"Diego Lass" <dlISCool@gmail.com> wrote in message <h171ga$inm$1@fred.mathworks.com>...
> Hi
> I want to replace NaN in the current matrix with 0, what is the fastest way?
>
> A = [ 1 2 NaN; NaN 2 3; NaN NaN 100 ]
>
> A =
>
> 1 2 NaN
> NaN 2 3
> NaN NaN 100
>
> Thanks
> Diego
"Diego Lass" <dlISCool@gmail.com> wrote in message <h171ga$inm$1@fred.mathworks.com>...
> Hi
> I want to replace NaN in the current matrix with 0, what is the fastest way?
>
> A = [ 1 2 NaN; NaN 2 3; NaN NaN 100 ]
>
> A =
>
> 1 2 NaN
> NaN 2 3
> NaN NaN 100
>
> Thanks
> Diego
In article <h1731i$pue$1@fred.mathworks.com>, dlISCool@gmail.com says...
> OK I got it
> k = find(isnan(A))';
> A(k) = 0;
>
> Diego
>
>
> "Diego Lass" <dlISCool@gmail.com> wrote in message <h171ga$inm$1@fred.mathworks.com>...
> > Hi
> > I want to replace NaN in the current matrix with 0, what is the fastest way?
> >
> > A = [ 1 2 NaN; NaN 2 3; NaN NaN 100 ]
> >
> > A =
> >
> > 1 2 NaN
> > NaN 2 3
> > NaN NaN 100
> >
> > Thanks
> > Diego
>
No need for the overhead with find. Follow John D's advice.
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