Use lambdify().
Aaron Meurer
On Wed, Jun 12, 2013 at 1:06 PM, Shawn Garbett <shawn.garb...@gmail.com> wrote:
> I'm struggling with the most direct route to use a time series numpy array
> to get a time series back from a SymPy equation.
>
> Ideally I'd like to call something like sol.evalf(x), but that doesn't work.
> I have a loop structure that does it at present, but the continual
> reconstructing a dict causes the processor to heat up and the fans come on.
> I figure, there has to be a shorter / better / direct method. I'm at a loss
> to find it.
>
> Here's what the data looks like:
>
> In [212]: sol
> Out[212]: 1.0e-6*s4 + 1.0e-12*s6
>
> In [213]: x
> Out[213]:
> array([(1.0, 0.33, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0),
> (7.332991703456984e-07, 0.06344978343322456, 1.0,
> 1.9622076748233104e-10, 0.7332991703050197, 0.266668723224782,
> 3.1373201206573815e-05, 1.1995787763723961e-07),
> (6.853150238285165e-07, 0.015615017796943445, 1.0,
> 2.359623438104902e-09, 0.6853150237422331, 0.31460014764904676,
> 8.41433121893229e-05, 6.888955138586781e-07),
> ...,
> (7.447520789346351e-07, 0.00010069599712919727, 1.0,
> 1.5661950866577716, 0.7447520734699161, 0.2497844169697847,
> 0.005464700864489391, 0.00815509952413137),
> (7.446567577748946e-07, 0.0001007088880961995, 1.0,
> 1.5662440343737307, 0.7446567523064337, 0.24987599614231418,
> 0.005468443007273509, 0.008160817601891841),
> (7.445614740433947e-07, 0.00010072177731066373, 1.0,
> 1.5662930164082658, 0.7445614685711894, 0.24996753549603745,
> 0.005472187539255912, 0.008166538792436515)],
> dtype=[('s0', '<f8'), ('s1', '<f8'), ('s2', '<f8'), ('s3', '<f8'),
> ('s4', '<f8'), ('s5', '<f8'), ('s6', '<f8'), ('s7', '<f8')])
>
>
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