abstract syntax tree

[krastano...@gmail.com]

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))))

[krastano...@gmail.com]

This:

import ast
ast.parse(repr(expression))

 will do the trick if repr is well coded.

How much faith should I put in the repr strings in sympy? Or there is another way?

[Alexey U. Gudchenko]

12.11.2011 21:42, krastano...@gmail.com пишет:

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

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.

[Aaron Meurer]

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, krastano...@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, krastano...@gmail.com <
>> krastano...@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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[krastano...@gmail.com]

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?

[Mateusz Paprocki]

Hi,

On 12 November 2011 17:00, krastano...@gmail.com <krastano...@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).

Mateusz

[krastano...@gmail.com]

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)

Is this difference between function and integral expected?

[Mateusz Paprocki]

Hi,

On 12 November 2011 17:16, krastano...@gmail.com <krastano...@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).

Mateusz

[krastano...@gmail.com]

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?

[Mateusz Paprocki]

Hi,

On 12 November 2011 17:25, krastano...@gmail.com <krastano...@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.

[krastano...@gmail.com]

Thank you for the information.