Replacing Values Between a 0 and a 1 in a Vector
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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
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Robert U
il 18 Mag 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
Più risposte (1)
Stephen23
il 18 Mag 2020
Modificato: Stephen23
il 22 Mag 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 Commenti
Stephen23
il 21 Mag 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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