Skipping nan's for computation and starting the vector where first value appears

Dear Matlab users,

I have two vectors of the same length, but one which has some missing values at the beginning:

x1= [1; 3; 4; 6; 8; 1; 3];
x2= [nan; nan; 5; 3; 7; 4; 2; 6];

of these two vectors I take the first difference to get the growth rates:

dx1=diff(x1);
dx2=diff(x2);

In a next step I would transform these growth rates (... with a break in the mean test, which is not described here as it does not matter for the example at hand) and add them back to the initial series. I use the following loop:

T=length(x1); 
for tt=2:T
    x1(tt,1)=x1(tt-1,1)+dx1(tt-1,1);
    x2(tt,1)=x2(tt-1,1)+dx2(tt-1,1);
end

While this works fine if the vector does not contain any NAN, such as for x1:

x1= [1; 3; 4; 6; 8; 1; 3];

my problem is that for x2 there now appears the following:

x2=[NaN; NaN; NaN; NaN; NaN; NaN; NaN; 6];

How can I transform the loop such that it detects for x2 that the first two elements are nan and yet starts at x2(3,1)? This might be a silly and simple question but I was not able to find an answer within a reasonable amount of time and would appreciate very much your support.

Best and many thanks, Philipp

Answer 1

[Stephen23]

The function isnan is used to identify NaN values. Even better would be to use isfinite, assuming that the NaN values only occur consecutively at the start of the vector:

idx = isfinite(x1);
x1 = x1(idx);

If the NaN values are non-consecutive, then you could locate the last one using find, something like this:

idy = [find(isnan(x1),1,'last'),1];
x1 = x1(idy(1):end);

Of course if you want both vectors to be of the same length then you need to apply this indexing to both of them.

Note that the for-loop can be replaced by much simpler and faster vectorized code:

x1 = x1(1)+cumsum([0;dx1])
x2 = x2(1)+cumsum([0;dx2])

Answer 2

[Philipp]

Thank you very much for your detailed answer which helped me a lot.

[Stephen23]

My pleasure!

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