Re: vectorization problem

[Bård Skaflestad]

On Tue, 2011-11-01 at 18:47 +0100, John W. Eaton wrote:
> Is there an efficient way to replace all NaNs in each column of the
> following matrix by the last non-NaN value in the column?
> 
> Example matrix:
> 
>   M = [1   3   7
>        2   4   NaN
>        NaN 5   NaN
>        NaN 6   NaN];
> 
> Desired result:
> 
>       [1   3   7
>        2   4   7
>        2   5   7
>        2   6   7];
> 
> For my purposes, I think it is safe to assume that all columns will
> have at least one value that is not NaN, but, as above, there may be
> some columns that do not have any NaN values at all.

Here is a possible solution using the run-length decoding technique
described in Peter J. Acklam's note "MATLAB array manipulation tips and
tricks" available at

   http://home.online.no/~pjacklam/matlab/doc/mtt/index.html

Below, this is represented by the 'rldecode' function.

        [m, n] = size (M);
        i      = isnan (M);
        last   = sum (~i); # or, maybe, sum (double (~i))
        M(i)   = rldecode (M ((0:n-1)*m + last), m - last, 2);

I attach the function that we use in our MRST package.  It was authored
by Jostein R. Natvig at SINTEF ICT.  We've found it to be a invaluable
tool when handling variable-length (e.g., CSR-type) data.

I cannot judge the applicability of this technique in a core function
like 'patch'.  There is a possibility that it uses too much memory or
that there is no performance gain with respect to the loop-based
solution you posted earlier.  I'd say careful testing and measurement is
advised.


Sincerely,
-- 
Bård Skaflestad <address@hidden>
SINTEF ICT, Applied Mathematics

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