Take a look at https://github.com/sympy/sympy/pull/7318. I remember that
this PR didn't work for all matrices (I guess matices including expressions
with sqrt(2)). If we find a better way to determine the common factor, I'll
update the PR.
On Saturday, November 28, 2015 at 8:08:52 PM UTC+1, Riccardo Rossi wrote:
>
> Dear Mateusz,
>
> what you suggest is exactly what i was hoping for and could not find by
> googling
>
> i'll definitely try that out :-)
>
> Dzienkuje Bardzo! (hope spelling is correct and...that i guessed correctly
> your nationalty)
>
> cheers
> Riccardo
>
> On Saturday, November 28, 2015 at 11:07:20 AM UTC+1, Mateusz Paprocki
> wrote:
>>
>> Hi,
>>
>> On 27 November 2015 at 19:34, Riccardo Rossi <roug...@gmail.com> wrote:
>> > Dear list,
>> >
>> > i am a newby to sympy, and i should say that i liked what i found, so
>> ...
>> > first of all kudos to the developers.
>> >
>> > as of now i can succesfully generate my finite element matrices using
>> sympy,
>> > which saves me quite a lot of work.
>> >
>> > the point is that now i would like to optimize a bit what i did, and i
>> would
>> > like to collect some common factors between the entries of a matrix.
>> >
>> > for example imagine that i have (pseudocode and just an example, no
>> physics
>> > behind)
>> >
>> > a,b = symbols('a b')
>> >
>> > A = Matrix(2,1)
>> > A[0] = a*(exp(a+b)+exp(b^2))
>> > A[1] = b*(exp(a+b)+exp(b^2))
>> >
>> > i would like a way to detect that the term
>> > (exp(a+b)+exp(b^2))
>> >
>> > is common to the different entries and eventually later on do something
>> of
>> > the type
>> >
>> > aux = (exp(a+b)+exp(b^2))
>> > A[0] = a*aux
>> > A[1] = b*aux
>> >
>> > note that later on for me it would be still interesting to do something
>> > similar on SOME of the entries of the matrix
>> >
>> > for example if i had
>> >
>> > A = Matrix(3,1)
>> > A[0] = a*(exp(a+b)+exp(b^2))
>> > A[1] = b*(exp(a+b)+exp(b^2))
>> > A[2] = a+b
>> >
>> > i would still love to have
>> >
>> >
>> > aux = (exp(a+b)+exp(b^2))
>> > A[0] = a*aux
>> > A[1] = b*aux
>> > A[2] = a+b
>> >
>>
>> you could use cse() (common subexpression elimination) for this, e.g.:
>>
>> In [1]: from sympy import *
>>
>> In [2]: var('a,b')
>> Out[2]: (a, b)
>>
>> In [3]: aux = exp(a + b) + exp(b**2)
>>
>> In [4]: Matrix([a*aux, b*aux, a + b])
>> Out[4]:
>> Matrix([
>> [a*(exp(b**2) + exp(a + b))],
>> [b*(exp(b**2) + exp(a + b))],
>> [ a + b]])
>>
>> In [5]: replacements, (M,) = cse(_)
>>
>> In [6]: M
>> Out[6]:
>> Matrix([
>> [a*x1],
>> [b*x1],
>> [ x0]])
>>
>> In [7]: replacements
>> Out[7]: [(x0, a + b), (x1, exp(b**2) + exp(x0))]
>>
>> In [8]: M.subs(list(reversed(replacements)))
>> Out[8]:
>> Matrix([
>> [a*(exp(b**2) + exp(a + b))],
>> [b*(exp(b**2) + exp(a + b))],
>> [ a + b]])
>>
>> However, this may not be exactly what you want, because it eliminates
>> `a + b` as well.
>>
>> Mateusz
>>
>> >
>> >
>> > thanks in advance for any suggestion.
>> >
>> > cheers
>> > Riccardo
>> >
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