您是否要使用高斯核进行图像平滑？如果是这样，则gaussian_filter()scipy中有一个函数：
这应该可以工作- 尽管仍不能100％准确，但它会尝试考虑网格每个像元内的概率质量。我认为在每个像元的中点使用概率密度的准确性稍差，尤其是对于小内核。有关示例，请参见https://homepages.inf.ed.ac.uk/rbf/HIPR2/gsmooth.htm。
import numpy as np
import scipy.stats as st
def gkern(kernlen=21, nsig=3):
"""Returns a 2D Gaussian kernel."""
x = np.linspace(-nsig, nsig, kernlen+1)
kern1d = np.diff(st.norm.cdf(x))
kern2d = np.outer(kern1d, kern1d)
return kern2d/kern2d.sum()
通过链接在图3的示例中对其进行测试：
gkern(5, 2.5)*273
给
array([[ 1.0278445 , 4.10018648, 6.49510362, 4.10018648, 1.0278445 ],
[ 4.10018648, 16.35610171, 25.90969361, 16.35610171, 4.10018648],
[ 6.49510362, 25.90969361, 41.0435344 , 25.90969361, 6.49510362],
[ 4.10018648, 16.35610171, 25.90969361, 16.35610171, 4.10018648],
[ 1.0278445 , 4.10018648, 6.49510362, 4.10018648, 1.0278445 ]])
。平方根是不必要的，并且间隔的定义不正确。
import numpy as np
import scipy.stats as st
def gkern(kernlen=21, nsig=3):
"""Returns a 2D Gaussian kernel array."""
interval = (2*nsig+1.)/(kernlen)
x = np.linspace(-nsig-interval/2., nsig+interval/2., kernlen+1)
kern1d = np.diff(st.norm.cdf(x))
kernel_raw = np.sqrt(np.outer(kern1d, kern1d))
kernel = kernel_raw/kernel_raw.sum()
return kernel
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