Re: [Numpy-discussion] Re: calculating on matrix indices

[Ryan K. <rya...@gm...>]

Brian's example works fine for me as long as
x=3Dzeros(len(t),'d')

for some reason the error seems to come from assignment of a double
array to a single precision array:

In [318]: x=3Dzeros(len(t),'f')

In [319]: t=3Darange(0,20,.1)

In [320]: idx=3Dwhere(t>5)

In [321]: x[idx]=3Dexp(-t[idx]/tau)
---------------------------------------------------------------------------
exceptions.TypeError                                 Traceback (most
recent call last)

/home/ryan/thesis/accuracy/<ipython console>

TypeError: array cannot be safely cast to required type

In [322]: x=3Dzeros(len(t),'d')

In [323]: x[idx]=3Dexp(-t[idx]/tau)

In [324]:

Ryan

On 2/16/06, Travis Oliphant <oli...@ie...> wrote:
> Brian Blais wrote:
>
> > Colin J. Williams wrote:
> >
> >> Brian Blais wrote:
> >>
> >>> In my attempt to learn python, migrating from matlab, I have the
> >>> following problem. Here is what I want to do, (with the wrong syntax)=
:
> >>>
> >>> from numpy import *
> >>>
> >>> t=3Darange(0,20,.1)
> >>> x=3Dzeros(len(t),'f')
> >>>
> >>> idx=3D(t>5)                # <---this produces a Boolean array,
> >>> probably not what you want.
> >>> tau=3D5
> >>> x[idx]=3Dexp(-t[idx]/tau)  # <---this line is wrong (gives a TypeErro=
r)
> >>>
> >> What are you trying to do?  It is most unlikely that you need Boolean
> >> values in x[idx]
> >>
> >
> > in this example, as in many that I would do in matlab, I want to
> > replace part of a vector with values from another vector.  In this
> > case, I want x to be zero from t=3D0 to 5, and then have a value of
> > exp(-t/tau) for t>5.  I could do it with an explicit for-loop, but
> > that would be both inefficient and unpython-like.  For those who know
> > matlab, what I am doing here is:
> >
> from numpy import *
>
>
> t =3D r_[0:20:0.1]
> idx =3D t>5
> tau =3D 5
> x =3D zeros_like(t)
> x[idx] =3D exp(-t[idx]/tau)
>
>
> Should do it.
>
> -Travis
>
>
>
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