Poul Riis wrote:
> Den onsdag den 30. marts 2016 kl. 17.59.49 UTC+2 skrev Steven D'Aprano:
>> On Thu, 31 Mar 2016 02:23 am, Poul Riis wrote:
>>
>> > What I intend to do is to let sympy find the derivative of some
>> > welldefined function and next define the foundation derivative as a
>> > normal function so that I can calculate numerical values or even make a
>> > graph.
>>
>>
>> I'm glad you explained what you *actually* wanted, because I was going to
>> guess that you wanted to evaluate the derivative at x = 3:
>>
>>
>> py> ftext.evalf(subs={x:3})
>> -0.0600000000000000
>>
>>
>>
>> --
>> Steven
>
> ... However, the sympy way seems to be about 70 times slower than using
> the derivative calculated 'by hand' (try the example below). Can it be
> done in a more efficient way?
Hm, the two functions fmsympy() and fm() do not return the same value:
$ python -i sympy_diff.py
10000 evaluations with sympy : dt1 = 0.7178411483764648
10000 evaluations without sympy: dt2 = 0.10177111625671387
>>> fm(42)
cos(42)
>>> fmsympy(42)
-0.399985314988351
What's the point of that benchmark?
> Poul Riis
>
>
>
> from sympy import *
> from time import *
> x=Symbol('x')
> ftext=diff(sin(x),x)
>
> def fmsympy(t):
> return ftext.evalf(subs={x:t})
>
> def fm(t):
> return cos(t)
>
> nloop=10000
> tstart=time()
> # nloop evaluations with sympy
> for i in range(0,nloop):
> a=fmsympy(1)
> dt1=time()-tstart
>
> # nloop evaluations without sympy
> tstart=time()
> for i in range(0,nloop):
> a=fm(1)
> dt2=time()-tstart
>
> print(nloop,' evaluations with sympy : dt1 =',dt1)
> print(nloop,' evaluations without sympy: dt2 =',dt2)
--
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