Change duplicate elements in an array with zero

hi, i have an array look like:

A = [1 1 0 0 -1 - 1 - 1 0 1 1 -1 ];

how can i turn it into:

A = [1 0 0 0 -1 0 0 0 1 0 -1];

Thanks

Answer 1

[Walter Roberson]

A([false,A(1:end-1)==A(2:end)]) = 0;

[Giorgio Proietti]

it is possible to do this using the 'unique' function?

[Walter Roberson]

[~, ~, uidx] = unique(A);
A([false, uidx(1:end-1)==uidx(2:end)]) = 0;

So you can do it -- it is just a waste of time to do so.

[Giorgio Proietti]

I tried, but it does not work. I insert the complete array:

A = [0, 1, 1, 0, -1, -1, -1, -1, -1, 0, 0, 0, 0, 0, -1, 0, 0, 0, -1, -1, -1, -1, -1, -1, -1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, -1, -1, -1, -1, 0, 1, 1, 1, 1, 0, 0, 0, 0, -1, 0, -1];

and it should be like this:

B = [0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0];

the array B is written by hand.

Anyway, thanks for your help!

[Ameer Hamza]

@Giorgio, your hand-written B vector is wrong. @Walter's code is giving correct output. Below is correct B

B = [0, 1, 0, 0, -1,  0,  0,  0,  0, 0, 0, 0, 0, 0, -1, 0, 0, 0, -1,  0,  0,  0,  0,  0,  0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1,  0,  0,  0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1];

[Walter Roberson]

I think there is a consistent interpretation: that a non-zero entry is to be zeroed out if it is the same as the most recent non-zero entry.

[Giorgio Proietti]

I'll explain better, referring to the A array of length 100. I would like every time that the sign is positive or negative, there are only zeroes. That is, if the first number is positive (+1) it must be followed by all zero until the sign change (-1) or otherwise. But this does not happen, if you look at the last three digits of the vector B (posted by @Ameer) are: ...- 1,0, -1]. Instead, I would like them to be: ...- 1,0,0]. The same happens in the central position to the array.

Answer 2

[Ameer Hamza]

@Giorgio as you mentioned in your comment, the new information change which you mentioned in your question. The question can now be interpreted as only retain first non-zero entry after a sign change. The following code will work for you

index = A==0;
A_ = A(~index);
A_ = [A_(1) diff(A_)/2];
A(~index) = A_;

[Giorgio Proietti]

Thank you all for your cooperation!