I am using SymPy 1.5.1. My package manager shows that as the newest.
On Thursday, April 30, 2020 at 7:43:28 PM UTC+2, Oscar wrote:
>
> What version of sympy are you using? I get an exception from eq3 = eq1 +
> eq2.
>
> We need a new Equation class that works properly with arithmetic
> operations etc. I have made written one but haven't added it to sympy
> yet. I got sidetracked at the time because deeper structural changes
> were needed like ensuring that Relational does not subclass Expr.
>
> Here's my new Equation class which supports arithmetic operations:
>
> from sympy.core.basic import Basic
> from sympy.core.sympify import _sympify
> from sympy.simplify import simplify
> from sympy.core.relational import Equality
>
> class Equation(Basic):
>
> def __new__(cls, lhs, rhs):
> lhs = _sympify(lhs)
> rhs = _sympify(rhs)
> return Basic.__new__(cls, lhs, rhs)
>
> @property
> def lhs(self):
> return self.args[0]
>
> @property
> def rhs(self):
> return self.args[1]
>
> def as_Boolean(self):
> return Equality(self.lhs, self.rhs)
>
> @property
> def reversed(self):
> return Equation(self.rhs, self.lhs)
>
> def applyfunc(self, func):
> return Equation(func(self.lhs), func(self.rhs))
>
> def applylhs(self, func):
> return Equation(func(self.lhs), self.rhs)
>
> def applyrhs(self, func):
> return Equation(self.lhs, func(self.rhs))
>
> @classmethod
> def _binary_op(cls, a, b, opfunc_ab):
> if isinstance(a, Equation) and not isinstance(b, Equation):
> return Equation(opfunc_ab(a.lhs, b), opfunc_ab(a.rhs, b))
> elif isinstance(b, Equation) and not isinstance(a, Equation):
> return Equation(opfunc_ab(a, b.lhs), opfunc_ab(a, b.rhs))
> elif isinstance(a, Equation) and isinstance(b, Equation):
> return Equation(opfunc_ab(a.lhs, b.lhs), opfunc_ab(a.rhs,
> b.rhs))
> else:
> raise TypeError('One of a or b should be an equation')
>
> def __add__(self, other):
> return self._binary_op(self, other, lambda a, b: a + b)
>
> def __radd__(self, other):
> return self._binary_op(other, self, lambda a, b: a + b)
>
> def __mul__(self, other):
> return self._binary_op(self, other, lambda a, b: a * b)
>
> def __rmul__(self, other):
> return self._binary_op(other, self, lambda a, b: a * b)
>
> def __sub__(self, other):
> return self._binary_op(self, other, lambda a, b: a - b)
>
> def __rsub__(self, other):
> return self._binary_op(other, self, lambda a, b: a - b)
>
> def __truediv__(self, other):
> return self._binary_op(self, other, lambda a, b: a / b)
>
> def __rtruediv__(self, other):
> return self._binary_op(other, self, lambda a, b: a / b)
>
> def __pow__(self, other):
> return self._binary_op(self, other, lambda a, b: a ** b)
>
> def __rpow__(self, other):
> return self._binary_op(other, self, lambda a, b: a ** b)
>
> def __str__(self):
> return '%r = %r' % self.args
>
> def expand(self, *args, **kwargs):
> return self.func(*(arg.expand(*args, **kwargs) for arg in
> self.args))
>
> def simplify(self, *args, **kwargs):
> return simplify(self, *args, **kwargs)
>
> def _eval_simplify(self, *args, **kwargs):
> return self.func(*(arg.simplify(*args, **kwargs) for arg in
> self.args))
>
> def factor(self, *args, **kwargs):
> return factor(self, *args, **kwargs)
>
> def evalf(self, *args, **kwargs):
> return self.func(*(arg.evalf(*args, **kwargs) for arg in
> self.args))
>
> def n(self, *args, **kwargs):
> return self.func(*(arg.n(*args, **kwargs) for arg in self.args))
>
> from sympy import symbols, cos, sin
>
> x, y = symbols('x, y')
> eq = Equation(x, y)
> print(eq)
> eq = Equation(1, 2)
> print(eq)
> eq = Equation(1, 2) + 3
> print(eq)
> eq = Equation(1, 2) + Equation(x, y)
> print(eq)
> eq = Equation(1, 2) - 3
> print(eq)
> eq = Equation(1, 2) - Equation(x, y)
> print(eq)
> eq = 1 - Equation(1, 2)
> print(eq)
> eq = 1 + Equation(x, y)
> print(eq)
>
> eq = Equation(cos(x)**2 + sin(x)**2 - 1, 2)
>
>
> Oscar
>
> On Thu, 30 Apr 2020 at 16:18, Thomas Ligon <[email protected]
> <javascript:>> wrote:
> >
> > I have a few small questions where I can solve the issues manually and
> have the feeling that in my attempts to solve them with SymPy I am
> overlooking something. The examples below are trivial and easy to solve
> manually, but the project I am working on involves much more complex
> expressions where it would be great to have a better solution in SymPy.
> > Basically, adding two equations does not produce the desired solution,
> and sums do not expand or simplify as expected.
> > PS. The print('end') statement is just there as a convenient place for a
> breakpoint in my debugger.
> >
> > from sympy import symbols, Eq, expand, simplify, latex, oo, Sum
> > x, y, p, q, a, j, Aj, Bj = symbols('x y p q a j A_j B_j')
> > eq1 = Eq(x, y)
> > print(latex(eq1)) # OK
> > eq2 = Eq(p, q)
> > print(latex(eq2)) # OK
> > eq3 = eq1 + eq2
> > print(latex(eq3)) # not what I want
> > eq4 = Eq(eq1.lhs + eq2.lhs, eq1.rhs + eq2.rhs)
> > print(latex(eq4)) # yes, what I want, but tedious
> > eq5 = a*eq1
> > print(latex(eq5)) # not exactly what I want
> > eq6 = Eq(a*eq1.lhs, a*eq1.rhs)
> > print(latex(eq6)) # yes, what I want, but tedious
> > ex1 = Sum(Aj + Bj, (j, -oo, oo))
> > print(latex(ex1)) # OK
> > ex2 = Sum(Aj - Bj, (j, -oo, oo))
> > print(latex(ex2)) # OK
> > ex3 = ex1 + ex2
> > print(latex(ex3)) # correct, but needs consolidation
> > ex4 = expand(ex3)
> > print(latex(ex4)) # latex looks bad
> > ex5 = simplify(ex3)
> > print(latex(ex5)) # latex completely wrong
> > print('end')
> >
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>
>
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