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PEP 572: Python assignment expressions has been accepted

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317 points est | 1 comments | 03 Jul 18 08:30 UTC | HN request time: 0.346s | source

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jkabrg ◴[03 Jul 18 14:06 UTC] No.17450249[source]▶

>>17448439 (OP) #

It would be nice if in Python you could define new operators instead of overloading existing ones. It would make matrix multiplication look nicer.

I'm thinking it could look like this:

  import numpy as np
  
  def M1 %*% M2 with same precedence as *:
    return M1.matmul(M2)

  foo_matrix = np.matrix([[1,1],[1,1]])
  bar_matrix = np.matrix([[2,2],[2,2]])
  print(foo_matrix %*% bar_matrix)

Also, it would be nice to have a pipe operator `%>%` such that

  foo %>% f()

is equivalent to

  f(foo)

The alternative is to make f a method of foo, and then you can write

  foo.f()

But what happens if I don't want f to be a method? I just want the style of writing the f after the foo, but I don't want the baggage of OOP. Is that anti-Pythonic?

replies(6): >>17450465 #>>17450468 #>>17450486 #>>17451059 #>>17451069 #>>17452494 #

1. joshuamorton ◴[03 Jul 18 18:17 UTC] No.17452494[source]▶

>Also, it would be nice to have a pipe operator `%>%` such that

You can absolutely define a pipe operator to apply functions. Something like

    class Pipeable(object):
        def __init__(self, f: Callable):
            self.f = f

        def __or__(self, other: Callable):
            return Pipeable(other(self.f))

Then use it like

    composed = Pipeable(foo) | f

Replace `|` with `>>` if you feel that conveys what you want better. Libraries already exist which provide this interface.

>Is that anti-Pythonic?

In a sense, yes. Operators and methods are defined in terms of objects, not in terms of abstract types.