python - 计算多元正态分布python的均值向量

我在将多元高斯分布拟合到我的数据集时遇到了麻烦，更具体地说，是找到均值向量（或多个均值向量）。我的数据集是一个N x 8矩阵，目前我正在使用此代码：

muVector = np.mean(Xtrain, axis=0)其中Xtrain是我的训练数据集。

对于协方差，我正在使用任意方差值（.5）构建它，并执行以下操作：

covariance = np.dot(.5, np.eye(N,N)其中N是观察数。

但是，当我构造我的Phi矩阵时，我得到的都是零。这是我的代码：

muVector = np.mean(Xtrain, axis=0)
# get covariance matrix from Xtrain
cov = np.dot(var, np.eye(N,N))
cov = np.linalg.inv(cov)

# build Xtrain Phi
Phi = np.ones((N,M))
for row in range(N):
  temp = Xtrain[row,:] - muVector
  temp.shape = (1,M)
  temp = np.dot((-.5), temp)
  temp = np.dot(temp, cov)
  temp = np.dot(temp, (Xtrain[row,:] - muVector))
  Phi[row,:] = np.exp(temp)

任何帮助表示赞赏。我想我可能必须使用np.random.multivariate_normal（）？但是我不知道在这种情况下如何使用它。

最佳答案

我认为“ Phi”是指您要估计的概率密度函数（pdf）。在这种情况下，协方差矩阵应为MxM，输出Phi将为Nx1：

# -*- coding: utf-8 -*-

import numpy as np

N = 1024
M = 8
var = 0.5

# Creating a Xtrain NxM observation matrix.
# Its muVector is [0, 1, 2, 3, 4, 5, 6, 7] and the variance for all
# independent random variables is 0.5.
Xtrain = np.random.multivariate_normal(np.arange(8), np.eye(8,8)*var, N)

# Estimating the mean vector.
muVector = np.mean(Xtrain, axis=0)

# Creating the estimated covariance matrix and its inverse.
cov = np.eye(M,M)*var
inv_cov = np.linalg.inv(cov)

# Normalization factor from the pdf.
norm_factor = 1/np.sqrt((2*np.pi)**M * np.linalg.det(cov))

# Estimating the pdf.
Phi = np.ones((N,1))
for row in range(N):
    temp = Xtrain[row,:] - muVector
    temp.shape = (1,M)
    temp = np.dot(-0.5*temp, inv_cov)
    temp = np.dot(temp, (Xtrain[row,:] - muVector))
    Phi[row] = norm_factor*np.exp(temp)

另外，您可以使用pdf中的scipy.stats.multivariate_normal方法：

# -*- coding: utf-8 -*-

import numpy as np
from scipy.stats import multivariate_normal

N = 1024
M = 8
var = 0.5

# Creating a Xtrain NxM observation matrix.
# Its muVector is [0, 1, 2, 3, 4, 5, 6, 7] and the variance for all
# independent random variables is 0.5.
Xtrain = np.random.multivariate_normal(np.arange(8), np.eye(8,8)*var, N)

# Estimating the mean vector.
muVector = np.mean(Xtrain, axis=0)

# Creating the estimated covariance matrix.
cov = np.eye(M,M)*var

Phi2 = multivariate_normal.pdf(Xtrain, mean=muVector, cov=cov)

Phi和Phi2输出数组将相等。

关于python - 计算多元正态分布python的均值向量，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/28402167/