我有许多元组（par1，par2），即通过多次重复实验获得的二维参数空间中的点。
我正在寻找一种计算和可视化置信椭圆的可能性（不确定这是否是正确的术语）。这是我在网上找到的一个示例图来说明我的意思：
来源：blogspot.ch/2011/07/classification-and-discrimination-with.html
因此，原则上，我猜必须将多元正态分布拟合到数据点的二维直方图。有人可以帮我吗？
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4 回答
37
听起来你只想要点散点的 2-sigma 椭圆？
如果是这样，请考虑这样的事情（来自这里的论文的一些代码：https ://github.com/joferkington/oost_paper_code/blob/master/error_ellipse.py ）：
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse
def plot_point_cov(points, nstd=2, ax=None, **kwargs):
"""
Plots an `nstd` sigma ellipse based on the mean and covariance of a point
"cloud" (points, an Nx2 array).
Parameters
----------
points : An Nx2 array of the data points.
nstd : The radius of the ellipse in numbers of standard deviations.
Defaults to 2 standard deviations.
ax : The axis that the ellipse will be plotted on. Defaults to the
current axis.
Additional keyword arguments are pass on to the ellipse patch.
Returns
-------
A matplotlib ellipse artist
"""
pos = points.mean(axis=0)
cov = np.cov(points, rowvar=False)
return plot_cov_ellipse(cov, pos, nstd, ax, **kwargs)
def plot_cov_ellipse(cov, pos, nstd=2, ax=None, **kwargs):
"""
Plots an `nstd` sigma error ellipse based on the specified covariance
matrix (`cov`). Additional keyword arguments are passed on to the
ellipse patch artist.
Parameters
----------
cov : The 2x2 covariance matrix to base the ellipse on
pos : The location of the center of the ellipse. Expects a 2-element
sequence of [x0, y0].
nstd : The radius of the ellipse in numbers of standard deviations.
Defaults to 2 standard deviations.
ax : The axis that the ellipse will be plotted on. Defaults to the
current axis.
Additional keyword arguments are pass on to the ellipse patch.
Returns
-------
A matplotlib ellipse artist
"""
def eigsorted(cov):
vals, vecs = np.linalg.eigh(cov)
order = vals.argsort()[::-1]
return vals[order], vecs[:,order]
if ax is None:
ax = plt.gca()
vals, vecs = eigsorted(cov)
theta = np.degrees(np.arctan2(*vecs[:,0][::-1]))
# Width and height are "full" widths, not radius
width, height = 2 * nstd * np.sqrt(vals)
ellip = Ellipse(xy=pos, width=width, height=height, angle=theta, **kwargs)
ax.add_artist(ellip)
return ellip
if __name__ == '__main__':
#-- Example usage -----------------------
# Generate some random, correlated data
points = np.random.multivariate_normal(
mean=(1,1), cov=[[0.4, 9],[9, 10]], size=1000
)
# Plot the raw points...
x, y = points.T
plt.plot(x, y, 'ro')
# Plot a transparent 3 standard deviation covariance ellipse
plot_point_cov(points, nstd=3, alpha=0.5, color='green')
plt.show()
于 2012-09-07T15:39:26.937 回答
6
请参阅帖子如何绘制协方差误差椭圆。
这是python的实现：
import numpy as np
from scipy.stats import norm, chi2
def cov_ellipse(cov, q=None, nsig=None, **kwargs):
"""
Parameters
----------
cov : (2, 2) array
Covariance matrix.
q : float, optional
Confidence level, should be in (0, 1)
nsig : int, optional
Confidence level in unit of standard deviations.
E.g. 1 stands for 68.3% and 2 stands for 95.4%.
Returns
-------
width, height, rotation :
The lengths of two axises and the rotation angle in degree
for the ellipse.
"""
if q is not None:
q = np.asarray(q)
elif nsig is not None:
q = 2 * norm.cdf(nsig) - 1
else:
raise ValueError('One of `q` and `nsig` should be specified.')
r2 = chi2.ppf(q, 2)
val, vec = np.linalg.eigh(cov)
width, height = 2 * sqrt(val[:, None] * r2)
rotation = np.degrees(arctan2(*vec[::-1, 0]))
return width, height, rotation
Joe Kington 的回答中标准差的含义是错误的。通常我们使用 1, 2 sigma 来表示 68%, 95% 的置信度，但他的答案中的 2 sigma 椭圆不包含总分布的 95% 概率。正确的方法是使用卡方分布来估计椭圆大小，如帖子所示。
于 2016-09-28T13:40:59.323 回答
4
我稍微修改了上面绘制错误或置信区域轮廓的示例之一。现在我认为它给出了正确的轮廓。
它给出了错误的轮廓，因为它应该将 scoreatpercentile 方法应用于联合数据集（蓝色 + 红色点），而它应该单独应用于每个数据集。
修改后的代码如下：
import numpy
import scipy
import scipy.stats
import matplotlib.pyplot as plt
# generate two normally distributed 2d arrays
x1=numpy.random.multivariate_normal((100,420),[[120,80],[80,80]],400)
x2=numpy.random.multivariate_normal((140,340),[[90,-70],[-70,80]],400)
# fit a KDE to the data
pdf1=scipy.stats.kde.gaussian_kde(x1.T)
pdf2=scipy.stats.kde.gaussian_kde(x2.T)
# create a grid over which we can evaluate pdf
q,w=numpy.meshgrid(range(50,200,10), range(300,500,10))
r1=pdf1([q.flatten(),w.flatten()])
r2=pdf2([q.flatten(),w.flatten()])
# sample the pdf and find the value at the 95th percentile
s1=scipy.stats.scoreatpercentile(pdf1(pdf1.resample(1000)), 5)
s2=scipy.stats.scoreatpercentile(pdf2(pdf2.resample(1000)), 5)
# reshape back to 2d
r1.shape=(20,15)
r2.shape=(20,15)
# plot the contour at the 95th percentile
plt.contour(range(50,200,10), range(300,500,10), r1, [s1],colors='b')
plt.contour(range(50,200,10), range(300,500,10), r2, [s2],colors='r')
# scatter plot the two normal distributions
plt.scatter(x1[:,0],x1[:,1],alpha=0.3)
plt.scatter(x2[:,0],x2[:,1],c='r',alpha=0.3)
于 2013-07-01T16:29:17.903 回答
0
我想您正在寻找的是计算Confidence Regions。
我不太了解它，但作为一个起点，我会检查python的sherpa应用程序。至少，在他们的 Scipy 2011 演讲中，作者提到您可以使用它来确定和获得置信区域（尽管您可能需要为您的数据建立一个模型）。
高温高压
于 2012-09-06T23:25:14.233 回答