numba cuda无法使用+=(需要减少gpu?)

是的,要将多个GPU线程的数据求和到单个变量,需要进行适当的并行缩减。

下面是一个简单的示例,说明如何从单个内核执行此操作:

$ cat t23.py
import math
def numba_example(number_of_maximum_loop,gs,ts,bs):
    from numba import cuda
    result = cuda.device_array([3,])

    @cuda.jit(device=True)
    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    @cuda.jit
    def cuda_kernel(number_of_maximum_loop,result,gs,ts,bs):
        i = cuda.grid(1)
        if i < number_of_maximum_loop:
            cuda.atomic.add(result, 0, BesselJ0(i/100+gs))
            cuda.atomic.add(result, 1, BesselJ0(i/100+ts))
            cuda.atomic.add(result, 2, BesselJ0(i/100+bs))

# Configure the blocks
    threadsperblock = 128
    blockspergrid = (number_of_maximum_loop + (threadsperblock - 1)) // threadsperblock

 # Start the kernel
    init = [0.0,0.0,0.0]
    result = cuda.to_device(init)
    cuda_kernel[blockspergrid, threadsperblock](number_of_maximum_loop,result,gs,ts,bs)

    return result.copy_to_host()

print(numba_example(1000,20,20,20))
$ python t23.py
[ 162.04299487  162.04299487  162.04299487]
$

你也可以直接用手指适当地减少麻木 reduce 如上所述的装饰者 here 虽然我不确定这样一个内核可以实现3个缩减。

最后,您可以使用numba cuda编写一个普通的cuda并行缩减,如图所示 here . 我认为,将其扩展到在单个内核中执行3次缩减应该并不困难。

$ cat t24.py
import math
def numpy_example(number_of_maximum_loop,gs,ts,bs):
    import numpy as np
    result = np.zeros([3,])

    def BesselJ0(x):
        return math.sqrt(2/math.pi/x)

    for i in range(number_of_maximum_loop):
        result[0] += BesselJ0(i/100+gs)
        result[1] += BesselJ0(i/100+ts)
        result[2] += BesselJ0(i/100+bs)

    return result

print(numpy_example(1000,20,20,20))
$ python t24.py
[ 162.04299487  162.04299487  162.04299487]
$