There is also a Matrix.jacobian method that might interest you. On Thu, Aug 23, 2012 at 2:34 PM, Matthew Rocklin <[email protected]> wrote:
> J = Matrix(6,6,lambda i,j:0) # This is for the Jacobian > > vars = [x11,x12,x13,x22,x23,x33] >> > > I'm not totally clear what you're doing here > > >> >> In [38]: for i in [0,1,2]: >> ....: for j in [0,1,2]: >> ....: k = 3*i+j >> ....: for l in [0,1,2,3,4,5]: >> ....: J[k,l] = diff(Y[i,j], vars[l]) >> ....: >> >> --------------------------------------------------------------------------- >> IndexError Traceback (most recent call >> last) >> <ipython-input-38-46a5a73fca3d> in <module>() >> 3 k = 3*i+j >> 4 for l in [0,1,2,3,4,5]: >> ----> 5 J[k,l] = diff(Y[i,j], vars[l]) >> > In this line Y is of shape 3, 3 but i goes from 0 to 5. So you're going > outside the bounds of the Matrix. You're asking for the 3rd, 4th, and 5th, > row of a 3x3 matrix. > > > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
