Targeting specific subexpression for transformations

[Francesco Bonazzi]

Consider this use case

In [97]: expr = 1/(1-x) + 1/(1+x)

In [98]: e2 = Integral(expr, x)

In [99]: e2
Out[99]:
⌠                    
⎮ ⎛  1       1   ⎞  
⎮ ⎜───── + ──────⎟ dx
⎮ ⎝x + 1   -x + 1⎠  
⌡                    

Suppose now I want to act on the expression inside the integral by applying together and expand on it, is there a simple way to do so?

In [102]: expr.together().expand()
Out[102]:
   2    
────────
   2    
- x  + 1

More accurately, is there an easy way to select a subexpression, apply some transformations only on that subexpression, and returning the entire expression with the applied transformations?

In this case one could extract the integral argument by e2.args[0], and then rebuild e2.func(new_arg_0, e2.args[1:]), but imagine if the tree expression is much more complicated and it is hard/uncomfortable to select the subexpression by accessing the args, what can one do?

[Mateusz Paprocki]

Hi,

You could use epath(), e.g.:

In [1]: expr = 1/(1-x) + 1/(1+x)

In [2]: e2 = Integral(expr, x)

In [3]: epath("/[0]", e2, lambda e: e.together().expand())
Out[3]:
⌠
⎮ 2
⎮ ──────── dx
⎮ 2
⎮ - x + 1
⌡

If you know XPath, then this approach should be familiar. See the
docstring for details. If unsure what expressions will be selected,
then skip the lambda part and epath() will return matching
expressions.

Mateusz

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[Francesco Bonazzi]

Interesting, I didn't know of epath, it looks like it supports type-matching, which current wildcards do not support.

[Chris Smith]

see also the docstring of replace for instructions on how it might be used

>>> expr=1/(1-x)+1/(1+x)

>>> i=Integral(expr,x)

>>> i.replace(lambda arg: arg.is_Add, lambda arg: arg.together().expand())

Integral(2/(-x**2 + 1), x)