Replacing Values Between a 0 and a 1 in a Vector
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Daniel Steyer
el 18 de Mayo de 2020
Comentada: Robert U
el 25 de Mayo de 2020
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
0 comentarios
Respuesta aceptada
Robert U
el 18 de Mayo de 2020
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
3 comentarios
Robert U
el 25 de Mayo de 2020
Thanks Stephen, I did not see that. I like your suggested improvement.
Más respuestas (1)
Stephen23
el 18 de Mayo de 2020
Editada: Stephen23
el 22 de Mayo de 2020
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
3 comentarios
Stephen23
el 21 de Mayo de 2020
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
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