Replacements on certain locations of expression tree
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Francesco Bonazzi
Sep 5, 2015, 2:35:35 PM9/5/15
to sympy
Hi,
in Mathematica one can access the expression tree by the [[ ]] operator:
Now one could change the x for a q:
This works by replacing a part in the expression tree (position [[1, 1]]), there is no pattern matching.
I was wondering, is there something like that in SymPy?
Maybe, instead of accessing e.args[0].args[0] (Mathematica's arrays are 1-starting), have some
e.<some method>(0, 0)
and
new_e = e.<some other method>(<new value>, 0, 0)
Is there some easy way to avoid writing a traversal function?
in Mathematica one can access the expression tree by the [[ ]] operator:
In[1]:= e = x*y + z
Out[1]= x y + z
In[2]:= e[[1]]
Out[2]= x y
In[3]:= e[[1, 1]]
Out[3]= x
Now one could change the x for a q:
In[4]:= e[[1, 1]] = q
Out[4]= q
In[5]:= e
Out[5]= q y + z
This works by replacing a part in the expression tree (position [[1, 1]]), there is no pattern matching.
I was wondering, is there something like that in SymPy?
Maybe, instead of accessing e.args[0].args[0] (Mathematica's arrays are 1-starting), have some
e.<some method>(0, 0)
and
new_e = e.<some other method>(<new value>, 0, 0)
Is there some easy way to avoid writing a traversal function?
Aaron Meurer
Sep 5, 2015, 3:00:47 PM9/5/15
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expression from the bottom up with the remaining args (remember that
expr == expr.func(*expr.args)).
It would probably be useful to have some utilities functions for
dealing with tuples like (1, 0) meaning expr.args[1].args[0].
Aaron Meurer
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Francesco Bonazzi
Sep 10, 2015, 7:44:10 AM9/10/15
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On Saturday, 5 September 2015 21:00:47 UTC+2, Aaron Meurer wrote:
I don't think you need traversal. You just need to rebuild the
expression from the bottom up with the remaining args (remember that
expr == expr.func(*expr.args)).
That still requires unpacking the expression tree.
An API like expr.args[2] = new_third_arg_value would violate the object's immutability.
Mathematica's expression trees are not immutable. It has some kind of mechanism by which any modification gets notified to the reference counter, which updates all usage instances.
It would probably be useful to have some utilities functions for
dealing with tuples like (1, 0) meaning expr.args[1].args[0].
Maybe expr.args[1, 2] ?
Some extensions like those used by Mathematica?
Cfr: https://reference.wolfram.com/language/tutorial/PartsOfExpressions.html
That is:
- Range: expr.args[0, :3] ==> picks 1st arg, then returns its first 3 subargs.
- Groups: expr.args[(1, 3)] ==> picks 2nd and 4th args.
What about if I add such options on this line:
Is it reasonable?
Francesco Bonazzi
Sep 10, 2015, 7:47:57 AM9/10/15
to sympy
What about if I add such options on this line:
Is it reasonable?
I mean, making expr.args return a proxy of expr._args with such indexing options. Or better add a new method?
Aaron Meurer
Sep 10, 2015, 4:35:04 PM9/10/15
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implications of making args not a tuple.
What you want to do should definitely be possible without modifying the core.
Aaron Meurer
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Francesco Bonazzi
Sep 25, 2015, 1:12:37 PM9/25/15
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A possible way is this:
https://gist.github.com/Upabjojr/3718fd700f123fa899ee
Could this be useful if inserted to the API of Basic?
https://gist.github.com/Upabjojr/3718fd700f123fa899ee
Could this be useful if inserted to the API of Basic?
Aaron Meurer
Sep 26, 2015, 6:34:20 PM9/26/15
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What is PartReplace supposed to do? For me it just returns the replacement.
Aaron Meurer
> https://groups.google.com/d/msgid/sympy/21e5ff28-42f6-4e28-b568-fce93be51c02%40googlegroups.com.
Aaron Meurer
Francesco Bonazzi
Sep 27, 2015, 5:39:46 PM9/27/15
to sympy
On Sunday, 27 September 2015 00:34:20 UTC+2, Aaron Meurer wrote:
What is PartReplace supposed to do? For me it just returns the replacement.
I updated the gist to make it act as a replacement:
In [14]: PartReplace(x*y+z, w, 1,1)
Out[14]: w⋅x + z
This should basically act to replace element at position .args[1].args[1] with w. Of couse I avoided using the syntax expr.args[1].args[1] = w to comply with SymPy's immutability of objects.
In [15]: (x*y+z).args[1].args[1]
Out[15]: y
This gets element at (1,1), or the equivalent form in my gist:
In [18]: Part((x*y+z), 1, 1)
Out[18]: y
I think that Basic could be equipped with these two methods.
Francesco Bonazzi
Sep 27, 2015, 5:45:14 PM9/27/15
to sympy
It may also support slices, as in:
Which would be roughly something equivalent to the intended usage of this wrong python expression: expr.args[:].args[1] = o
In [20]: Part(x*y+w*z, slice(None, None, None), 1)
Out[20]: y + z
In [21]: PartReplace(x*y + w*z, o, slice(None, None, None), 1)
Out[21]: o⋅w + o⋅x
Which would be roughly something equivalent to the intended usage of this wrong python expression: expr.args[:].args[1] = o
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