.g.:

B = A;
X = B<0;
B(X) = 999;
...
C = A;
X = C>0;
C(X) = sqrt(C(X)); % note the indexing on the RHS!
...

Notes:

• the indexing on the RHS for C: you don't want the square root of all array elements (like you did), only those one that you want to change (which is why you need to use indexing).
• MATLAB does not have ternary logical functions like X<=Y<=Z. MATLAB has binary logical function like X<=Y. So you will need to combine two binary functions using logical AND or OR, e.g.: X<=Y & Y<=Z
• To create a column vector out of any array X use X(:), or extract a subset of the elements using indexing.

[Emily Stewart]

Thanks for such a quick answer! I understand now and the code works, thank you

[Ningbo Hu]

THANK YOU!

[Hamish McAlpine]

i'm not entirely clear how how to write the code for D. my indexing seems like where i'm going wrong