Question

我正在尝试在Python中实现矩阵过滤，到目前为止，实现似乎非常缓慢且效率低下。我想知道是否存在执行这种过滤的有效方法。

提供一个大矩阵A和一个过滤矩阵M，该函数应返回一个“重新混合”矩阵R，该矩阵是通过将A的每个元素（i，j）乘以M获得的，然后将结果叠加/插入到R中在位置（i，j）。请在下面找到所需的代码。

下面的示例在我的计算机上耗时约68秒（！），这似乎效率很低。

如果您能推荐加速此功能的方法，我将不胜感激。提前非常感谢！


import numpy as np
import time

nx = ny = 1500
n_mix = 50

# matrix to be filtered
A = np.random.random_sample( (nx, ny) )

# filter to be applied to each point:
M = np.random.random_sample( (2*n_mix+1, 2*n_mix+1) )

# the result is stored in "remix":
remix = np.zeros_like(A)

start = time.time()

for i in range(n_mix, nx-n_mix):
    for j in range(n_mix, ny-n_mix):
        remix[i - n_mix:i + n_mix + 1, j - n_mix:j + n_mix + 1 ] += M * A[i,j]

print remix

duration = time.time() - start
print(round(duration))

更新

实际上，scipy中的ndimage包具有完成此工作的常规卷积功能。我将在尊重您的时间的情况下，发布进行过滤的3种方法。最快的是ndimage.convolution（24秒对比其他方法的56和68）。但是，它似乎仍然很慢...

import numpy as np
from scipy import ndimage
import time
import sys


def remix_function(A, M):
    n = (np.shape(M)[0]-1)/2
    R = np.zeros_like(A)

    for k in range(-n, n+1):
        for l in range(-n, n+1):
            # Ak  = np.roll(A, -k, axis = 0)
            # Akl = np.roll(Ak, -l, axis = 1)
            R += np.roll(A, (-k,-l), axis = (0,1) ) * M[n-k, n-l]
    return R

if __name__ == '__main__':
    np.set_printoptions(precision=2)

    nx = ny = 1500
    n_mix = 50
    nb = 2*n_mix+1

    # matrix to be filtered
    A = np.random.random_sample( (nx, ny) )
    # filter to be applied to each point:
    M = np.random.random_sample( (nb, nb) )


    # the result is stored in "remix":
    remix1 = np.zeros_like(A)
    remix2 = np.zeros_like(A)
    remix3 = np.zeros_like(A)


#------------------------------------------------------------------------------
# var 1
#------------------------------------------------------------------------------
    start = time.time()

    remix1 = remix_function(A, M)

    duration = time.time() - start
    print('time for var1 =', round(duration))

#------------------------------------------------------------------------------
# var 2
#------------------------------------------------------------------------------
    start = time.time()

    for i in range(n_mix, nx-n_mix):
        for j in range(n_mix, ny-n_mix):
            remix2[i - n_mix:i + n_mix + 1, j - n_mix:j + n_mix + 1 ] += M * A[i,j]


    duration = time.time() - start
    print('time for var2  =', round(duration))

#------------------------------------------------------------------------------
# var 3
#------------------------------------------------------------------------------
    start = time.time()

    remix3 = ndimage.convolve(A, M)

    duration = time.time() - start
    print('time for var3 (convolution) =', round(duration))

Answer 1

我还不能评论帖子，但是您的double for循环是问题所在。您是否尝试过定义一个函数然后使用np.vectorize？

在python中执行2D矩阵过滤的有效方法是什么？

1 个答案: