I think you should use a Symbol for f in your expression, and then add
{'f': mpf(...)} to the namespace dict used by lambdify.
Aaron Meurer
On Dec 1, 2012, at 4:14 PM, Freddie Witherden <fred...@witherden.org> wrote:
> Hi all,
>
> Consider (with mp.dps = 30):
>
> f = mp.mpf('0.1234123094834543252345098')
> q = sy.Symbol('q')
> x = f*q
>
> Now, running lambdastr over the expression we find:
>
> 'lambda q: (0.1234123094834543252345098*q)'
>
> which is curtailed by the fact that our mpf float will be interpreted as
> a Python float. Often times (especially when q is of type mpf) this is
> not sufficient.
>
> Hence, what is the best way to hack/fudge the lambdify mechanism into
> either wrapping mpf constants suitably to yield:
>
> 'lambda q: (mp.mpf('constant')*q)'
>
> or better still having them constructed once (rather than per each
> evaluation)?
>
> Polemically yours, Freddie.
>
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