根據相鄰值

Question

我有一個數組（N，M）中的2D陣列替換值：

[255 100 255] 
[100 255 100] 
[255 100 255] 

我需要創建一個這樣的新的數組，其中neigboring值進行測試的，如果北，東，南，西都等於100我的值設置爲100：

[255 100 255] 
[100 100 100] 
[255 100 255] 

我有一個簡單的解決方案，在1環：N和1：米，但它顯然是很慢的，我想知道是否有更快地做到這一點。 我發現有幾個鏈接正在討論滑動窗口來計算平均值，但我沒有看到如何跟蹤我的索引來創建新的數組。 Using strides for an efficient moving average filter

在此先感謝您的意見。

Answer 1

假設A作爲輸入陣列，這裏是使用slicing和boolean indexing一種方法 -

# Get west, north, east & south elements for [1:-1,1:-1] region of input array 
W = A[1:-1,:-2] 
N = A[:-2,1:-1] 
E = A[1:-1,2:] 
S = A[2:,1:-1] 

# Check if all four arrays have 100 for that same element in that region 
mask = (W == 100) & (N == 100) & (E == 100) & (S == 100) 

# Use the mask to set corresponding elements in a copy version as 100s 
out = A.copy() 
out[1:-1,1:-1][mask] = 100 

樣品運行 -

In [90]: A 
Out[90]: 
array([[220, 93, 205, 82, 23, 210, 22], 
     [133, 228, 100, 27, 210, 186, 246], 
     [196, 100, 73, 100, 86, 100, 53], 
     [195, 131, 100, 142, 100, 214, 100], 
     [247, 73, 117, 116, 24, 100, 50]]) 

In [91]: W = A[1:-1,:-2] 
    ...: N = A[:-2,1:-1] 
    ...: E = A[1:-1,2:] 
    ...: S = A[2:,1:-1] 
    ...: mask = (W == 100) & (N == 100) & (E == 100) & (S == 100) 
    ...: 
    ...: out = A.copy() 
    ...: out[1:-1,1:-1][mask] = 100 
    ...: 

In [92]: out 
Out[92]: 
array([[220, 93, 205, 82, 23, 210, 22], 
     [133, 228, 100, 27, 210, 186, 246], 
     [196, 100, 100, 100, 86, 100, 53], 
     [195, 131, 100, 142, 100, 100, 100], 
     [247, 73, 117, 116, 24, 100, 50]]) 

---

這樣的問題被認爲主要是在信號處理和/或圖像處理域。所以，你可以使用2D convolution過一個替代的解決方案，像這樣 -

from scipy import signal 
from scipy import ndimage 

# Use a structuring elements with north, west, east and south elements as 1s 
strel = ndimage.generate_binary_structure(2, 1) 

# 2D Convolve to get 4s at places that are surrounded by 1s 
mask = signal.convolve2d((A==100).astype(int),strel,'same')==4 

# Use the mask to set corresponding elements in a copy version as 100 
out = A.copy() 
out[mask] = 100 

採樣運行 -

In [119]: A 
Out[119]: 
array([[108, 184, 0, 176, 131, 86, 201], 
     [ 22, 47, 100, 78, 151, 196, 221], 
     [185, 100, 142, 100, 121, 100, 24], 
     [201, 101, 100, 138, 100, 20, 100], 
     [127, 227, 217, 19, 206, 100, 43]]) 

In [120]: strel = ndimage.generate_binary_structure(2, 1) 
    ...: mask = signal.convolve2d((A==100).astype(int),strel,'same')==4 
    ...: 
    ...: out = A.copy() 
    ...: out[mask] = 100 
    ...: 

In [121]: out 
Out[121]: 
array([[108, 184, 0, 176, 131, 86, 201], 
     [ 22, 47, 100, 78, 151, 196, 221], 
     [185, 100, 100, 100, 121, 100, 24], 
     [201, 101, 100, 138, 100, 100, 100], 
     [127, 227, 217, 19, 206, 100, 43]]) 

一個更直接的方法是用ndimage.binary_closing，這正是預期運行closing這裏。所以，另一種替代方式獲得面具將是 -

strel = ndimage.generate_binary_structure(2, 1) 
mask = ndimage.binary_closing(A==100, structure=strel)