在二進制numpy矩陣中將連續的1的塊翻轉到特定大小

Question

我正在做一個圖像分析項目。 我已經將感興趣的圖片（NxM numpy數組）轉換為二進制格式。 矩陣中的“ 1”是關注區域。 存在感興趣的區域，並且存在無法代表圖像特征的噪點。 例如，在圖像的水平快照中，我不關心孤立的1或2個組，最多5個連續的1。 我想找到一種快速的方法來翻轉它們（即使它們= 0）。

我的MWE用於翻轉孤立的1：

import numpy as np
img = np.random.choice([0,1],size=(1000,1000), p=[1./2,1./2])

#now we take the second derivative of the matrix in the horizontal axis
#since we have a binary matrix, an isolated 1, that is [...010...] is captured
#by a second derivative entry equal to -2
#because ([...010...]->dx->[...1,-1,...]->dx->[...-2...]

ddx_img = np.diff(np.diff(img,1),1)
to_flip = np.where(ddx_img==-2) #returns a tuple of [x,y] matrix entries

# the second derivative eats up an index position on horizontally, so I need to add
# +1 to the horizontal axis of the tuple

temp_copy = to_flip[1].copy() #cannot modify tuple directly, for some reason its read only
temp_copy+=1
to_flip = (to_flip[0],temp_copy)

#now we can flip the entries by adding +1 to the entries to flip and taking mod 2
img[to_flip]=mod(img[to_flip]+1,2)

這在我的機器上花費大約9毫秒。 我可以使用長達1秒的例程。

我歡迎任何對代碼的批評（我不是一個優秀的python程序員），以及任何有關如何有效擴展此過程以消除連續的1的孤立小島到通用大小S的小島的想法。

提前致謝

編輯：我意識到國防部是不必要的。 當時，我還想翻轉太小的0島。 一個人可以用== 0代替= mod ....

Answer 1

特定問題的情況

編輯之后，似乎可以使用一些slicing ，從而避免制作中間副本以提高性能。 這是兩行代碼，可實現所需的輸出-

# Calculate second derivative
ddx_img = np.diff(np.diff(img,1),1)

# Get sliced version of img excluding the first and last columns 
# and use mask with ddx elements as "-2" to zeros
img[:,1:-1][ddx_img==-2] = 0

運行時測試並驗證結果-

In [42]: A = np.random.choice([0,1],size=(1000,1000), p=[1./2,1./2])

In [43]: def slicing_based(A):
    ...:    img = A.copy()
    ...:    ddx_img = np.diff(np.diff(img,1),1)
    ...:    img[:,1:-1][ddx_img==-2] = 0
    ...:    return img
    ...: 
    ...: 
    ...: def original_approach(A):
    ...: 
    ...:    img = A.copy()
    ...: 
    ...:    ddx_img = np.diff(np.diff(img,1),1)
    ...:    to_flip = np.where(ddx_img==-2)
    ...: 
    ...:    temp_copy = to_flip[1].copy()
    ...:    temp_copy+=1
    ...:    to_flip = (to_flip[0],temp_copy)
    ...: 
    ...:    img[to_flip] = 0
    ...: 
    ...:    return img
    ...: 

In [44]: %timeit slicing_based(A)
100 loops, best of 3: 15.3 ms per loop

In [45]: %timeit original_approach(A)
10 loops, best of 3: 20.1 ms per loop

In [46]: np.allclose(slicing_based(A),original_approach(A))
Out[46]: True

---

一般情況

為了使該解決方案通用，可以使用一些信號處理，特別是2D convolution ，如下所示-

# Define kernel
K1 = np.array([[0,1,1,0]]) # Edit this for different island lengths
K2 = 1-K1

# Generate masks of same shape as img amd based on TRUE and inverted versions of 
# kernels being convolved and those convolved sums being compared against the 
# kernel sums indicating those spefic positions have fulfiled both the ONES 
# and ZEROS criteria
mask1 = convolve2d(img, K1, boundary='fill',fillvalue=0, mode='same')==K1.sum()
mask2 = convolve2d(img==0, K2, boundary='fill',fillvalue=0, mode='same')==K2.sum()

# Use a combined mask to create that expanses through the kernel length 
# and use it to set those in img to zeros
K3 = np.ones((1,K1.size))
mask3 = convolve2d(mask1 & mask2, K3, boundary='fill',fillvalue=0, mode='same')>0
img_out = img*(~mask3)

樣本輸入，輸出-

In [250]: img
Out[250]: 
array([[0, 1, 1, 1, 0, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 0, 1],
       [1, 0, 1, 1, 1, 1, 0, 0],
       [1, 1, 1, 1, 0, 1, 0, 1],
       [1, 1, 0, 1, 1, 0, 1, 1],
       [1, 0, 1, 1, 1, 1, 1, 1],
       [1, 1, 0, 1, 1, 0, 1, 0],
       [1, 1, 1, 0, 1, 1, 1, 1]])

In [251]: img_out
Out[251]: 
array([[0, 1, 1, 1, 0, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 0, 1],
       [1, 0, 1, 1, 1, 1, 0, 0],
       [1, 1, 1, 1, 0, 1, 0, 1],
       [1, 1, 0, 0, 0, 0, 0, 1],
       [1, 0, 1, 1, 1, 1, 1, 1],
       [1, 1, 0, 0, 0, 0, 0, 0],
       [1, 1, 1, 0, 1, 1, 1, 1]])