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?
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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