numVars = ( n + 1 ) * 3 > for i in range ( numVars ) : > eqnList [ i ] = eqnList [ i ] . expand () > print ( eqnList [ i ] ) > > # the linear equation system is expressed in matrix form as: eqnCoeffs > * eqnVars = eqnConstants > > eqnCoeffs = zeros ( numVars, numVars ) > eqnVars = Matrix ( numVars, 1, [ Ep ( i ) for i in fromto ( 0, n ) ] + > [ De ( i ) for i in fromto ( 0, n ) ] + [ La ( i ) for i in fromto ( > 0, n ) ] )
ok, that's being explicit > > eqnConstants = zeros ( numVars, 1 ) > > for myrow in range ( numVars ) : > eqn = eqnList [ myrow ] > for mycol in range ( numVars ) : > found = eqn . coeff ( eqnVars [ mycol ] ) > eqnCoeffs [ myrow, mycol ] = found > eqn -= found * eqnVars [ mycol ] > eqnConstants [ myrow ] = - eqn > > That's the verbose way to say eqnCoeffs = Matrix(eqn).jacobian(Matrix(eqnVars)) and eqnConstants = eqnCoeffs - Matrix(eqn). Here's a toy system: >>> eqns = (2*x+y+3,x-y-2) >>> E = Matrix(eqns) >>> X = Matrix([x, y]) >>> A = E.jacobian(X) >>> B = A*X-E >>> A [2, 1] [1, -1] >>> X [x] [y] >>> B [-3] [ 2] > print ( eqnCoeffs ) > print ( eqnVars . T ) > print ( eqnConstants . T ) > > I hope this approach is more straightforward even though it is not as > power-user as yours i.e. using as_independent(*X) etc :-) > > It's fine...hopefully the above, however, looks a little more text-bookish and easier on the fingers. > > -- You received this message because you are subscribed to the Google Groups "sympy" group. To view this discussion on the web visit https://groups.google.com/d/msg/sympy/-/dgB0cj253QUJ. To post to this group, send email to sympy@googlegroups.com. To unsubscribe from this group, send email to sympy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
