马尔可夫链平稳分布的Python代码解释

我有这个密码：

import numpy as np
from scipy.linalg import eig 
transition_mat = np.matrix([
    [.95, .05, 0., 0.],\
    [0., 0.9, 0.09, 0.01],\
    [0., 0.05, 0.9, 0.05],\
    [0.8, 0., 0.05, 0.15]])

S, U = eig(transition_mat.T)
stationary = np.array(U[:, np.where(np.abs(S - 1.) < 1e-8)[0][0]].flat)
stationary = stationary / np.sum(stationary)

>>> print stationary
[ 0.34782609  0.32608696  0.30434783  0.02173913]

我得到了输出：

:  [ 0.6144763   0.57607153  0.53766676  0.03840477]

但它并没有给出相同的输出。为什么

stational=np.数组（U[：，np.其中（np.abs（S-1.）<1e-8）[0][0]]平坦）
stationary = np.array(U[:,np.where(np.abs(S-1.) < 1e-8)[0][0]].flat)

这段代码正在搜索U中的元素，其对应的特征值-1小于1e-8

stational=np.array（U[：，np.where（np.abs（s-1.）<1e-8）[0][0]]

这段代码正在搜索U中的元素，该元素对应的特征值-1小于1e-8

如何找到平稳分布。 好的，我来这篇文章是想看看是否有一个内置的方法来找到平稳分布。看起来没有。因此，对于来自谷歌的任何人来说，在这种情况下，我会这样发现固定分布：

import numpy as np

#note: the matrix is row stochastic.
#A markov chain transition will correspond to left multiplying by a row vector.
Q = np.array([
    [.95, .05, 0., 0.],
    [0., 0.9, 0.09, 0.01],
    [0., 0.05, 0.9, 0.05],
    [0.8, 0., 0.05, 0.15]])

#We have to transpose so that Markov transitions correspond to right multiplying by a column vector.  np.linalg.eig finds right eigenvectors.
evals, evecs = np.linalg.eig(Q.T)
evec1 = evecs[:,np.isclose(evals, 1)]

#Since np.isclose will return an array, we've indexed with an array
#so we still have our 2nd axis.  Get rid of it, since it's only size 1.
evec1 = evec1[:,0]

stationary = evec1 / evec1.sum()

#eigs finds complex eigenvalues and eigenvectors, so you'll want the real part.
stationary = stationary.real

那条奇怪的线在做什么。 让我们把这条线分成几部分：

#Find the eigenvalues that are really close to 1.
eval_close_to_1 = np.abs(S-1.) < 1e-8

#Find the indices of the eigenvalues that are close to 1.
indices = np.where(eval_close_to_1)

#np.where acts weirdly.  In this case it returns a 1-tuple with an array of size 1 in it.
the_array = indices[0]
index = the_array[0]

#Now we have the index of the eigenvector with eigenvalue 1.
stationary = U[:, index]

#For some really weird reason, the person that wrote the code
#also does this step, which is completely redundant.
#It just flattens the array, but the array is already 1-d.
stationary = np.array(stationary.flat)

然后，您的平稳分布将匹配。

如何找到平稳分布。 好的，我来这篇文章是想看看是否有一个内置的方法来找到平稳分布。看起来没有。因此，对于来自谷歌的任何人来说，在这种情况下，我会这样发现固定分布：

import numpy as np

#note: the matrix is row stochastic.
#A markov chain transition will correspond to left multiplying by a row vector.
Q = np.array([
    [.95, .05, 0., 0.],
    [0., 0.9, 0.09, 0.01],
    [0., 0.05, 0.9, 0.05],
    [0.8, 0., 0.05, 0.15]])

#We have to transpose so that Markov transitions correspond to right multiplying by a column vector.  np.linalg.eig finds right eigenvectors.
evals, evecs = np.linalg.eig(Q.T)
evec1 = evecs[:,np.isclose(evals, 1)]

#Since np.isclose will return an array, we've indexed with an array
#so we still have our 2nd axis.  Get rid of it, since it's only size 1.
evec1 = evec1[:,0]

stationary = evec1 / evec1.sum()

#eigs finds complex eigenvalues and eigenvectors, so you'll want the real part.
stationary = stationary.real

那条奇怪的线在做什么。 让我们把这条线分成几部分：

#Find the eigenvalues that are really close to 1.
eval_close_to_1 = np.abs(S-1.) < 1e-8

#Find the indices of the eigenvalues that are close to 1.
indices = np.where(eval_close_to_1)

#np.where acts weirdly.  In this case it returns a 1-tuple with an array of size 1 in it.
the_array = indices[0]
index = the_array[0]

#Now we have the index of the eigenvector with eigenvalue 1.
stationary = U[:, index]

#For some really weird reason, the person that wrote the code
#also does this step, which is completely redundant.
#It just flattens the array, but the array is already 1-d.
stationary = np.array(stationary.flat)

---

然后，您的固定分布将匹配。

Hm。。。它看起来像是从U中抓取元素，其中s中的对应元素的值为1。。。但是我不确定那是做什么用的。然后，它会将整个过程展平为一维数组。嗨，凯文，谢谢你的回答。你能告诉我这个代码：where（np.abs（S-1.）<1e-8）应该如何工作吗？你在一个简单的例子中试过吗？阅读任何文档？到底是什么让你困惑？@Jornsharpe，是的，我还在努力。我已经用我的代码更新了这个问题。请查收。谢谢。这里的页边空白太小，不能全部写下来。只要仔细阅读

numpy.where

的文档，并尝试一下示例！；）陛下它看起来像是从U中抓取元素，其中s中的对应元素的值为1。。。但是我不确定那是做什么用的。然后，它会将整个过程展平为一维数组。嗨，凯文，谢谢你的回答。你能告诉我这个代码：where（np.abs（S-1.）<1e-8）应该如何工作吗？你在一个简单的例子中试过吗？阅读任何文档？到底是什么让你困惑？@Jornsharpe，是的，我还在努力。我已经用我的代码更新了这个问题。请查收。谢谢。这里的页边空白太小，不能全部写下来。只要仔细阅读

numpy.where

的文档，并尝试一下示例！；）

stationary = matrix/matrix.sum()