Python 有效地计算平方差之和

这是一个循环，用于提取两幅图像的RGB值，并计算所有三个通道的平方差之和。 直接在my main.py中运行此代码需要0.07秒。如果我在这个.pyx文件中运行它，速度会降低到1秒。我已经读过cdef函数，但是我没有成功地传递数组。如能帮助您将此函数转换为cdef函数，我们将不胜感激。我真的需要这个循环越快越好

from cpython cimport array
import array
import numpy as np
cimport numpy as np

def fittnes(Orginal, Mutated):

    Fittnes = 0

    for x in range(0, 299):

        for y in range(0, 299):

            DeltaRed   = (Orginal[x][y][0] - Mutated[x][y][0])
            DeltaGreen = (Orginal[x][y][1] - Mutated[x][y][1])
            DeltaBlue  = (Orginal[x][y][2] - Mutated[x][y][2])

            Fittnes += (DeltaRed * DeltaRed + DeltaGreen * DeltaGreen + DeltaBlue * DeltaBlue)

    return Fittnes

My Main.py函数调用

 NewScore = cythona.fittnes(numpy.array(Orginal), numpy.array(MutatedImage))

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让我有兴趣了解加速数字，所以我将此作为解决方案发布。因此，如注释中所述/讨论的，如果输入是NumPy数组，您可以使用本机NumPy工具，在本例中也是如此-

out = ((Orginal - Mutated)**2).sum()

sub = Orginal - Mutated
out = np.einsum('ijk,ijk->',sub,sub)

您也可以将非常高效的方法用于相同的任务，如下所示-

out = ((Orginal - Mutated)**2).sum()

sub = Orginal - Mutated
out = np.einsum('ijk,ijk->',sub,sub)

运行时测试

定义函数-

def org_app(Orginal,Mutated):
    Fittnes = 0
    for x in range(0, Orginal.shape[0]):
        for y in range(0, Orginal.shape[1]):
            DR = (Orginal[x][y][0] - Mutated[x][y][0])
            DG = (Orginal[x][y][1] - Mutated[x][y][1])
            DB  = (Orginal[x][y][2] - Mutated[x][y][2])
            Fittnes += (DR * DR + DG * DG + DB * DB)
    return Fittnes

def einsum_based(Orginal,Mutated):
    sub = Orginal - Mutated
    return np.einsum('ijk,ijk->',sub,sub)

def dot_based(Orginal,Mutated): # @ali_m's suggestion
    sub = Orginal - Mutated
    return np.dot(sub.ravel(), sub.ravel())

def vdot_based(Orginal,Mutated):  # variant of @ali_m's suggestion
    sub = Orginal - Mutated
    return np.vdot(sub, sub)

时间安排-

In [14]: M,N = 100,100
    ...: Orginal = np.random.rand(M,N,3)
    ...: Mutated = np.random.rand(M,N,3)
    ...: 

In [15]: %timeit org_app(Orginal,Mutated)
    ...: %timeit ((Orginal - Mutated)**2).sum()
    ...: %timeit einsum_based(Orginal,Mutated)
    ...: %timeit dot_based(Orginal,Mutated)
    ...: %timeit vdot_based(Orginal,Mutated)
    ...: 
10 loops, best of 3: 54.9 ms per loop
10000 loops, best of 3: 112 µs per loop
10000 loops, best of 3: 69.8 µs per loop
10000 loops, best of 3: 86.2 µs per loop
10000 loops, best of 3: 85.3 µs per loop

In [16]: # Inputs
    ...: M,N = 1000,1000
    ...: Orginal = np.random.rand(M,N,3)
    ...: Mutated = np.random.rand(M,N,3)
    ...: 

In [17]: %timeit org_app(Orginal,Mutated)
    ...: %timeit ((Orginal - Mutated)**2).sum()
    ...: %timeit einsum_based(Orginal,Mutated)
    ...: %timeit dot_based(Orginal,Mutated)
    ...: %timeit vdot_based(Orginal,Mutated)
    ...: 
1 loops, best of 3: 5.49 s per loop
10 loops, best of 3: 63 ms per loop
10 loops, best of 3: 23.9 ms per loop
10 loops, best of 3: 24.9 ms per loop
10 loops, best of 3: 24.9 ms per loop

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你的cython密码在哪里？这是cython密码。从my main.py调用它如果输入数组是NumPy数组，你可以只使用

（（原始-变异）**2）.sum（）

来实现超高速。什么是（**2）意味着我在你的代码中看不到任何cython特定的东西，但正如Divakar建议的那样，你可以单独使用NumPy来显著加快代码速度…或者只使用

np.dot（sub.ravel）（），sub.ravel（））

@ali_m Nice！那也行！把它作为解决方案？无论哪种方式，都想将其添加到基准测试中。与基于einsum_的测试相比，np.dot（sub.ravel（），sub.ravel（））的速度有多快？@Funktiona啊，是的！我认为如果您不使用

float64

，则使用

np.einsum

的值会有所不同。您的输入数据类型是什么？@Funktiona您可以通过

np.einsum

了解值不一致的情况。