Vectorized method to sum missed one values

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Hi there,
I am trying to achieve something, but I can't think of a vectorized way of doing this. The problem is as follows.
Say I have a vector of 0's and 1's, e.g. [0, 1, 1, 0, 0, 1, 0, 1]. Then I want to manipulate it in such a way that I get the following vector: [0, 2, 1, 0, 0, 3, 0, 2]. Hence, from left to right, every time a 1 occurs, it adds the number of consecutive preceding zero's, if any.
This can easily be done in a loop, but because of the vast number of computations, I am looking for a vectorized way to achieve this.
Any help is appreciated!
Best, Robert

Accepted Answer

Amir Xz
Amir Xz on 19 Sep 2018
Edited: Amir Xz on 19 Sep 2018
A=[0, 1, 1, 0, 0, 1, 0, 1];
[~,NonZr] = find(A~=0);
A(NonZr) = [NonZr(1),NonZr(2:end)-NonZr(1:end-1)];
Result:
A =
0 2 1 0 0 3 0 2
  4 Comments
Robert Vullings
Robert Vullings on 2 Nov 2018
Any ideas on how to expand this (or another method) to a 2D array?
For example, say
A=[0, 1, 1, 0, 0, 1, 0, 1;
1, 0, 1, 0, 1, 1, 0, 1];
would then become
A=[0, 2, 1, 0, 0, 3, 0, 2;
1, 0, 2, 0, 2, 1, 0, 2];

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More Answers (2)

Guillaume
Guillaume on 19 Sep 2018
Edited: Guillaume on 19 Sep 2018
You can replace the earlier part of this answer by the compiled version of rcumsumc for speed
v = [0, 1, 1, 0, 0, 1, 0, 1];
rcum = double(~v);
csum = cumsum(rcum);
rcum(v == 1) = -diff([0, csum(v == 1)]);
rcsum = cumsum(rcum) + 1;
%all the above can be replaced by rcumsum
%rcsum = rcumsum(~v) + 1;
reploc = diff(v) == 1
v([false, reploc]) = rcsum([reploc, false])
Note that I'm not convinced that it will be faster than a well written loop (which can do the job in only one pass over the data).
  1 Comment
Christopher Wallace
Christopher Wallace on 19 Sep 2018
This is about 20x faster than my answer when run on my machine. Nice work!

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Christopher Wallace
Christopher Wallace on 19 Sep 2018
startingData = [0, 1, 1, 0, 0, 1, 0, 1];
stringArr = sprintf('%d', startingData ); % Convert to string for use with regexp
zerosLoc = regexp(stringArr , '(0*)'); % Find starting index of groups of 0's
onesLoc = regexp(stringArr , '(1*)'); % Find starting index of groups of 1's
startingData(onesLoc) = (onesLoc - zerosLoc) + 1; The difference in the starting location of the ones and the starting location of the zeros which will result in the number of zeros leading up to the 1.
  1 Comment
Guillaume
Guillaume on 19 Sep 2018
Conversions from numbers to strings are never fast, but
stringArr = char(startingData + '0');
will be a lot faster than using sprintf.
However, you don't need regexp and strings to find the start of the sequences.
zerosLoc = find(diff([1, startingData]) == -1); %find [1 0] transitions
onesLoc = find(diff([0, startingData]) == 1); %find [0 1] transitions
With this it may actually be faster than my solution.

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