You can get all the variables in an expression using
expr.free_symbols. I hope that helps.
Aaron Meurer
On Mon, Jun 24, 2013 at 3:37 PM, Shawn Garbett <shawn.garb...@gmail.com> wrote:
> Here's a simpler example:
>
> from numpy import *
> from sympy import *
>
> x = array([(1.1, 2.2), (3.3, 4.4), (5.5, 6.6)], dtype=[('s0', '<f8'), ('s1',
> '<f8')])
>
> eq = 2.0 * Symbol('s0') - Symbol('s1')
>
> 2.0*x['s0']-x['s1'] # Gives correct result at command prompt
>
> eq.evalf(x) # Fails
>
> The constraints are that the symbols in the equation are unknown, and there
> is a random number of them that can be huge, like thousands of variables as
> symbols. What is provided is the time course data x and the equation eq
> which contains matching variables. I.e., The code has no way of knowing
> beforehand what the variables are going to be, nor how many so a predefined
> lambda will not solve the problem.
>
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