how to generate new matrix with if statment

I have two matrices first one is:

test = [5;6;0;-1;0;5;0;6;0;8];

and the second one is:

test5 = [2;6;8;-1;0;7;8;6;8;8];

how to generate third matrix which is the result after the condition (if statment)...

the condition is if the value of test is equal 0 then the value of the new matrix is 0 , else if the value of the first matrix isn't equal 0 do some calculations on the second matrix which is test5 like (test5*7+5).

so the third matrix values depends on the two matrix before...

Answer 1

[Stanislao Pinzón]

Maybe like this?

test = [5;6;0;-1;0;5;0;6;0;8]+2;
test5 = [2;6;8;-1;0;7;8;6;8;8];

if ismember(0,test)
    Matrix3 = 0;
else
    Matrix3 = test5*7+5;
end

[Stanislao Pinzón]

This assuming that "if the value of test is equal 0" means you need to check if any of the elements of test is 0.

[Ibrahim AlZoubi]

Matrix3 =

19

47

61

-2

5

54

61

47

61

61

but it doesn't take zero in consideration, it should be

Matrix3 =

19

47

0

-2

0

54

0

47

0

61

[Ibrahim AlZoubi]

Okay, What I really need is checking if test equal zero or not:

if yes(test=0): then the value of test at this point is zero.

if no(test~=0): then do the calculations.

after that the 3rd matrix is combination of this condition.

so how to write a code like this?

[Ibrahim AlZoubi]

ex:

test = [5;6;0;-1;0;5;0;6;0;8];
test5 = [2;6;8;-1;0;7;8;6;8;8];

calculate :Matrix3 = test5*7+5

Matrix3 = [ 19;47;0;-2; 0;54;0;47;0;61];

[Stanislao Pinzón]

test = [5;6;0;-1;0;5;0;6;0;8];
test5 = [2;6;8;-1;0;7;8;6;8;8];

Matrix3 = zeros(length(test),1);
for i=1:length(test)
    if test(i,1)==0
        Matrix3(i,1) = 0;
    else
        Matrix3(i,1) = test5(i,1)*7+5;
    end
end

[Stanislao Pinzón]

or instead

test = [5;6;0;-1;0;5;0;6;0;8];
test5 = [2;6;8;-1;0;7;8;6;8;8];

Matrix3 = test5*7+5;
a = find(test==0);
Matrix3(a) = 0;

[Stephen23]

find is superfluous. Get rid of it.

Answer 2

[Image Analyst]

Try masking:

test = [5;6;0;-1;0;5;0;6;0;8];
test5 = [2;6;8;-1;0;7;8;6;8;8];
% Now multiply by 7 and add 5 only.
output = test5 * 7 + 5;
% Find indexes where test is zero.
mask = (test == 0)
% Erase where test was 0.
output(mask) = 0

Answer 3

[Yundie Zhang]

test = [5;6;0;-1;0;5;0;6;0;8];
test5 = [2;6;8;-1;0;7;8;6;8;8];
if test ==0
newMAT = 0
elseif test ~=0 
newMAT = (test5*7)+5
end

[Ibrahim AlZoubi]

the result is:

Undefined function or variable 'newMAT'.

when I want to know the new matrix "newMAT"

[Walter Roberson]

Remember that when you apply if or while to a non-scalar, that the result is only considered true if every item being tested is non-zero.

if test ==0

Only some of the items in test are 0, so that fails

elseif test ~=0 

Only some of the items in test are non-zero, so that fails.

Answer 4

[Walter Roberson]

Create the new matrix by applying the calculation to all of the entries in the second matrix, as if the rule about 0 was not present. This can be done in vectorized form in a single statement.

Now, everywhere that there is a 0 in the first matrix, replace the content of the third matrix with 0. This can be done in vectorized form in a single statement, using logical indexing.

[Ibrahim AlZoubi]

so you mean to edit the 3rd matrix manually ?

[Stephen23]

"so you mean to edit the 3rd matrix manually ?"

No. Use logical indexing:

https://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html

which could be generated very simply using ==.

[Walter Roberson]

For example:

A = randi(10, 5, 8);
A(A>7) = -1;