numba的协程

本文介绍了numba的协程的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在研究需要快速协程的东西，我相信numba可以加快我的代码的速度。

I'm working on something that requires fast coroutines and I believe numba could speed up my code.

这是一个愚蠢的例子：一个将输入平方的函数，

Here's a silly example: a function that squares its input, and adds to it the number of times its been called.

def make_square_plus_count(): i = 0 def square_plus_count(x): nonlocal i i += 1 return x**2 + i return square_plus_count

您甚至不能 nopython = False 这样做，大概是由于非本地关键字。

You can't even nopython=False JIT this, presumably due to the nonlocal keyword.

但是如果您使用课程，则不需要非本地相反：

But you don't need nonlocal if you use a class instead:

def make_square_plus_count(): @numba.jitclass({'i': numba.uint64}) class State: def __init__(self): self.i = 0 state = State() @numba.jit() def square_plus_count(x): state.i += 1 return x**2 + state.i return square_plus_count

这至少有效，但是如果您会执行 nopython = True 。

This at least works, but it breaks if you do nopython=True.

是否存在可以使用编译的解决方案nopython = True ？

推荐答案

如果您要使用状态类，则可以还使用方法而不是闭包（应该是非python编译的）：

If you're going to use a state-class anyway you could also use methods instead of a closure (should be no-python compiled):

import numba @numba.jitclass({'i': numba.uint64}) class State(object): def __init__(self): self.i = 0 def square_plus_count(self, x): self.i += 1 return x**2 + self.i square_with_call_count = State().square_plus_count # using the method print([square_with_call_count(i) for i in range(10)]) # [1, 3, 7, 13, 21, 31, 43, 57, 73, 91]

但是时间显示，这实际上比纯python闭包实现要慢tion。我希望只要您不使用 nonlocal numpy-arrays或在您的方法（或闭包）中对数组进行操作，效率就会降低！

However timings show that this is actually slower than a pure python closure implementation. I expect that as long as you don't use nonlocal numpy-arrays or do operations on arrays in your method (or closure) this will be less efficient!