flip the sign at zero crossing point

I have an array vector (signal ) contains vales ones and zeros. e.g data set as follow

Data1 = [1,0,1,0,1,0,1,0,1,0];

I want to reverse the sign (phase) of values at zero-crossing points to become as follow

Data1=[1, 0,-1, 0, 1, 0,-1, 0, 1, 0]

do you know an elegant way to do that on a large data set?

Thanks in advance.

Answer 1

[Stephen23]

>> Data1 = [1,0,1,0,1,0,1,0,1,0]
Data1 =
   1   0   1   0   1   0   1   0   1   0
>> Data2 = Data1.*cumprod([1,diff(Data1==0)])
Data2 =
   1   0  -1   0   1   0  -1   0   1   0

[nashat sawai]

Dear Stephen , thank you so much, it works perfect,

you made my day..

[Vladimir Sovkov]

This algorithm works with only this input vector, for which case it can be done much simpler. If you have two equal elements in a sequence, it produces wrong results. Try Data1 = [1,0,1,0,1,1,0,1,0,1] or Data1 = [1,0,1,0,0,1,0,1,0,1,0].

[Stephen23]

@Vladimir Sovkov: you are right, the diff can return zero. Luckily this is easily fixed:

>> Data1 = [1,0,0,1,1,0,1,0,1,1,0,0,0,1,0,0]
Data1 =
     1     0     0     1     1     0     1     0     1     1     0     0     0     1     0     0
>> D = [1,diff(Data1==0)];
>> D(D==0) = 1;
>> Data2 = Data1 .* cumprod(D)
Data2 =
     1     0     0    -1    -1     0     1     0    -1    -1     0     0     0     1     0     0

[Vladimir Sovkov]

In my understanding, the sign must alternate between every zero elements. Hence, it must change after the odd number of zeros in a sequence but must not after the even number of zeros. It is not fulfilled yet.

Answer 2

[Vladimir Sovkov]

Data1 = [1,0,1,1,0,1,1,0,0,1,0,1,0];
k=find(~Data1);
if ~isempty(k)
    if k(end)<numel(Data1)
        k(end+1)=numel(Data1)+1;
    end
    for n=1:2:numel(k)-1
        Data1(k(n)+1:k(n+1)-1)=-1;
    end
end
disp(Data1);

[Stephen23]

Simpler:

>> Data1 = [1,0,1,1,0,1,1,0,0,1,0,1,0]
Data1 =
     1     0     1     1     0     1     1     0     0     1     0     1     0
>> Data2 = Data1 .* cumprod(1-2*(Data1==0))
Data2 =
     1     0    -1    -1     0     1     1     0     0     1     0    -1     0

[Vladimir Sovkov]

Yes!