Question

我正在尝试使用numba来加速一个将另一个函数作为参数的函数。最小的例子如下：

import numba as nb

def f(x):
    return x*x

@nb.jit(nopython=True)
def call_func(func,x):
    return func(x)

if __name__ == '__main__':
    print(call_func(f,5))

然而，这并不起作用，因为显然numba不知道如何处理该函数参数。回溯很长：

Traceback (most recent call last):
  File "numba_function.py", line 15, in <module>
    print(call_func(f,5))
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/dispatcher.py", line 330, in _compile_for_args
    raise e
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/dispatcher.py", line 307, in _compile_for_args
    return self.compile(tuple(argtypes))
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/dispatcher.py", line 579, in compile
    cres = self._compiler.compile(args, return_type)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/dispatcher.py", line 80, in compile
    flags=flags, locals=self.locals)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 740, in compile_extra
    return pipeline.compile_extra(func)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 360, in compile_extra
    return self._compile_bytecode()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 699, in _compile_bytecode
    return self._compile_core()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 686, in _compile_core
    res = pm.run(self.status)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 246, in run
    raise patched_exception
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 238, in run
    stage()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 452, in stage_nopython_frontend
    self.locals)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/compiler.py", line 841, in type_inference_stage
    infer.propagate()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/typeinfer.py", line 773, in propagate
    raise errors[0]
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/typeinfer.py", line 129, in propagate
    constraint(typeinfer)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/typeinfer.py", line 380, in __call__
    self.resolve(typeinfer, typevars, fnty)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/numba/typeinfer.py", line 402, in resolve
    raise TypingError(msg, loc=self.loc)
numba.errors.TypingError: Failed at nopython (nopython frontend)
Invalid usage of pyobject with parameters (int64)
No type info available for pyobject as a callable.
File "numba_function.py", line 10
[1] During: resolving callee type: pyobject
[2] During: typing of call at numba_function.py (10)

This error may have been caused by the following argument(s):
- argument 0: cannot determine Numba type of <class 'function'>

有没有办法解决这个问题？

Answer 1

这取决于您传递给func的{{1}}是否可以call_func模式进行编译。

如果它不能以nopython模式编译，那么它是不可能的，因为numba不支持nopython函数内的python调用（这就是它的原因）叫做nopython）。

但是如果它可以在nopython模式下编译，你可以使用闭包：

nopython

这种方法有一些明显的缺点，因为每次调用import numba as nb def f(x): return x*x def call_func(func, x): func = nb.njit(func) # compile func in nopython mode! @nb.njit def inner(x): return func(x) return inner(x) if __name__ == '__main__': print(call_func(f,5))时都需要编译func和inner。这意味着只有通过编译函数的加速比编译成本大，它才有可行。如果多次使用相同的函数调用call_func，则可以减轻这种开销：

call_func

只是一般性说明：我不会创建带有函数参数的numba函数。如果你不能对函数进行硬编码，那么numba不能生成非常快的函数，如果你还包括闭包的编译成本，那么它大多不值得。

Answer 2

根据错误消息的建议，Numba无法处理table1类型的值。您可以查看the documentation Numba可以使用的类型。原因是Numba一般不能在function模式下优化（jit-compile）任意函数，它们基本上被认为是一个黑盒子（实际上，传递的函数甚至可以是原生函数！）。

通常的方法是让Numba改为优化被调用函数。如果您无法将装饰器添加到该函数中（例如，因为它不是源代码的一部分），您仍然可以手动使用它，如：

noptyhon

显然，如果Numba无法编译import numba as nb def f(x): return x*x if __name__ == '__main__': f_opt = nb.jit(nopython=True)(f) print(f_opt(5))，它仍会失败，但在这种情况下，你无论如何都无法做到。

加速函数，将函数作为numba的参数

2 个答案: