在Numpy数组中创建缓冲区 [英] Create buffer zone within a Numpy array
问题描述
我有一个二进制图像,如下所示:
I have a binary image as follows:
data = np.array([[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0]])
对于具有1s值的像素,我想使两个像素的缓冲区分别在每个四个方向上具有1s值.预期结果将是:
For pixels having 1s values, I want to make buffer zone of two pixels with value 1s surrounded in every four directions. The expected result would be:
result=np.array([[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1],
[1, 1 , 1 , 1 , 1 , 1 , 1 , 1]])
我该怎么办?
推荐答案
如果输入和输出数组上只有1和0,则可以使用2D卷积来完成,这很简单而且可行.
If you only have ones and zeros on the input and output array, you can do it with a 2D convolution, which is simple and works.
from scipy.signal import convolve2d
data = np.array([[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 1 , 1 , 1 , 1 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0],
[0, 0 , 0 , 0 , 0 , 0 , 0 , 0]])
# the kernel doesn't need to be ones, it just needs to be positive and
# non-zero.
kernel = np.ones((5, 5))
result = np.int64(convolve2d(data, kernel, mode='same') > 0)
这将为您提供所需的输出.您需要定义要在边缘发生的事情-在此版本中,输出数组的大小与输入数组的大小相同.
Which gives you the output you want. You need to define what you want to happen at the edges - in this version, the output array is the same size as the input array.
如果数组稀疏,则可能可以更快地执行某些操作.
It might be possible you can do something faster if you have a sparse array.
如果数组中的值不是1和0,则需要更多考虑.
If you have other values than one and zero in your array, more thought would be needed.
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