How to ignore values in an array

I'm creating an array [array 1] that fulfills the formula (A - B/C), where A and B are matrices with different elements and C is a matrix with a constant value. I then want to make another array [array 2] which is dependent on the values obtained from the formula in array 1.

So in array 2, I want it to check for values greater than 0.5 and set those to 0, and values less than or equal to 0 to be set to 1.

However, as A and B both begin at a value of 0; the first element in array 1 will be 0, making the first element of array 2 1.

How could I ignore this value?

I've currently made array 2 by doing:

A_2 = (A_1<= 0.5 & A_1>=0)

[dpb]

I couldn't tell what you want the end result to be, sorry. Give us a really small example input/output as illustration.

Answer 1

[dpb]

Having read the question multiple times now, the crystal ball is suggesting you simply wanted to treat the first element uniquely. The simplest way to do this is to just add the additional rule after done creating A_2--

A_2 = (A_1<= 0.5 & A_1>=0); A_2(1) = 0;

The ways to write the addressing to only operate on the elements of A_1 excluding (1,1) end up either being quite messy or returning the result as a vector of N-1 elements that has to be reshaped so it's easier to just brute-force it since it's just the one case/element.

ADDENDUM To generalize if there were ever a case that wanted other than the one constant would be to write

A_2 = (A_1<= 0.5 & A_1>=0); A_2(1) = A_1(1);

[Image Analyst]

He wanted " to check for values greater than 0.5 and set those to 0, and values less than or equal to 0 to be set to 1." and that code changes the values in between 0 and 0.5 whereas my code leaves them alone and does exactly what was asked. Since he accepted your answer, all I can guess is that mathman must have meant "less than or equal to 0.5" even though he did not write that. I'll have to ask for a copy of that Crystal Ball Toolbox.

[mathman]

Sorry that was an error. I did write it in the code I provided though, which is what dpb used.

[dpb]

Well, actually I just divined what OP meant the question to be regarding the one value and presumed that his code otherwise was providing the result desired so really paid no attention to that portion...addressing only how to treat the one location uniquely explicitly. :)

Answer 2

[Image Analyst]

Try this:

a2 = A1; % Initialize
zerosMask = a2 > 0.5;
a2(zerosMask) = 0;
onesMask = a2 < 0;
a2(onesMask) = 1;