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## Re: Easy about Matrices

 From: avraham Subject: Re: Easy about Matrices Date: Tue, 19 Apr 2005 14:47:30 +0300 User-agent: Mutt/1.5.5.1+cvs20040105i

```On Mon, Apr 18, 2005 at 02:18:46PM -0700, Henry F. Mollet wrote:
> I don't understand fully. As outlined x will end up to be a zero matrix,
> whereas I thought one should end up with 2 matrices, pos values replaced
> with zeros in one, negative values replaced with zeros in the other?
> Henry
> octave:18> x = randn(2,4)
> x =
>
>   -0.273788  -0.671445  -1.522762   0.170152
>   -2.135687   1.040917   0.017514  -0.413427
>
> octave:19> x_positive = x(x<0)=0
> x_positive = 0
> octave:20> x
> x =
>
>   0.00000  0.00000  0.00000  0.17015
>   0.00000  1.04092  0.01751  0.00000
>
> octave:21> x_negative = x(x>=0)=0
> x_negative = 0
> octave:22> x
> x =
>
>   0  0  0  0
>   0  0  0  0
>
> % Trying the following:
> octave:27> x=xpos=xneg=randn(2,4)
> x =
>
>   -0.77668  -0.21781   0.45580  -0.34135
>   -0.57917   0.78780   1.97339  -0.79024
>
> octave:28> x_positive = xpos(xpos<0)=0
> x_positive = 0
> octave:29> xpos
> xpos =
>
>   0.00000  0.00000  0.45580  0.00000
>   0.00000  0.78780  1.97339  0.00000
>
> octave:30> x_negative = xneg(xneg>=0)=0
> x_negative = 0
> octave:31> xneg
> xneg =
>
>   -0.77668  -0.21781   0.00000  -0.34135
>   -0.57917   0.00000   0.00000  -0.79024
>
> octave:32> whos
>
> *** local user variables:
>
> prot  type                       rows   cols  name
> ====  ====                       ====   ====  ====
>  rwd  matrix                        2      4  x
>  rwd  scalar                        1      1  x_negative
>  rwd  scalar                        1      1  x_positive
>  rwd  matrix                        2      4  xneg
>  rwd  matrix                        2      4  xpos
>
>
> on 4/18/05 11:06 AM, avraham at address@hidden wrote:
...
> > An alternative could be (example):
> > 684~> x = randn(12,100);
> > x_positive = x(x<0)=0;
> > x_negative = x(x>=0)=0;
> > The two arrays have the size of the original one, but the terms
> > of "wrong sign" are replaced by zeros.
> > Check:
> > y=x_positive+x_negative;
> > 694~> min(min(y-x))
> > ans = 0
> > 694~> max(max(y-x))
> > ans = 0
> >
> > In order to avoid plotting the points at y=0, one can use NaN
> > instead of 0: x_positive = x(x<0)=NaN; etc
> >
> > Cheers, Avraham
> >
> >
> >
> > -------------------------------------------------------------
> > Octave is freely available under the terms of the GNU GPL.
> >
> > Octave's home on the web:  http://www.octave.org
> > How to fund new projects:  http://www.octave.org/funding.html
> > Subscription information:  http://www.octave.org/archive.html
> > -------------------------------------------------------------
> >
Thanks for the correction. The funny part is that, as I felt very
tired, I opened an octave terminal to check myself, before I
write. And then I copy-pasted from it.
I guess that I wrote the wrong answer and then checked
my answer in a way that did not reveal the mistake, but lulled me
in the belief, that despite tiredness, I wrote the right thing.
Next time I shall know better (which probably means that I'll
make a different mistake!).
Cheers, Avraham

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