2D numpy的基于阵列检查，看看是否所有相邻方面都是平等的

问题描述：

我开始用n×m个布尔数组，定义regions-真，如果是在该地区，否则为false。例如：

I am starting with a nxm boolean array, that defines regions- true if it is in the region, false if not. For example:

r = np.array([[ 0,  0,  1,  1,  1],
              [ 0,  1,  1,  0,  0],
              [ 0,  1,  1,  1,  0],
              [ 1,  1,  1,  0,  0],
              [ 1,  1,  0,  0,  0]])

区域之间的线可以被定义为n 1×m-1个阵列中，这将重新present在每组四个值之间的点。如果所有4个周边值是相同的，你是不是在区域之间的边缘。如果任何的4个值不同，则是。上述R：

The line between regions can be defined as an n-1 x m-1 array, which would represent a 'point' in between each set of four values. If all 4 surrounding values are the same, you are not on the edge between regions. If any of the 4 values are different, you are. For r above:

l = np.array([[ 1, 1, 1, 1],
              [ 1, 0, 1, 1],
              [ 1, 0, 1, 1],
              [ 0, 1, 1, 0]])

这是如何能够高效地完成有什么想法？我曾尝试做两个方向的一个差异，但这双打。是否有一个二维差异函数？这样做的其他方式？

Any thoughts on how this can be done efficiently? I have tried doing a diff in both directions, but this doubles up. Is there a 2D diff function? Some other way of doing this?

答

这将做的百分点的由真包围测试，

tmp = r[1:] & r[:-1]
l = np.logical_not(tmp[:, 1:] & tmp[:, :-1])

然后，你可以做的百分点的包围测试假以同样的方式将它们结合起来，

Then you can do the test for points surrounded by False the same way and combine them,

r = np.logical_not(r)
tmp = r[1:] & r[:-1]
l &= np.logical_not(tmp[:, 1:] & tmp[:, :-1])

print l.astype(int)
# [[1 1 1 1]
#  [1 0 1 1]
#  [1 0 1 1]
#  [0 1 1 0]]

2D numpy的基于阵列检查，看看是否所有相邻方面都是平等的

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