how to generate new matrix with if statment
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Ibrahim AlZoubi
el 17 de Mayo de 2020
Comentada: Walter Roberson
el 17 de Mayo de 2020
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...
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Stanislao Pinzón
el 17 de Mayo de 2020
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
7 comentarios
Stanislao Pinzón
el 17 de Mayo de 2020
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;
Más respuestas (3)
Image Analyst
el 17 de Mayo de 2020
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
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Yundie Zhang
el 17 de Mayo de 2020
Editada: Walter Roberson
el 17 de Mayo de 2020
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
2 comentarios
Walter Roberson
el 17 de Mayo de 2020
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.
Walter Roberson
el 17 de Mayo de 2020
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.
3 comentarios
Stephen23
el 17 de Mayo de 2020
"so you mean to edit the 3rd matrix manually ?"
No. Use logical indexing:
which could be generated very simply using ==.
Walter Roberson
el 17 de Mayo de 2020
For example:
A = randi(10, 5, 8);
A(A>7) = -1;
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