Question

matlab中有函数用于获取每个条目的n个最近条目的平均值。

x=np.array([[0.1,0.8,.2],
            [0.5,0.2,np.nan],
            [0.7,0.2,0.9],
            [0.4,0.7,1],
            [np.nan,0.14,1]])

在matlab中创建平均过滤器：

x=[[0.1,0.8,.2],
   [0.5,0.2,nan],
   [0.7,0.2,0.9],
   [0.4,0.7,1],
   [nan,0.14,1]]

fspecial('average',[3,3])
filter2(ave1,x)

[[ 0.17777778         nan         nan]
 [ 0.27777778         nan         nan]
 [ 0.3                nan         nan]
 [        nan         nan  0.43777778]
 [        nan         nan  0.31555556]]

我想将其转换为python。我发现了这个： uniform filter 和： skimage.filters.rank.mean

但结果与matlab不同。均匀过滤器：

x=np.array([[0.1,0.8,.2],
            [0.5,0.2,np.nan],
            [0.7,0.2,0.9],
            [0.4,0.7,1],
            [np.nan,0.14,1]])

print(uniform_filter(x, size=3, mode='constant'))

[[ 0.17777778         nan         nan]
 [ 0.27777778         nan         nan]
 [ 0.3                nan         nan]
 [        nan         nan         nan]
 [        nan         nan         nan]]

skimage过滤器：

from skimage.filters.rank import mean
from skimage.morphology import square
from skimage import img_as_float
x=np.array([[0.1,0.8,.2],
        [0.5,0.2,np.nan],
        [0.7,0.2,0.9],
        [0.4,0.7,1],
        [np.nan,0.14,1]])

print(mean(x, square(3)))

[[102  76  76]
 [106 102  97]
 [114 130 127]
 [ 90 142 167]
 [ 79 137 181]]

print(img_as_float(mean(x, square(3))))

[[ 0.4         0.29803922  0.29803922]
 [ 0.41568627  0.4         0.38039216]
 [ 0.44705882  0.50980392  0.49803922]
 [ 0.35294118  0.55686275  0.65490196]
 [ 0.30980392  0.5372549   0.70980392]]

最后我自己也不得不这样做，但表现并不成熟：

x=np.array([[0.1,0.8,.2],
            [0.5,0.2,np.nan],
            [0.7,0.2,0.9],
            [0.4,0.7,1],
            [np.nan,0.14,1]])



Winsize=3
adder=int(Winsize/2)
result=np.zeros_like(x)
nan_window_index=np.array([])
for i in range(x.shape[0]):
    for j in range(x.shape[1]):
        top_left_r= int(i-adder)
        top_left_c= int(j-adder)
        bottom_right_r=int(i+adder)
        bottom_right_c=int(j+adder)
        sum_list=np.array([])

        for r_counter in range(top_left_r, bottom_right_r+1):
            if r_counter<0 or r_counter > x.shape[0]-1:
                continue
            for c_counter in range(top_left_c, bottom_right_c+1):
                if c_counter<0 or c_counter > x.shape[1]-1:
                    continue
                if not np.isnan(x[r_counter, c_counter]):
                    sum_list=np.append(sum_list, x[r_counter, c_counter])
                else:
                    nan_window_index=np.append(nan_window_index, [[r_counter, c_counter]])

        result[i,j]= np.sum(sum_list)/(Winsize*Winsize)

nan_window_index=np.unique(nan_window_index.reshape(int(len(nan_window_index)/2),2), axis=0)

for i,j in nan_window_index:
    top_left_r= int(i-adder)
    top_left_c= int(j-adder)
    bottom_right_r=int(i+adder)
    bottom_right_c=int(j+adder)

    for r_counter in range(top_left_r, bottom_right_r+1):
        if r_counter<0 or r_counter > x.shape[0]-1:
            continue
        for c_counter in range(top_left_c, bottom_right_c+1):
            if c_counter<0 or c_counter > x.shape[1]-1:
                continue
            result[r_counter, c_counter]=np.nan

print(result)

，结果与matlab相同：

[[ 0.17777778         nan         nan]
 [ 0.27777778         nan         nan]
 [ 0.3                nan         nan]
 [        nan         nan  0.43777778]
 [        nan         nan  0.31555556]]

有关更好表现的任何建议吗？

Answer 1

您可以使用scipy.signal.convolve（或者scipy.signal.convolve2d，因为它可能会更快）：

import numpy as np
# from scipy.signal import convolve
from scipy.signal import convolve2d

x=np.array([[0.1,0.8,.2],
            [0.5,0.2,np.nan],
            [0.7,0.2,0.9],
            [0.4,0.7,1],
            [np.nan,0.14,1]])

core = np.full((3,3),1/3**2)

# convolve(x, core, mode='same')
convolve2d(x, core, mode='same')

具有均匀值的卷积与均匀滤波器相同。请注意，这将自动＆＃34;假设＆＃34;在矩阵之外的零，但这符合您的要求，因此它将适用于您当前的设置。

在python中创建matlab平均过滤器

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