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From: Young Ryu on 5 Aug 2010 02:08
I'd like to replace NaN by using linear interpolation in the below example. Could you recommend fast way as I am dealing very large data actually.

A=[1 2 3 NaN 5 NaN 5 4 6];

I tried this:
ind=1:length(A);
A1=interp1(ind(~isnan(A)), A(~isnan(A)), ind, 'linear');

However, I need faster way.

Thank you!
From: us on 5 Aug 2010 03:24
"Young Ryu" <ryuyr77(a)gmail.com> wrote in message <i3dkg4$lus$1(a)fred.mathworks.com>...
> I'd like to replace NaN by using linear interpolation in the below example. Could you recommend fast way as I am dealing very large data actually.
>
> A=[1 2 3 NaN 5 NaN 5 4 6];
>
> I tried this:
> ind=1:length(A);
> A1=interp1(ind(~isnan(A)), A(~isnan(A)), ind, 'linear');
>
> However, I need faster way.
>
> Thank you!

one of the solutions

ix=~isnan(A); % <- compute the logical array once...
a1=interp1(ind(ix),A(ix),ind,'linear');

us
From: TideMan on 5 Aug 2010 05:14
On Aug 5, 6:08 pm, "Young Ryu" <ryuy...(a)gmail.com> wrote:
> I'd like to replace NaN by using linear interpolation in the below example. Could you recommend fast way as I am dealing very large data actually.
>
> A=[1 2 3 NaN 5 NaN 5 4 6];
>
> I tried this:
> ind=1:length(A);
> A1=interp1(ind(~isnan(A)), A(~isnan(A)), ind, 'linear');
>
> However, I need faster way.
>
> Thank you!

Here's a very elegant routine that is part of the (free) tidal
analysis toolbox t_tide:

function y=fixgaps(x);
% FIXGAPS Linearly interpolates gaps in a time series
% YOUT=FIXGAPS(YIN) linearly interpolates over NaN
% in the input time series (may be complex), but ignores
% trailing and leading NaN.
%

% R. Pawlowicz 6/Nov/99

y=x;

bd=isnan(x);
gd=find(~bd);

bd([1:(min(gd)-1) (max(gd)+1):end])=0;


y(bd)=interp1(gd,x(gd),find(bd));
From: Greg Heath on 5 Aug 2010 08:42
On Aug 5, 2:08 am, "Young Ryu" <ryuy...(a)gmail.com> wrote:
> I'd like to replace NaN by using linear interpolation in the below example. Could you recommend fast way as I am dealing very large data actually.
>
> A=[1 2 3 NaN 5 NaN 5 4 6];
>
> I tried this:
> ind=1:length(A);
> A1=interp1(ind(~isnan(A)), A(~isnan(A)), ind, 'linear');

As us has indicated, if A is long only compute ~isnan(A)
once.

However, if the gaps are long, you might consider linear
prediction instead of linear interpolation.

Hope this helps.

Greg
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