If I have a vector of either 1\'s or NaN\'s like this:
[1 1 1 NaN 1 1 NaN 1 1 1 1]
How can I reset the cumsum to zero at the location of th
I can only think of a few-pass solution:
v = [1 1 1 NaN 1 1 1 1 NaN 1];
a = v==v; %% convert the values first to [1 1 1 0 1 1 1 1 0 1] format
n = a==0; %% positions of the NaNs
c = cumsum(a); %% your intermediate result
d = diff([0 c(n)]); %% runs of ones
v(n) = -d; %% replace Nans by -3, -4 [1 1 1 -3 1 1 1 1 -4 1]
cumsum(v) %% the answer [1 2 3 0 1 2 3 4 0 1]
Note: haven't checked extreme conditions (NaN
in first/Last position, consecutive NaN
s etc.)