python - CUDA GPU处理: TypeError: compile_kernel() got an unexpected keyword argument 'boundscheck'

Question

今天，我开始使用CUDA和GPU处理。我找到了本教程:
https://www.geeksforgeeks.org/running-python-script-on-gpu/

不幸的是，我第一次运行gpu代码的尝试失败了:

from numba import jit, cuda 
import numpy as np 
# to measure exec time 
from timeit import default_timer as timer 

# normal function to run on cpu 
def func(a):                                 
    for i in range(10000000): 
        a[i]+= 1    

# function optimized to run on gpu 
@jit(target ="cuda")                         
def func2(a): 
    for i in range(10000000): 
        a[i]+= 1
if __name__=="__main__": 
    n = 10000000                            
    a = np.ones(n, dtype = np.float64) 
    b = np.ones(n, dtype = np.float32) 

    start = timer() 
    func(a) 
    print("without GPU:", timer()-start)     

    start = timer() 
    func2(a) 
    print("with GPU:", timer()-start) 

输出:

/home/amu/anaconda3/bin/python /home/amu/PycharmProjects/gpu_processing_base/gpu_base_1.py
without GPU: 4.89985659904778
Traceback (most recent call last):
  File "/home/amu/PycharmProjects/gpu_processing_base/gpu_base_1.py", line 30, in <module>
    func2(a)
  File "/home/amu/anaconda3/lib/python3.7/site-packages/numba/cuda/dispatcher.py", line 40, in __call__
    return self.compiled(*args, **kws)
  File "/home/amu/anaconda3/lib/python3.7/site-packages/numba/cuda/compiler.py", line 758, in __call__
    kernel = self.specialize(*args)
  File "/home/amu/anaconda3/lib/python3.7/site-packages/numba/cuda/compiler.py", line 769, in specialize
    kernel = self.compile(argtypes)
  File "/home/amu/anaconda3/lib/python3.7/site-packages/numba/cuda/compiler.py", line 785, in compile
    **self.targetoptions)
  File "/home/amu/anaconda3/lib/python3.7/site-packages/numba/core/compiler_lock.py", line 32, in _acquire_compile_lock
    return func(*args, **kwargs)
TypeError: compile_kernel() got an unexpected keyword argument 'boundscheck'

Process finished with exit code 1

我已经在pycharm的anaconda环境中安装了教程中提到的 numba和 cudatoolkit。

Answer 1

添加答案以使此答案脱离未答复的队列。

该示例中的代码已损坏。您的numba或CUDA安装没有任何问题。问题中的代码(或从其复制博客的博客)无法发出博客帖子声明的结果。

有很多方法可以将其修改为起作用。一个会是这样的:

from numba import vectorize, jit, cuda 
import numpy as np 
# to measure exec time 
from timeit import default_timer as timer 

# normal function to run on cpu 
def func(a):                                 
    for i in range(10000000): 
        a[i]+= 1    

# function optimized to run on gpu 
@vectorize(['float64(float64)'], target ="cuda")                         
def func2(x): 
    return x+1

if __name__=="__main__": 
    n = 10000000                            
    a = np.ones(n, dtype = np.float64) 

    start = timer() 
    func(a) 
    print("without GPU:", timer()-start)     

    start = timer() 
    func2(a) 
    print("with GPU:", timer()-start) 

在这里， func2变成为设备编译的ufunc。然后，它将在GPU的整个输入阵列上运行。这样做是这样的:

$ python bogoexample.py 
without GPU: 4.314514834433794
with GPU: 0.21419800259172916

因此速度更快，但请记住，GPU时间包括编译GPU ufunc所需的时间

另一种选择是实际编写GPU内核。像这样:

from numba import vectorize, jit, cuda 
import numpy as np 
# to measure exec time 
from timeit import default_timer as timer 

# normal function to run on cpu 
def func(a):                                 
    for i in range(10000000): 
        a[i]+= 1    

# function optimized to run on gpu 
@vectorize(['float64(float64)'], target ="cuda")                         
def func2(x): 
    return x+1

# kernel to run on gpu
@cuda.jit
def func3(a, N):
    tid = cuda.grid(1)
    if tid < N:
        a[tid] += 1


if __name__=="__main__": 
    n = 10000000                            
    a = np.ones(n, dtype = np.float64) 

    for i in range(0,5):
         start = timer() 
         func(a) 
         print(i, " without GPU:", timer()-start)     

    for i in range(0,5):
         start = timer() 
         func2(a) 
         print(i, " with GPU ufunc:", timer()-start) 

    threadsperblock = 1024
    blockspergrid = (a.size + (threadsperblock - 1)) // threadsperblock
    for i in range(0,5):
         start = timer() 
         func3[blockspergrid, threadsperblock](a, n) 
         print(i, " with GPU kernel:", timer()-start) 

像这样运行:

$ python bogoexample.py 
0  without GPU: 4.885275377891958
1  without GPU: 4.748716968111694
2  without GPU: 4.902181145735085
3  without GPU: 4.889955999329686
4  without GPU: 4.881594380363822
0  with GPU ufunc: 0.16726416163146496
1  with GPU ufunc: 0.03758022002875805
2  with GPU ufunc: 0.03580896370112896
3  with GPU ufunc: 0.03530424740165472
4  with GPU ufunc: 0.03579768259078264
0  with GPU kernel: 0.1421878095716238
1  with GPU kernel: 0.04386183246970177
2  with GPU kernel: 0.029975440353155136
3  with GPU kernel: 0.029602501541376114
4  with GPU kernel: 0.029780613258481026

在这里，您可以看到内核的运行速度比ufunc快，并且缓存(这是JIT编译函数的缓存，而不是调用的内存)大大提高了GPU上的调用速度。