问题描述
我目前正在尝试找出如何有效使用CuPy流的方法。以下代码通过重复的矩阵乘法计算矩阵幂。我希望下面的代码将其大部分时间花费在同步行上,但似乎将其大部分时间花费在matmul行上。这是CuPy中的错误,还是我在滥用CuPy流?
#!/usr/bin/env python
"""
stream_example.py
Inefficiently calculates a matrix power through repeated matrix multiplication.
"""
import numpy as np
import cupy
import sys
import time
def main(N,power):
compute_stream = cupy.cuda.stream.Stream(non_blocking=True)
with compute_stream:
d_mat = cupy.random.randn(N*N,dtype=cupy.float64).reshape(N,N)
d_ret = d_mat
cupy.matmul(d_ret,d_mat)
start_time = time.time()
for i in range(power - 1):
d_ret = cupy.matmul(d_ret,d_mat)
end_time = time.time()
print(f"Time spent on cupy.matmul for loop: {end_time - start_time}")
start_time = time.time()
compute_stream.synchronize()
end_time = time.time()
print(f"Time spent compute_stream.synchronize(): {end_time - start_time}")
if __name__ == "__main__":
main(int(sys.argv[1]),int(sys.argv[2]))
结果表明,大部分时间都花在循环的重复乘法中,而不是stream.synchronize()。 cupy.matmul()不能异步使用吗?
$ python3 stream_example.py 16384 1024
Time spent on cupy.matmul for loop: 2.667935609817505
Time spent compute_stream.synchronize(): 4.2438507080078125e-05
解决方法
似乎可以解决此问题并添加以下内容。我将绿色复选标记保留给可以提出一种不太hacky解决方案的人:
import cupy_backends.cuda.libs.cublas
from cupy.cuda import device
handle = device.get_cublas_handle()
...
cupy_backends.cuda.libs.cublas.setStream(handle,compute_stream.ptr)
$ python3 stream_example.py 16384 4
Time spent on cupy.matmul for loop: 0.007548093795776367
Time spent compute_stream.synchronize(): 5.099333047866821