[SciPy-User] re[SciPy-user] moving for loops...
Benjamin Root
ben.root@ou....
Thu Jun 17 11:33:27 CDT 2010
Well, there is numpy.tile() that can replicate a numpy array multiple times
in various dimensions, but I have to wonder if you need to step back and
think about why your dimensions aren't matching up. If you are pulling in
data where one of the dimensions has a length of 90, but the corresponding
array that denotes that dimension (timesteps) has a length of 30, then I
have to wonder if there needs to be some rethinking of the overall design.
Ben Root
On Thu, Jun 17, 2010 at 11:02 AM, mdekauwe <mdekauwe@gmail.com> wrote:
>
> So what happens if I need to extend in two directions at once? So for
> example
> I have 2 arrays...
>
> timesteps = np.arange(30)
> and
> y which has dimensions 90, 3, where each row has values of say 1.2, 3.4,
> 5.5,
>
> then I have some function I call "func", where the arguments are timesteps
> and each of the rows of y, e.g.
>
> func(x, y[0,:])
>
> However if I wanted to carry out this step for each of the rows (90) of y,
> I
> can't seem to broadcast this correctly.
>
> tmp = func(timesteps[:,np.newaxis], y)
>
> I can see why, as this will only stretch timesteps in one direction so that
> it becomes (30, 3), so is there a nice way to stretch the rows as well? I
> thought perhaps I needed to reshape timesteps first, but I didn't seem to
> solve it that way either.
>
> Incase none of this made sense... my original loop version
>
> tmp = np.zeros((90, 30))
> for i in xrange(len(y)):
> tmp[i,:] = func(timsteps, y[i])
>
> I have the feeling I am missing something very obvious here! Thanks!
>
> --
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> http://old.nabble.com/removing-for-loops...-tp28633477p28916343.html
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>
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