numba的协程 [英] Coroutines in numba
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!
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