如何在python(sklearn)中进行2d内核密度估计？

我很抱歉这个可能很愚蠢的问题,但我现在正试着用几个小时来估算一组2d数据的密度.我们假设我的数据由数组给出:sample = np.random.uniform(0,1,size=(50,2)).我只是想使用scipys scikit learn package来估算样本数组的密度(这当然是2d均匀密度),我正在尝试以下方法:

import numpy as np
from sklearn.neighbors.kde import KernelDensity
from matplotlib import pyplot as plt
sp = 0.01

samples = np.random.uniform(0,1,size=(50,2))  # random samples
x = y = np.linspace(0,1,100)
X,Y = np.meshgrid(x,y)     # creating grid of data , to evaluate estimated density on

kde = KernelDensity(kernel='gaussian', bandwidth=0.2).fit(samples) # creating density from samples

kde.score_samples(X,Y) # I want to evaluate the estimated density on the X,Y grid

但最后一步总是产生错误: score_samples() takes 2 positional arguments but 3 were given

所以.score_samples可能不会将网格作为输入,但是没有针对2d案例的教程/文档,所以我不知道如何解决这个问题.如果有人可以提供帮助,那真的很棒.

查看物种分布的核密度估计示例,您必须将x,y数据打包在一起(训练数据和新的样本网格).

下面是一个简化sklearn API的函数.

from sklearn.neighbors import KernelDensity

def kde2D(x, y, bandwidth, xbins=100j, ybins=100j, **kwargs): 
    """Build 2D kernel density estimate (KDE)."""

    # create grid of sample locations (default: 100x100)
    xx, yy = np.mgrid[x.min():x.max():xbins, 
                      y.min():y.max():ybins]

    xy_sample = np.vstack([yy.ravel(), xx.ravel()]).T
    xy_train  = np.vstack([y, x]).T

    kde_skl = KernelDensity(bandwidth=bandwidth, **kwargs)
    kde_skl.fit(xy_train)

    # score_samples() returns the log-likelihood of the samples
    z = np.exp(kde_skl.score_samples(xy_sample))
    return xx, yy, np.reshape(z, xx.shape)

import numpy as np
import matplotlib.pyplot as plt

m1 = np.random.normal(size=1000)
m2 = np.random.normal(scale=0.5, size=1000)

x, y = m1 + m2, m1 - m2

xx, yy, zz = kde2D(x, y, 1.0)

plt.pcolormesh(xx, yy, zz)
plt.scatter(x, y, s=2, facecolor='white')