Replacing Values Between a 0 and a 1 in a Vector

Hello,

I have a data set vector that I've reduced down to 0's, 1's, and 2's. What I want to do is replace any 2's with 0's if they are following a 0, up until the next 1 shows up. For example:

if the original vector: A = [0 0 0 2 2 2 1 2 1 2 2 0 0]

the new vector : B = [0 0 0 0 0 0 1 2 1 2 2 0 0]

My vectors will have about 25,000 to 500,000 data points. Any way that I've tried to do this ends up taking way too long. I'd be appreciative of any advice that you'd be willing to give. If it helps, 0's will never be followed directly by 1's, and any 2's following a 0 will always lead into a 1 before the next 0 shows up.

Daniel

Answer 1

[Robert U]

Hi Daniel Steyer,

this code snippet should provide the requested functionality.

cIn = cellfun(@num2str,num2cell(A),'UniformOutput',false);
strIn = [cIn{:}];

indToChange = regexp(strIn,'(?<=0)(2)+(?=1)','tokenExtents');

for indChanges = 1:numel(indToChange)
    dInput(indToChange{indChanges}(1):indToChange{indChanges}(2)) = 0;
end

B = dInput;

Kind regards,

Robert

[Daniel Steyer]

Hey Robert,

I think I've got it up and running, thanks for the help!

Best,

Daniel

[Stephen23]

Note that this

cIn = cellfun(@num2str,num2cell(A),'UniformOutput',false);
strIn = [cIn{:}];

should be replaced with this simpler and much more efficient code:

strIn = sprintf('%u',A);

[Robert U]

Thanks Stephen, I did not see that. I like your suggested improvement.

Answer 2

[Stephen23]

This should be reasonably efficient:

A = [0,0,0,2,2,2,1,2,1,2,2,0,0];
D = diff(A);
B = find([0,D]==2);
E = find([D==-1,true] & A==2);
for k = 1:numel(B)
    A(B(k):E(k)) = 0;
end

Giving:

A =
   0   0   0   0   0   0   1   2   1   2   2   0   0

Note that this approach relies on your statement "...any 2's following a 0 will always lead into a 1..."

EDIT: more robust end detection:

A = [0,0,0,2,2,2,1,2,1,2,2,0,0,0,0,0,2,2,2,1,2,1,2,2,0,0];
D = diff(A);
B = find([false,D==2]);
E = find([D==-1,true]);
for k = 1:numel(B)
    X = B(k):E(find(E>B(k),1));
    A(X) = 0;
end

[Daniel Steyer]

Hey Stephen,

Thanks for the feedback. The approach worked well on the tester set, but I'm having trouble applying it to larger data sets where multiple regions need to be acted on. For example, if I string two A's together in a vector, it doesn't recognize the second set of 2's to be replaced. E does recognize the final 2 in each chunk to be replaced, but it also recognizes any other 2's that are followed by a 1, which I think trips up the final step. For the below example, when k=2, E(k)=8 when it should equal 19 to make this work.

A=[0,0,0,2,2,2,1,2,1,2,2,0,0]
A2=[A,A]
D=diff(A2)
B=find([0,D]==2)
    %B=[4,17]
E=find([D==-1,true]&A==2)
    %E=[6,8,19,21]
for k=1:numel(B)
    A2(B(k):E(k))=0
end
% The final output A2=[0,0,0,0,0,0,1,2,1,2,2,0,0 , 0,0,0,2,2,2,1,2,1,2,2,0,0]

I think that this approach is close to working and just needs a more selective way of creating E values. I'll see if I can't make it work.

[Stephen23]

Yes you are right, detecting the end index was not very robust. I tried various methods, and this worked well:

A = [0,0,0,2,2,2,1,2,1,2,2,0,0,0,0,0,2,2,2,1,2,1,2,2,0,0];
D = diff(A);
B = find([false,D==2]);
E = find([D==-1,true]);
for k = 1:numel(B)
    X = B(k):E(find(E>B(k),1));
    A(X) = 0;
end

Giving:

A =
   0   0   0   0   0   0   1   2   1   2   2   0   0   0   0   0   0   0   0   1   2   1   2   2   0   0

[Daniel Steyer]

The new version worked like a charm! Thanks for your help.