使用函数对象作为 numba njit 函数的参数

Question

我想做一个通用的函数，它接受一个函数对象作为参数。

最简单的情况之一：

import numpy as np
import numba as nb
@nb.njit()
def test(a, f=np.median):
    return f(a)

test(np.arange(10), np.mean)

给出错误，尽管 test(np.arange(10)) 按预期工作。

错误：

TypingError: Failed in nopython mode pipeline (step: nopython frontend)
non-precise type pyobject
[1] During: typing of argument at <ipython-input-54-52cead0f097d> (5)

File "<ipython-input-54-52cead0f097d>", line 5:
def test(a, f=np.median):
    return f(a)
    ^

This error may have been caused by the following argument(s):
- argument 1: cannot determine Numba type of <class 'function'>

This is not usually a problem with Numba itself but instead often caused by
the use of unsupported features or an issue in resolving types.

这是不允许的还是我遗漏了什么？

Answer 1

使用函数作为参数对 numba 来说很棘手，而且非常昂贵。 Frequently Asked Questions: "1.18.1.1. Can I pass a function as an argument to a jitted function?":

中提到了这一点

1.18.1.1. Can I pass a function as an argument to a jitted function?

As of Numba 0.39, you can, so long as the function argument has also been JIT-compiled:
@jit(nopython=True)
def f(g, x):
    return g(x) + g(-x)
result = f(jitted_g_function, 1)
However, dispatching with arguments that are functions has extra overhead. If this matters for your application, you can also use a factory function to capture the function argument in a closure:
def make_f(g):
    # Note: a new f() is created each time make_f() is called!
    @jit(nopython=True)
    def f(x):
        return g(x) + g(-x)
    return f
f = make_f(jitted_g_function)
result = f(1)
Improving the dispatch performance of functions in Numba is an ongoing task.

这意味着您可以选择使用函数工厂：

import numpy as np
import numba as nb

def test(a, func=np.median):
    @nb.njit
    def _test(a):
        return func(a)
    return _test(a)

>>> test(np.arange(10))
4.5
>>> test(np.arange(10), np.min)
0
>>> test(np.arange(10), np.mean)
4.5

或者在将函数参数作为参数传递之前将函数参数包装为 jitted-function:

import numpy as np
import numba as nb

@nb.njit()
def test(a, f=np.median):
    return f(a)

@nb.njit
def wrapped_mean(a):
    return np.mean(a)

@nb.njit
def wrapped_median(a):
    return np.median(a)

>>> test(np.arange(10))
4.5
>>> test(np.arange(10), wrapped_mean)
4.5
>>> test(np.arange(10), wrapped_median)
4.5

这两个选项都有很多样板文件，并不像人们希望的那样直截了当。

函数工厂方法也会重复创建和编译函数，因此如果您经常使用与参数相同的函数调用它，您可以使用字典来存储已知的编译函数：

import numpy as np
import numba as nb

_precompiled_funcs = {}

def test(a, func=np.median):
    if func not in _precompiled_funcs:
        @nb.njit
        def _test(arr):
            return func(arr)
        result = _test(a)
        _precompiled_funcs[func] = _test
        return result
    return _precompiled_funcs[func](a)

另一种方法（使用 wrapped 和 jitted 函数）也有一些开销，但是只要您传入的数组具有大量元素（>1000），它就不会真正引人注意。

如果您展示的函数确实是您想要使用的函数，我根本不会在其上使用 numba。使用 Python + NumPy 这样简单的任务不会锻炼 numba 的强度（索引和迭代数组或大量数字运算）应该更快（或同样快）并且更容易调试和理解：

import numba as nb

def test(a, f=np.median):
    return f(a)

使用函数对象作为 numba njit 函数的参数

Using a function object as an argument for numba njit function

python

numba

1.18.1.1. Can I pass a function as an argument to a jitted function?