CUDA GPU 处理：TypeError: compile_kernel() 得到一个意外的关键字参数“boundscheck”

Question

今天我开始使用 CUDA 和 GPU 处理。我找到了这个教程：https : //www.geeksforgeeks.org/running-python-script-on-gpu/

不幸的是，我第一次尝试运行 GPU 代码失败了：

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numba并cudatoolkit在教程中提到过。

Answer 1

添加一个答案以将其从未答复的队列中移除。

该示例中的代码已损坏。您的 numba 或 CUDA 安装没有任何问题。您的问题（或您从中复制的博客）中的代码无法发出博客文章声称的结果。

可以通过多种方式对其进行修改以使其起作用。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) 

这里func2变成了一个为设备编译的 ufunc。然后它将在 GPU 上的整个输入数组上运行。这样做是为了：

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

所以它更快，但请记住，GPU 时间包括编译 GPU ufunc 所花费的时间

另一种选择是实际编写一个 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) 

它像这样运行：

$ 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运行速度比 ufunc 稍快，并且缓存（这是 JIT 编译函数的缓存，而不是调用的记忆）显着加快了 GPU 上的调用。