Replacing Values Between a 0 and a 1 in a Vector

15 Ansichten (letzte 30 Tage)
Daniel Steyer
Daniel Steyer am 18 Mai 2020
Kommentiert: Robert U am 25 Mai 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

Akzeptierte Antwort

Robert U
Robert U am 18 Mai 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 Kommentare
Stephen23
Stephen23 am 22 Mai 2020
Bearbeitet: Stephen23 am 22 Mai 2020
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
Robert U am 25 Mai 2020
Thanks Stephen, I did not see that. I like your suggested improvement.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Stephen23
Stephen23 am 18 Mai 2020
Bearbeitet: Stephen23 am 22 Mai 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 Kommentare
Stephen23
Stephen23 am 21 Mai 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
Daniel Steyer
Daniel Steyer am 22 Mai 2020
The new version worked like a charm! Thanks for your help.

Melden Sie sich an, um zu kommentieren.

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by