问题描述
我已遵循，用于核密度估计的应用.我的目标是为一个阵列组创建两个不同的组/集群或更多.以下代码适用于该数组以外的数组组的每个成员:
I have followed this link for the application of kernel density estimation. My aim is creating two different groups/clusters or more for an array group. The below code works for every members of array group except this array:
X = np.array([[77788], [77793],[77798], [77803], [92886], [92891], [92896], [92901]])
所以我的期望是看到两个不同的集群，例如:
So my expectation is seeing two different clusters such as:
first_group =([[77788]，[77793]，[77798]，[77803]])
first_group = ([[77788], [77793],[77798], [77803]])
second_group =([[92886]，[92891]，[92896]，[92901]])
second_group = ([[92886], [92891], [92896], [92901]])
我有一个动态列表，所以我无法固定linspace的值.因为此数组可能是0到10或100000到2000000.这就是为什么我将数组的最大和最小点放在linspace中的原因.
I have a dynamic list, so I can not fix a value for linspace. Because this array may be 0to 10 or 100000 to 2000000. That's why I have put max and min points of the array in the linspace.
毕竟，即使尝试了各种带宽，我也无法获得不同的群集.我的代码如下所示:
After all, I could not obtain different clusters even though I tried various bandwidths. My code can be seen below:
a = X.reshape(-1,1)
kde = KernelDensity(kernel='gaussian', bandwidth=8).fit(a)
s = linspace(min(a),max(a))
e = kde.score_samples(s.reshape(-1,1))
plot(s, e)
mi, ma = argrelextrema(e, np.less)[0], argrelextrema(e, np.greater)[0]
print("Minima:", s[mi]) # output: []
print("Maxima:", s[ma]) # output: []
s [mi]和s [ma]的值为空，这表示此数组没有两个不同的簇.在可视化中可以看到，我们至少有一个最小点.为什么看不到s [mi]输出的此值?
s[mi] and s[ma] values are empty which means there is no two different clusters for this array. In the visualization can be seen that we have at least one minimum point. why can not be seen this value for the s[mi] output?
我将相同的代码应用于不同的带宽，如下所示，但是，此群集没有最小值或最大值.所以知道我在做什么错吗?
And I applied the same code for different bandwidths which can be seen below, however, there is no minimum or maximum values for this cluster. so any idea what am I doing wrong?
bandwidth=0.008
bandwidth = 0.00002
推荐答案
尝试使用10000的带宽，或者尝试依靠启发式方法选择带宽.
Try a bandwidth of 10000, or try relying on heuristics for choosing the bandwidth.
为了使您的代码更健壮，还可以在连续的最小值处拆分群集.因为您的问题是这里没有唯一的最小值，而是一个间隔.
To make your code more robusty also split clusters at consecutive minima. Because your problem is that there is no unique minimum here, but an interval.
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