On Sep 19, 10:20 pm, "Aaron S. Meurer" <asmeu...@gmail.com> wrote: > So given Chris's tip in issue 2058 (which flew over my head the first time > for some reason), I think I might have this figured out, except for one > thing. Given M(i, j, n), how to I get the third argument, n? I tried M(i, > j, n)[2] and M(i, j, n).args, but neither works. >
>>> m = M(i,j,n) >>> m.args (M, i, j, n) >>> m.args[-1] n -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to sy...@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.
