Comment by librasteve

Comment by librasteve 5 days ago

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Very nice post - it's good to see benchmarks done for humans.

For fun, I tried this in Raku:

  (0, 1, *+* ... *)[40]    #0.10s user 0.03s system 63% cpu 0.214 total
lol

Seriously, Python is doing great stuff to squeeze out performance from a scripting language. Realistically, Raku has fewer native libraries (although there is Inline::Python) and the compiler still has a lot of work to get the same degree of optimisation (although one day it could compare).

EDIT: for those who have commented, yes you are correct … this is a “cheat” and does not seek to state that Raku is faster than Python - as I said Raku still has a lot of work to do to catch up.

lucb1e 5 days ago

I take it this is supposed to be the equivalent of fib(40), which ran on the author's system in Pyπ in 6.59 seconds and apparently on yours, with Raku, in 0.21?

Do you have the same hardware as the author or should one of you run the other's variant to make this directly comparable?

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  • OskarS 5 days ago

    No, this is very much not the same. The Raku version is like writing this in Python:

        def fibonacci():
        a, b = 0, 1

        while True:
            yield a
            a, b = b, a+b
    And taking the 40th element. It's not comparable at all to the benchmark, that's deliberately an extremely slow method of calculating fibonacci numbers for the purpose of the benchmark. For this version, it's so fast that the time is dominated by the time needed to start up and tear down the interpreter.
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zahlman 5 days ago

Well, sure; you're using dynamic programming, while the stress test Python Fibonacci code is deliberately using recursion without memoization — it makes function calls proportionate to the number computed. Most of the time you're seeing in the Raku code is the interpreter startup. Python doesn't have syntax strongly oriented towards that sort of trick (it's not as strong of a second-best APL as it is a second-best Lisp or Haskell), but:

  $ python -m timeit "x = (1, 0); [x[0] for _ in range(40) if (x := (x[0] + x[1], x[0]))][-1]"
  50000 loops, best of 5: 4 usec per loop
(Or a "lazy iterator" approach:)

  $ python -m timeit --setup 'from itertools import islice, count' 'x = (1, 0); next(islice((x[0] for _ in count() if (x := (x[0] + x[1], x[0]))), 40, None))'
  50000 loops, best of 5: 5.26 usec per loop
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