Interesting, I didn't know of epath, it looks like it supports
type-matching, which current wildcards do not support.
On Friday, October 24, 2014 4:50:16 PM UTC+2, Mateusz Paprocki wrote:
>
> Hi,
>
> On 24 October 2014 15:16, Francesco Bonazzi <franz....@gmail.com
> <javascript:>> wrote:
> > 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?
>
> 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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