支持 python 的双变量分布

Question

我正在生成下面的多元概率密度函数。这工作正常，但我希望标准化 Z 值，以便它得出 0 到 1 之间的值。

为了实现这一点，我想将分布值除以均值，因此它总是在均值处为 1 而在其他地方较低。 我知道所有值的总和将大于 1。

我正在潜水 Z 但是 Z 的 sum 但是当打印这些值时，它们仍然在我预期的标准化范围之外。

import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D

# Our 2-dimensional distribution will be over variables X and Y
N = 60
X = np.linspace(-3, 3, N)
Y = np.linspace(-3, 4, N)
X, Y = np.meshgrid(X, Y)

# Mean vector and covariance matrix
mu = np.array([0., 1.])
Sigma = np.array([[ 1. , -0.5], [-0.5,  1.5]])

# Pack X and Y into a single 3-dimensional array
pos = np.empty(X.shape + (2,))
pos[:, :, 0] = X
pos[:, :, 1] = Y

def multivariate_gaussian(pos, mu, Sigma):
    """Return the multivariate Gaussian distribution on array pos.

    pos is an array constructed by packing the meshed arrays of variables
    x_1, x_2, x_3, ..., x_k into its _last_ dimension.

    """

    n = mu.shape[0]
    Sigma_det = np.linalg.det(Sigma)
    Sigma_inv = np.linalg.inv(Sigma)
    N = np.sqrt((2*np.pi)**n * Sigma_det)
    # This einsum call calculates (x-mu)T.Sigma-1.(x-mu) in a vectorized
    # way across all the input variables.
    fac = np.einsum('...k,kl,...l->...', pos-mu, Sigma_inv, pos-mu)

    return np.exp(-fac / 2) / N

# The distribution on the variables X, Y packed into pos.
Z = multivariate_gaussian(pos, mu, Sigma)

#normalise Z so range is 0-1
Z = Z/sum(Z)
print(Z)

# Create a surface plot and projected filled contour plot under it.
fig = plt.figure()
ax = fig.gca(projection='3d')
ax.plot_surface(X, Y, Z, rstride=3, cstride=3, linewidth=1, antialiased=True,
            cmap=cm.magma)

cset = ax.contourf(X, Y, Z, zdir='z', offset=-0.15, cmap=cm.magma)

# Adjust the limits, ticks and view angle
ax.set_zlim(-0.15,1)
ax.view_init(27, -21)

plt.show()

Answer 1

如果要标准化 Z，您需要将其除以总和，而是除以所有值的最大值。因此，您确保新的最大值为 1:

# normalise Z so range is 0-1
Z = Z / np.max(Z)

# show summary statistics for the result
import pandas as pd
print(pd.Series(Z.flatten()).describe())

count    3.600000e+03
mean     1.605148e-01
std      2.351826e-01
min      6.184228e-08
25%      7.278911e-03
50%      4.492385e-02
75%      2.135538e-01
max      1.000000e+00
dtype: float64

由于您有一个高斯分布并且您只更改了缩放比例，因此最大值仍将处于平均值 x 和 y 值。但是请注意，Z 现在不再是概率密度函数。

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