Because I want to build the name of the expressions to check by
concatenating other strings, and then evaluate.
All this within loops, etc.
On Monday, January 9, 2023 at 1:42:26 PM UTC-3 gu...@uwosh.edu wrote:
> I am not clear why you are working with strings and not just the
> expressions? I think the following may be what you want:
> ```
> >>> import sympy as sp
> >>> a,b = sp.symbols('a b', positive = True)
> >>> expr = 1.01*a**1.01*b**0.99
> >>> expr
> 1.01*a**1.01*b**0.99
> >>> expr1 = 2*expr
> >>> expr1
> 2.02*a**1.01*b**0.99
> >>> expr1/expr
> 2.00000000000000
> ```
> Your "verify ratio function" would then just take the two expressions
> directly. If you are trying to do something where you need to have
> intermediate strings, we will need more explicit details to provide some
> direction.
>
> On Monday, January 9, 2023 at 10:00:04 AM UTC-6 Santiago S wrote:
>
>> I have the following code
>>
>> import sympy as sp
>> a, b = sp.symbols('a,b', real=True, positive=True)
>> expr2 = 1.01 * a**1.01 * b**0.99
>> print(type(expr2), '->', expr2)
>>
>>
>> Now I want a function that takes the string `'expr2'` and returns the
>> expression `1.01 * a**1.01 * b**0.99`.
>> The ultimate objective is to put together the strings for two different
>> expressions `'expr2'` and `'expr3'`, which should presumably give the same
>> result, and verify their ratio, as in
>>
>> def verify_ratio(vstr1, vstr2):
>> """Compare the result of two different computations of the same
>> quantity"""
>> ratio = sp.N(sp.parsing.sympy_parser.parse_expr(vstr1)) /
>> sp.parsing.sympy_parser.parse_expr(vstr2)
>> print(vstr1 + ' / ' + vstr2, '=', sp.N(ratio))
>> return
>>
>> which does not work, as per what I tried:
>>
>> expr2 = 1.01 * a**1.01 * b**0.99
>> print(type(expr2), '->', expr2)
>>
>> expr2b = sp.parsing.sympy_parser.parse_expr('expr2')
>> print(type(expr2b), '->', expr2b)
>>
>> expr2c = sp.N(sp.parsing.sympy_parser.parse_expr('expr2'))
>> print(type(expr2c), '->', expr2c)
>> #print(sp.N(sp.parsing.sympy_parser.parse_expr('expr2')))
>>
>> expr2d = sp.sympify('expr2')
>> print(type(expr2d), '->', expr2d)
>>
>> with output
>>
>> <class 'sympy.core.mul.Mul'> -> 1.01*a**1.01*b**0.99
>> <class 'sympy.core.symbol.Symbol'> -> expr2
>> <class 'sympy.core.symbol.Symbol'> -> expr2
>> <class 'sympy.core.symbol.Symbol'> -> expr2
>>
>> None of my attempts achieved the objective.
>> Questions or links which did not help (at least for me):
>>
>> 1.
>> https://stackoverflow.com/questions/33606667/from-string-to-sympy-expression
>> 2.
>> https://docs.sympy.org/latest/tutorials/intro-tutorial/basic_operations.html
>> 3. https://docs.sympy.org/latest/modules/parsing.html
>> 4.
>> https://docs.sympy.org/latest/modules/core.html#sympy.core.sympify.sympify
>> 5.
>> https://docs.sympy.org/latest/tutorials/intro-tutorial/manipulation.html
>>
>>
>> **Note**:
>> Besides the practical aspects of my objective, I don't know if there is
>> any formal difference between `Symbol` (which is a specific class) and
>> *expression*. From the sources I read (e.g., [this][1]) I did not arrive to
>> a conclusion.
>> This understanding may help in solving the question.
>>
>>
>> [1]:
>> https://docs.sympy.org/latest/tutorials/intro-tutorial/manipulation.html
>>
>
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