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我试图优化下面的代码，但我不知道如何提高计算速度。我尝试了 Cthon，但性能就像在 python 中一样。

是否可以在不重写 C/C++ 的所有内容的情况下提高性能？

谢谢你的帮助

import numpy as np

heightSequence = 400
widthSequence = 400
nHeights = 80

DOF = np.zeros((heightSequence, widthSequence), dtype = np.float64)
contrast = np.float64(np.random.rand(heightSequence, widthSequence, nHeights))

initDOF = np.zeros([heightSequence, widthSequence], dtype = np.float64)
initContrast = np.zeros([heightSequence, widthSequence, nHeights], dtype = np.float64)
initHeight = np.float64(np.r_[0:nHeights:1.0])
initPixelContrast = np.array(([0 for ii in range(nHeights)]), dtype = np.float64)


# for each row
for row in range(heightSequence):
    # for each col
    for col in range(widthSequence):

        # initialize variables            
        height = initHeight # array ndim = 1
        c = initPixelContrast # array ndim = 1

        # for each height            
        for indexHeight in range(0, nHeights):
            # get contrast profile for current pixel
            tempC = contrast[:, :, indexHeight]
            c[indexHeight] = tempC[row, col]

        # save original contrast            
        # originalC = c
        # originalHeight = height                

        # remove profile before maximum and after minumum contrast
        idxMaxContrast = np.argmax(c)
        c = c[idxMaxContrast:]
        height = height[idxMaxContrast:]

        idxMinContrast = np.argmin(c) + 1
        c = c[0:idxMinContrast]
        height = height[0:idxMinContrast]              

        # remove some refraction
        if (len(c) <= 1) | (np.max(c) <= 0):
            DOF[row, col] = 0                  

        else:

            # linear fitting of profile contrast                                             
            P = np.polyfit(height, c, 1)
            m = P[0]
            q = P[1]

            # remove some refraction               
            if m >= 0:
                DOF[row, col] = 0

            else:
                DOF[row, col] = -q / m

    print 'row=%i/%i' %(row, heightSequence)

# set range of DOF
DOF[DOF < 0] = 0
DOF[DOF > nHeights] = 0
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1 回答 1

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通过查看代码，您似乎可以完全摆脱两个外部循环，将代码转换为矢量化形式。然而，np.polyfit调用必须被其他表达式替换，但线性拟合的系数很容易找到，也是向量化的形式。if-else然后可以将最后一个转换为np.where呼叫。

于 2013-02-28T11:53:09.447 回答