Hi,
On 12 November 2011 17:25, krastanov.ste...@gmail.com <
krastanov.ste...@gmail.com> wrote:
> Can the default printer be changed so lambdarepr(SomeClass(args)) and
> lambdarepr(SomeClass(args).func) give always the same thing.
>
> lambdarepr is using LambdaPrinter. What should be done there (or to a
> subclass of LambdaPrinter) to ensure the behavior that I desire?
So, the difference comes from StrPrinter.emptyPrinter and
FunctionClass.__repr__. You can override _print_FunctionClass or
emptyPrinter in LambdaReprPrinter to fix the problem.
>
>
> On 13 November 2011 02:21, Mateusz Paprocki <matt...@gmail.com> wrote:
>
>> Hi,
>>
>> On 12 November 2011 17:16, krastanov.ste...@gmail.com <
>> krastanov.ste...@gmail.com> wrote:
>>
>>> Then I suppose I can just call a function that returns a string. Here is
>>> the problem I have:
>>>
>>> In [42]: lambdarepr(Integral(x, (x,0,1)).func)
>>> Out[42]: <class 'sympy.integrals.integrals.Integral'>
>>>
>>> In [43]: lambdarepr(Integral(x, (x,0,1)))
>>> Out[43]: Integral(x, (x, 0, 1))
>>>
>>> Why is out[42] not the same style as out[43]. This causes problems with
>>> lambdify, so I would be very happy if somebody explains this to me. With
>>> 'sin' it gives the same style for both inputs:
>>>
>>> In [46]: lambdarepr(sin(x).func)
>>> Out[46]: sin
>>>
>>> In [47]: lambdarepr(sin(x))
>>> Out[47]: sin(x)
>>>
>>
>> Because sin is a Function and Function has a custom printer, which is not
>> true for Integral, which is just an ordinary type (default printer).
>>
>>
>>>
>>> Is this difference between function and integral expected?
>>>
>>>
>>>
>>> On 13 November 2011 02:09, Mateusz Paprocki <matt...@gmail.com> wrote:
>>>
>>>> Hi,
>>>>
>>>> On 12 November 2011 17:00, krastanov.ste...@gmail.com <
>>>> krastanov.ste...@gmail.com> wrote:
>>>>
>>>>> Ok, but why are those printed differently (ipython --profile=sympy):
>>>>>
>>>>> In [34]: sin(x).func
>>>>> Out[34]: sympy.functions.elementary.trigonometric.sin
>>>>>
>>>>> In [35]: str(sin(x).func)
>>>>> Out[35]: sin
>>>>>
>>>>> In [39]: Integral(x, (x,0,1)).func
>>>>> Out[39]: sympy.integrals.integrals.Integral
>>>>>
>>>>> In [40]: str(Integral(x, (x,0,1)).func)
>>>>> Out[40]: <class 'sympy.integrals.integrals.Integral'>
>>>>>
>>>>> What should I do to have them print in the same manner?
>>>>
>>>>
>>>> This seems to be a problem with IPython's printing hooks. In CPython I
>>>> get:
>>>>
>>>> >>> from sympy import *
>>>> >>> init_printing()
>>>> >>> var('x')
>>>> x
>>>> >>> sin(x).func
>>>> sin
>>>> >>> str(sin(x).func)
>>>> sin
>>>> >>> Integral(x, (x, 0, 1)).func
>>>> <class 'sympy.integrals.integrals.Integral'>
>>>> >>> str(Integral(x, (x, 0, 1)).func)
>>>> <class 'sympy.integrals.integrals.Integral'>
>>>>
>>>> There reason for different output in IPython is that IPython not always
>>>> uses SymPy's pretty printer, but sometimes uses its own printing hooks.
>>>> It's better visible in the notebook where latex output is used (e.g. try to
>>>> print data structures).
>>>>
>>>>
>>>>>
>>>>>
>>>>> On 13 November 2011 00:42, Aaron Meurer <asmeu...@gmail.com> wrote:
>>>>>
>>>>>> Hi.
>>>>>>
>>>>>> As far as I know, we don't have a function that does exactly that,
>>>>>> though I could be wrong. It would be nice to have one, though.
>>>>>>
>>>>>> On Sat, Nov 12, 2011 at 11:01 AM, Alexey U. Gudchenko <
>>>>>> pr...@goodok.ru> wrote:
>>>>>> > 12.11.2011 21:42, krastanov.ste...@gmail.com пишет:
>>>>>> >> This:
>>>>>> >>
>>>>>> >> import ast
>>>>>> >> ast.parse(repr(expression))
>>>>>>
>>>>>> If you want a repr() representation, you should instead use srepr().
>>>>>> (repr() is the same as str()).
>>>>>>
>>>>>> >>
>>>>>> >> will do the trick if repr is well coded.
>>>>>>
>>>>>> str() is coded so that it returns the same thing back from sympify(),
>>>>>> but it may not give the same thing directly, because you can have
>>>>>> int/int in an expression. srepr() should always give the same thing
>>>>>> back.
>>>>>>
>>>>>> >>
>>>>>> >> How much faith should I put in the repr strings in sympy? Or there
>>>>>> is
>>>>>> >> another way?
>>>>>> >>
>>>>>> >> On 12 November 2011 18:20, krastanov.ste...@gmail.com <
>>>>>> >> krastanov.ste...@gmail.com> wrote:
>>>>>> >>
>>>>>> >>> Is there any way to get the expression tree from an expression
>>>>>> (either
>>>>>> >>> using the python abstract syntax tree module or just some tuples):
>>>>>> >>>
>>>>>> >>> for example
>>>>>> >>>
>>>>>> >>> get_tree( x+y*sin(z) ) would return
>>>>>> >>>
>>>>>> >>> (Add, x, (Mul, y, (Sin z)))
>>>>>> >>>
>>>>>> >>> or
>>>>>> >>>
>>>>>> >>> (BinOp, Add, ((Symbol, x), (BinOp, Mul, (blah blah blah))))
>>>>>> >>>
>>>>>> >>
>>>>>> >
>>>>>> > I know only how to obtain the childes:
>>>>>> >
>>>>>> >>>> e = x+y*sin(z) + z
>>>>>> >>>> e.args
>>>>>> > (y*sin(z), z, x)
>>>>>> >
>>>>>> >>>> e.args[0]
>>>>>> >>>> y*sin(z)
>>>>>> >
>>>>>> >>>> e.args[0].args
>>>>>> > (y, sin(z))
>>>>>> >
>>>>>> >
>>>>>> >
>>>>>> > And test the classes:
>>>>>> >
>>>>>> >>>> e.is_Add
>>>>>> > True
>>>>>> >
>>>>>>
>>>>>> You can get the class name by using .func:
>>>>>>
>>>>>> In [25]: e = x + y
>>>>>>
>>>>>> In [26]: e.func
>>>>>> Out[26]: sympy.core.add.Add
>>>>>>
>>>>>> In [27]: e.func(*e.args)
>>>>>> Out[27]: x + y
>>>>>>
>>>>>> The invariant in [27] should always hold (except for possibly some
>>>>>> differences in assumptions).
>>>>>>
>>>>>> Aaron Meurer
>>>>>>
>>>>>> >
>>>>>> >
>>>>>> >
>>>>>> > In other words, the somewhat tree of the expressions exists.
>>>>>> >
>>>>>> > How to represent expression-tree in other formats (strings or
>>>>>> > structures), I do not know.
>>>>>> >
>>>>>> > Regards.
>>>>>> >
>>>>>> > --
>>>>>> > Alexey U.
>>>>>> >
>>>>>> > --
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>>>>>> >
>>>>>> >
>>>>>>
>>>>>> --
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>>>>>>
>>>>>>
>>>>> --
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>>>>>
>>>>
>>>> Mateusz
>>>>
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>>>>
>>>
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>>
>> Mateusz
>>
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>
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