Hello, I need help with: Replace elements in Vector A with those of vector B of the same position, only if they meet a certain condition, otherwise replace by a zero. Thanks

I have Vectors A and B of the same length and a threshhold value:

Threshold = 4;
A = [3     7    1     3    10     3     4     1     5     5];
B = [5     3     6     1     6     3     5     4     9     4];

I want replace all values in A that are greater than threshold (A > Threshold) with values in B if they are less than the threshold (B < Threshold), otherwise make them zeros. In thi case, the new vector of A should be:

Thank you.

[MuluMe]

Am so grateful team, for your brilliant ideas. They have worked.

Answer 1

[Jon]

I like @Torsten's one liner and use of the multiplication times the logical zeros to null out the values where B is greater than the threshold. Alternatively, you could do it like this in two lines and the logic is a little more obvious

Threshold = 4;
A = [3     7    1     3    10     3     4     1     5     5];
B = [5     3     6     1     6     3     5     4     9     4];
A(A>Threshold) = B(A>Threshold)

A = 1×10

3 3 1 3 6 3 4 1 9 4

A(A>Threshold) = 0 % in case any of the B's we put in are over the threshold set them to zero

A = 1×10

3 3 1 3 0 3 4 1 0 4

[Torsten]

The last "4" is wrong. I think the exchange of values between A and B must be done in one step.

[Jon]

Good catch @Torsten You're right this approach does not work - thanks

[Jon]

Thanks for accepting my answer, but I think actually @Torsten approach looks the closest to what you describe. I think you can click on the accepted answer and change it if you want.

Answer 2

[Torsten]

I kept the values in A unchanged that are less or equal to Threshold. If you also want to replace them by 0, take David's answer.

Threshold = 4;
A = [3     7    1     3    10     3     4     1     5     5];
B = [5     3     6     1     6     3     5     4     9     4];
A(A>Threshold) = B(A>Threshold).*(B(A>Threshold) < Threshold)

A = 1×10

3 3 1 3 0 3 4 1 0 0

Answer 3

[David Hill]

t=4;
A = [3     7    1     3    10     3     4     1     5     5];
B = [5     3     6     1     6     3     5     4     9     4];
idx=A>t&B<t;
A(idx)=B(idx);
A(~idx)=0

A = 1×10

0 3 0 0 0 0 0 0 0 0