[Numpy-discussion] element wise help

[josef.pktd at gmail.com]

On Thu, May 7, 2009 at 2:11 PM, Chris Colbert <sccolbert at gmail.com> wrote:
> alright I got it working. Thanks!
>
> This version is an astonishingly 1900x faster than my original
> implementation which had two for loops. Both versions are below:
>
> thanks again!
>
> ### new fast code ####
>
>     b = 4.7
>     n = arange(1, N+1, 1.0).reshape(N, -1)
>     n1 = (-1)**n
>     prefix = exp(b) / timearray
>
>     arg1 = {'S': b / timearray}
>     exec('from numpy import *', arg1)
>     term1 = (0.5) * eval(transform, arg1)
>
>     temp1 = b + (1J * pi * n)
>     temp2 = temp1 / timearray
>     arg2 = {'S': temp2}
>     exec('from numpy import *', arg2)
>     term2 = (eval(transform, arg2) * n1).sum(axis=0).real
>
>     f = prefix * (term1 + term2)
>
>     return f

If you don't do code generation and have control over transform, then,
I think, it would be more readable to replace the exec and eval by a
function call to transform.

I haven't found a case yet where eval is necessary, except for code
generation as in sympy.

Josef



>
> ##### old slow code ######
>     b = 4.7
>     f = []
>
>     for t in timearray:
>         rsum = 0.0
>         for n in range(1, N+1):
>             arg1 = {'S': ((b/t) + (1J*n*pi/t))}
>             exec('from numpy import *', arg1)
>             tempval = eval(transform, arg1)*((-1)**n)
>             rsum = rsum + tempval.real
>         arg2 = {'S': b/t}
>         exec('from numpy import *', arg2)
>         tempval2 = eval(transform, arg2)*0.5
>         fval = (exp(b) / t) * (tempval2 + rsum)
>         f.append(fval)
>
>    return f
>
>
>
>
> On Thu, May 7, 2009 at 1:41 PM, Chris Colbert <sccolbert at gmail.com> wrote:
>>
>> its part of a larger program for designing PID controllers. This
>> particular function numerical calculates the inverse laplace transform using
>> riemann sums.
>>
>> The exec statements, from what i gather, allow the follow eval statement
>> to be executed in the scope of numpy and its functions. I don't get how it
>> works either, but it doesnt work without it.
>>
>> I've just about got something working using broadcasting and will post it
>> soon.
>>
>> chris
>>
>> On Thu, May 7, 2009 at 1:37 PM, <josef.pktd at gmail.com> wrote:
>>>
>>> On Thu, May 7, 2009 at 1:08 PM, Chris Colbert <sccolbert at gmail.com>
>>> wrote:
>>> > let me just post my code:
>>> >
>>> > t is the time array and n is also an array.
>>> >
>>> > For every value of time t, these operations are performed on the entire
>>> > array n. Then, n is summed to a scalar which represents the system
>>> > response
>>> > at time t.
>>> >
>>> > I would like to eliminate this for loop if possible.
>>> >
>>> > Chris
>>> >
>>> > #### code ####
>>> >
>>> > b = 4.7
>>> > f = []
>>> > n = arange(1, N+1, 1)
>>> >
>>> > for t in timearray:
>>> >         arg1 = {'S': ((b/t) + (1J*n*pi/t))}
>>> >         exec('from numpy import *', arg1)
>>> >         tempval = eval(transform, arg1)*((-1)**n)
>>> >         rsum = tempval.real.sum()
>>> >         arg2 = {'S': b/t}
>>> >         exec('from numpy import *', arg2)
>>> >         tempval2 = eval(transform, arg2)*0.5
>>> >         fval = (exp(b) / t) * (tempval2 + rsum)
>>> >         f.append(fval)
>>> >
>>> >
>>> > #### /code #####
>>> >
>>>
>>> I don't understand what the exec statements are doing, I never use it.
>>> what is transform?
>>> Can you use regular functions instead or is there a special reason for
>>> the exec and eval?
>>>
>>> In these expressions ((b/t) + (1J*n*pi/t)),  (exp(b) / t)
>>> broadcasting can be used.
>>>
>>> Whats the size of t and n?
>>>
>>> Josef
>>> _______________________________________________
>>> Numpy-discussion mailing list
>>> Numpy-discussion at scipy.org
>>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>>
>
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