Comment #5 on issue 1966 by [email protected]: Add evaluate option to sympify
http://code.google.com/p/sympy/issues/detail?id=1966
Here is a little routine that now works pretty much as expected with a
sample session.
def lg():
"""Prompt for an equation and then prompt for 'change patterns' which
use a variable in an expression to indicate what should happen to the
equation, e.g. i + 1 would mean to add 1 to both sides of the equation.
A function can be used as `expand(i)` (or simply `expand`) to expand
both
sides of the equation.
"""
from sympy.parsing.sympy_parser import standard_transformations,\
implicit_multiplication_application, parse_expr
transformations = (standard_transformations +
(implicit_multiplication_application,))
s = raw_input("equation: ")
eq = Eq(*[parse_expr(a, transformations=transformations, evaluate=False)
for a in s.split('=')])
while 1:
print eq
if not isinstance(eq, Equality):
break
s = raw_input('> ')
if s == '.':
break
try:
e = parse_expr(s, transformations=transformations,
evaluate=False)
if isinstance(e, (FunctionClass, Lambda, type(lambda:1))):
eq = eq.func(*[e(a) for a in eq.args])
else:
i = e.free_symbols - eq.free_symbols
if len(i) != 1:
raise ValueError
i = i.pop()
f = lambda side: e.subs(i, side)
eq = eq.func(*[f(a) for a in eq.args])
except:
print('not understood: did you use a variable to represent each
side?')
lg()
equation: 2(x+1)=4(1-2)+x
2*(x + 1) == x + 4*(-2 + 1)
i/2
x + 1 == x/2 + 2*(-2 + 1)
i-1
x == x/2 + 2*(-2 + 1) - 1
expand
x == x/2 - 3
expand(i)
x == x/2 - 3
i-x/2
x/2 == -3
i*2
x == -6
.
lg()
equation: x=2(1+2)
x == 2*(1 + 2)
i/2
x/2 == 1 + 2
i-3
x/2 - 3 == 0
i+3
x/2 == 3
i*2
x == 6
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