Hi Everyone, I need your help on my problem.

I have a 3D matrix (abc), and I need to do if and elseif conditions to this matrix. However, the only the first expression(aa=abc-5) is applied to all elements. %

 %  
ab=[10,2,2;4,12,6;7,5,9];
abc=repmat(ab,1,1,3);
if 1<abc<=5
    aa=abc-5;
elseif 6<abc<=10
    aa=abc+10;
end

How can I make the elseif expression work?

Thank you.

Chane

Answer 1

[Jan]

1 < abc <= 5

This will not do, what you expect. Matlab processes this expression from left to right:

1. 1 < abc. This is either TRUE or FALSE, which is interpreted as 1 or 0
2. 1 <= 5 or 0 <= 5. This is TRUE in both cases.

You want:

1 < abc && abc <= 5

Then your if and elseif will work.

For productive code using logical indexing as suggested by Birdman is usually more efficient and nicer. But understanding why a<b<c fails is essential.

[Engdaw Chane]

Jan Saimon, Thank you for your explanatory answer. However,I am getting the error ” Operands to the and && operators must be convertible to logical scalar values.” on the following.

    % code 
if (0<Hx20_50 && Hx20_50<=0.5);
t20_50=sqrt(log(1./(Hx20_50.^2)));
elseif (0.5<Hx20_50 && Hx20_50<=1);
t20_50=sqrt(log(1./(1-Hx20_50).^2));
end

Thank you.

[Jan]

If your Hx20_50 is a vector, you cannot use && but need &. But then the condition of the if command is a vector also. Because if requires a scalar condition, a all() is inserted automatically. This is most likely not, what you want.

Either use a loop:

t20_50 = zeros(size(Hx20_50)); % Pre-allocate!
for k = 1:numel(Hx20_50)
  if 0<Hx20_50(k) && Hx20_50(k)<=0.5
    t20_50(k) = sqrt(log(1 ./ (Hx20_50(k) .^ 2)));
  elseif 0.5<Hx20_50(k) && Hx20_50(k)<=1
    t20_50(k) = sqrt(log(1 ./ (1-Hx20_50(k)) .^ 2));
  end
end

or use the logical indexing shown by Walter.

[Engdaw Chane]

@Jan This was what I have been looking for. The others also work but I wasn't getting the dimensions I needed because of a reason in my code. Thank you.

Answer 2

[Birdman]

Actually, you do not need a ifelse statement. Simple logical indexing will give you what you want.

ab=[10,2,2;4,12,6;7,5,9];
abc=repmat(ab,1,1,3);
abc(abc>1 & abc<=5)=abc(abc>1 & abc<=5)-5;
abc(abc>6 & abc<=10)=abc(abc>6 & abc<=10)+10

[Engdaw Chane]

Birdman, Thank you for replying. The logical indexing works. But when I adapt the LOGICAL INDEXING to my matrix I am still getting error “In an assignment A(:) = B, the number of elements in A and B must be the same. “

 % end
t20_50(Hx20_50>0 & Hx20_50<=0.5)=sqrt(log(1./(Hx20_50.^2)));

Where Hx20_50 is a 3D matrix, and t20_50 is the resulting matrix;

Can I use the LOGICAL INDEXING here? What went wrong in the above?

Kindly,

Chane

[Steven Lord]

The expression on the left side of the equals sign refers to only those elements in t20_50 whose corresponding elements in Hx20_50 are greater than 0 and less than or equal to 0.5. The expression on the right side operates on all the elements in Hx20_50.

Use the logical mask on both sides so you're working with the same sized "chunk" of t20_50 and Hx20_50.

ind = Hx20_50>0 & Hx20_50<=0.5;
t20_50(ind) = sqrt(log(1./(Hx20_50(ind).^2)));

[Engdaw Chane]

@Steven Lord, Thank you for helping. When I do so: 1, the dimension disappears.

2, I have four conditions to consider.

Please take a look at the whole code here.

     % code
if (0<Hx20_50 && Hx20_50<=0.5);
t20_50=sqrt(log(1./(Hx20_50.^2)));
elseif (0.5<Hx20_50 && Hx20_50<=1);
t20_50=sqrt(log(1./(1-Hx20_50).^2));
end

Thank you

[Walter Roberson]

mask = 0<Hx20_50 & Hx20_50<=0.5;
t20_50(mask) = sqrt(log(1./(Hx20_50(mask).^2)));
mask = 0.5<Hx20_50 & Hx20_50<=1;
t20_50(mask) = sqrt(log(1./(1-Hx20_50(mask)).^2));

What do you want to do for the cases where Hx20_50 <= 0 or Hx20_50 > 1 ?

[Engdaw Chane]

@Walter Hx20_50 is probability of elements, and the values are just between 0 and 1. just the dimensions are missing. Thank you.

[Walter Roberson]

If the inputs are certain to be in that range then you can simplify to

mask = Hx20_50 <= 0.5;
t20_50(mask) = sqrt(log(1./(Hx20_50(mask).^2)));
t20_50(~mask) = sqrt(log(1./(1-Hx20_50(~mask)).^2));