Python's equivalent to F#'s yield! (yield bang)?

Question

Let's say I have the following code in F#:

let rec all_numbers_from k =
  yield k
  yield! all_numbers_from (k+1)

This code can be called as all_numbers_from 0, generating all numbers from 0 to infinity. I know Python also has a yield keyword, that seems to behave very much like F#'s yield operator. My question is if there's also some sort of equivalent to F#'s yield! operator?

It might help if you explain what F# `yield!` does. Taking a shot in the dark here: are you looking for [`yield from`](https://stackoverflow.com/questions/9708902/in-practice-what-are-the-main-uses-for-the-new-yield-from-syntax-in-python-3)? — Jonathon Reinhart, Jan 05 '20 at 15:56
@JonathonReinhart: it can be inferred from the above snippet, but I'll explain: `yield` returns a value, `yield! X` is roughly equivalent to doing an iteration over `X`, yielding all its values (lazily, of course). — devoured elysium, Jan 05 '20 at 15:57
@devouredelysium There's [an edit button](https://stackoverflow.com/posts/59601574/edit) you can use to add the explanation to your actual post. — glennsl, Jan 05 '20 at 16:49

score 5 · Answer 1 · answered Jan 05 '20 at 16:01

5

While yield from is the construct you are looking for, defining this function (co)recursively is horribly inefficient in Python. Use iteration

def all_numbers_from(k):
    while True:
        yield k
        k = k + 1

or just use itertools.count(k).

answered Jan 05 '20 at 16:01

chepner

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Jonathon Reinhart · Accepted Answer · 2020-01-05T16:13:45.420

4

The equivalent Python feature would be yield from.

def all_numbers_from(k):
  yield k
  yield from all_numbers_from(k+1)

However this would create a large number of chained generators, and is a poor implementation in Python. See chepner's answer for a better alternative.

edited Jan 05 '20 at 16:13

answered Jan 05 '20 at 16:00

Jonathon Reinhart

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Python's equivalent to F#'s yield! (yield bang)?

2 Answers2