ok, I oversimplified my example. In fact my computations are in vectors. 
The problem seems to be that I cannot pass the results of 
[sympy.exp(x[i]) for i in range(dim)] to Theano (for x[i] small enough). 
Theano evaluate them as zero but here I need the real values of 
exp([x[i]]), where only
sympy.exp can give me.
Costas  

On Wednesday, May 20, 2015 at 4:57:18 PM UTC+3, Frédéric Bastien wrote:
>
> The execution by Theano should be faster when you work big not too small 
> inputs size. If you work only on scalar, Theano is probably not faster then 
> sympy.
>
> But if you work on vector/tensor big enough (this size depend of the 
> function...) then Theano get faster.
>
> Fred
>
> On Wed, May 20, 2015 at 9:48 AM, <[email protected] <javascript:>> wrote:
>
>> Thank you for your answer.
>> So, in what sense theano cooperates with sympy (or the inverse)?
>> Since, we can not have the advantage of sympy functions in theano.
>> Costas
>>
>> On Wednesday, May 20, 2015 at 1:41:06 AM UTC+3, [email protected] wrote:
>>>
>>> Hi,
>>>
>>> I was experimenting with the following code:
>>>
>>> import sympy
>>> from sympy.abc import x
>>> expr = sympy.exp(-x ** 2)
>>> from sympy.printing.theanocode import theano_function
>>> fn_theano  = theano_function([x], [expr], dims={x: 1}, dtypes={x: 
>>> 'float64'})
>>> print fn_theano([29.]),sympy.exp(-29.**2)
>>>
>>> I got 
>>>
>>>
>>> [ 0.] 5.73245586032578e-366
>>>
>>>
>>> My question is why fn_theano does not  give the same value?
>>>
>>> Thank you,
>>>
>>> Costas
>>>
>>>
>>>
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