我们可以利用numexpr
module有效地执行所有后面的算术运算作为一个评估表达式。
因此,下列步骤操作:
C = cv2.subtract(ml, mrd)
C = cv2.pow(C,2)
C = np.divide(C, sigma_m)
C = p0 + (1-p0)**(-C)
可以通过一个表达式替换 -
import numexpr as ne
C = ne.evaluate('p0 +(1-p0)**(-((ml-mrd)**2)/sigma_m)')
让我们确定的事情。原来的做法是FUNC -
def original_app(ml, mrd, sigma_m, p0):
C = cv2.subtract(ml, mrd)
C = cv2.pow(C,2)
C = np.divide(C, sigma_m)
C = p0 + (1-p0)**(-C)
return C
验证 - 整个数据集的大小不同
In [28]: # Setup inputs
...: S = 1024 # Size parameter
...: ml = np.random.randint(0,255,(S,S))/255.0
...: mrd = np.random.randint(0,255,(S,S))/255.0
...: sigma_m = 0.45
...: p0 = 0.56
...:
In [29]: out1 = original_app(ml, mrd, sigma_m, p0)
In [30]: out2 = ne.evaluate('p0 +(1-p0)**(-((ml-mrd)**2)/sigma_m)')
In [31]: np.allclose(out1, out2)
Out[31]: True
计时 -
In [19]: # Setup inputs
...: S = 1024 # Size parameter
...: ml = np.random.randint(0,255,(S,S))/255.0
...: mrd = np.random.randint(0,255,(S,S))/255.0
...: sigma_m = 0.45
...: p0 = 0.56
...:
In [20]: %timeit original_app(ml, mrd, sigma_m, p0)
10 loops, best of 3: 67.1 ms per loop
In [21]: %timeit ne.evaluate('p0 +(1-p0)**(-((ml-mrd)**2)/sigma_m)')
100 loops, best of 3: 12.9 ms per loop
In [22]: # Setup inputs
...: S = 512 # Size parameter
In [23]: %timeit original_app(ml, mrd, sigma_m, p0)
100 loops, best of 3: 15.3 ms per loop
In [24]: %timeit ne.evaluate('p0 +(1-p0)**(-((ml-mrd)**2)/sigma_m)')
100 loops, best of 3: 3.39 ms per loop
In [25]: # Setup inputs
...: S = 256 # Size parameter
In [26]: %timeit original_app(ml, mrd, sigma_m, p0)
100 loops, best of 3: 3.65 ms per loop
In [27]: %timeit ne.evaluate('p0 +(1-p0)**(-((ml-mrd)**2)/sigma_m)')
1000 loops, best of 3: 878 µs per loop
围绕5x
跨越各种尺寸加速与更大的阵列更好的加速!
此外,作为一个侧面说明,我会建议使用初始化数组,而不是像最后一步那样追加。因此,我们可以在进入循环之前初始化out = np.zeros((len(d), width, height))
/np.empty
之类的内容,并在最后一步将输入数组分配到:out[iteration_ID] = C
。
它也取决于你如何构建OpenCV,所以你可以发布'getBuildInformation()'的输出。 –
@MarkSetchell'cv2.getBuildInformation()'的输出太大而无法在评论中写入。你是否在想这个输出的具体内容? – Mira