[Python-ideas] Function composition (was no subject)

[Stephen J. Turnbull]

Andrew Barnert writes:
 > On May 8, 2015, at 19:58, Stephen J. Turnbull <stephen at xemacs.org> wrote:
 > > 
 > > Koos Zevenhoven writes:
 > > 
 > >> As a random example, (root @ mean @ square)(x) would produce the right 
 > >> order for rms when using [2].
 > > 
 > > Hardly interesting. :-)  The result is an exception, as root and square
 > > are conceptually scalar-to-scalar, while mean is sequence-to-scalar.
 > 
 > Unless you're using an elementwise square and an array-to-scalar
 > mean, like the ones in NumPy,

Erm, why would square be elementwise and root not?  I would suppose
that everything is element-wise in Numpy (not a user yet).

 > in which case it works perfectly well...

But that's an aspect of my point (evidently, obscure).  Conceptually,
as taught in junior high school or so, root and square are scalar-to-
scalar.  If you are working in a context such as Numpy where it makes
sense to assume they are element-wise and thus composable, the context
should provide the compose operator(s).  Without that context, Koos's
example looks like a TypeError.

 > But Koos's example, even if it was possibly inadvertent, shows that
 > I may be wrong about that. Maybe compose together with element-wise
 > operators actually _is_ sufficient for something beyond toy
 > examples.

Of course it is!<wink />  I didn't really think there was any doubt
about that.  I thought the question was whether there's enough
commonality among such examples to come up with a Pythonic generic
definition of compose, or perhaps a sufficiently compelling example to
enshrine its definition as the "usual" interpretation in Python (and
let other interpretations overload some operator to get that effect in
their contexts).

 > Of course the fact that we have two groups of people each arguing
 > that obviously the only possible reading of @ is compose/rcompose
 > respectively points out a whole other problem with the idea.

I prefer fgh = f(g(h(-))), but I hardly think it's obvious.  Unless
you're *not* Dutch.  (If it were obvious to a Dutchman, we'd have it
already. <wink />)