Question

我遇到了与this question中提到的相同的问题，即我希望2D相关的标量输出为matlab cgi.reg，但是在python中。我通过编写自己的算法解决了这个问题，但它存在性能问题。

它基于matlab docs中使用的公式：

Windows Registry Editor Version 5.00

[HKEY_CLASSES_ROOT\.cgi\Shell\ExceCGI\Command]
@="C:\\Perl\\bin\\perl.exe -T"

它有效，但速度非常慢。任何想法如何让它更优雅，更快？

提前致谢！

Answer 1

这种vectorized方法应该为您加速 -

# Vectorized versions of c,d,e
c_vect = (a-AM)*(b-BM)
d_vect = (a-AM)**2
e_vect = (b-BM)**2

# Finally get r using those vectorized versions
r_out = np.sum(c_vect)/float(np.sqrt(np.sum(d_vect)*np.sum(e_vect)))

运行时测试并验证结果

In [70]: M = 100
    ...: N = 100
    ...: a = np.random.rand(M,N)
    ...: b = np.random.rand(M,N)
    ...: 
    ...: k = np.shape(a)
    ...: H=k[0]
    ...: W=k[1]
    ...: AM=np.mean(a)
    ...: BM=np.mean(b)
    ...: 

In [71]: %%timeit
    ...: c = np.zeros((H,W))
    ...: d = np.zeros((H,W))
    ...: e = np.zeros((H,W))
    ...: 
    ...: for ii in range(0,H):
    ...:   for jj in range(0,W):
    ...:     c[ii,jj]=(a[ii,jj]-AM)*(b[ii,jj]-BM)
    ...:     d[ii,jj]=(a[ii,jj]-AM)**2
    ...:     e[ii,jj]=(b[ii,jj]-BM)**2
    ...: 
    ...: r = np.sum(c)/float(np.sqrt(np.sum(d)*np.sum(e)))
    ...: 
10 loops, best of 3: 24.7 ms per loop

In [72]: %%timeit
    ...: c_vect = (a-AM)*(b-BM)
    ...: d_vect = (a-AM)**2
    ...: e_vect = (b-BM)**2
    ...: r_out = np.sum(c_vect)/float(np.sqrt(np.sum(d_vect)*np.sum(e_vect)))
    ...: 
10000 loops, best of 3: 64.7 µs per loop

In [73]: np.allclose(r,r_out)
Out[73]: True

matlab中的corr2等效于python - 性能问题

1 个答案: