在Python中生成马尔可夫转移矩阵 [英] Generating Markov transition matrix in Python
问题描述
想象一下,我有一系列四个可能的马尔可夫状态(A,B,C,D):
Imagine I have a series of 4 possible Markovian states (A, B, C, D):
X = [A, B, B, C, B, A, D, D, A, B, A, D, ....]
如何使用Python生成Markov变换矩阵?矩阵必须为4乘4,显示从每个状态迁移到其他3个状态的概率. 我一直在网上查看许多示例,但在所有示例中,都是给出矩阵,而不是根据数据计算得出的. 我也研究了hmmlearn,但是在任何地方都没有读到如何使它吐出过渡矩阵.我可以为此目的使用一个库吗?
How can I generate a Markov transformation matrix using Python? The matrix must be 4 by 4, showing the probability of moving from each state to the other 3 states. I've been looking at many examples online but in all of them, the matrix is given, not calculated based on data. I also looked into hmmlearn but nowhere I read on how to have it spit out the transition matrix. Is there a library that I can use for this purpose?
以下是我要在Python中尝试执行的确切操作的R代码: https://stats.stackexchange.com/questions/26722/calculate-transition- matrix-markov-in-r
Here is an R code for the exact thing I am trying to do in Python: https://stats.stackexchange.com/questions/26722/calculate-transition-matrix-markov-in-r
推荐答案
这可能会给您一些想法:
This might give you some ideas:
transitions = ['A', 'B', 'B', 'C', 'B', 'A', 'D', 'D', 'A', 'B', 'A', 'D']
def rank(c):
return ord(c) - ord('A')
T = [rank(c) for c in transitions]
#create matrix of zeros
M = [[0]*4 for _ in range(4)]
for (i,j) in zip(T,T[1:]):
M[i][j] += 1
#now convert to probabilities:
for row in M:
n = sum(row)
if n > 0:
row[:] = [f/sum(row) for f in row]
#print M:
for row in M:
print(row)
输出:
[0.0, 0.5, 0.0, 0.5]
[0.5, 0.25, 0.25, 0.0]
[0.0, 1.0, 0.0, 0.0]
[0.5, 0.0, 0.0, 0.5]
编辑时:以下功能实现了上述想法:
On Edit Here is a function which implements the above ideas:
#the following code takes a list such as
#[1,1,2,6,8,5,5,7,8,8,1,1,4,5,5,0,0,0,1,1,4,4,5,1,3,3,4,5,4,1,1]
#with states labeled as successive integers starting with 0
#and returns a transition matrix, M,
#where M[i][j] is the probability of transitioning from i to j
def transition_matrix(transitions):
n = 1+ max(transitions) #number of states
M = [[0]*n for _ in range(n)]
for (i,j) in zip(transitions,transitions[1:]):
M[i][j] += 1
#now convert to probabilities:
for row in M:
s = sum(row)
if s > 0:
row[:] = [f/s for f in row]
return M
#test:
t = [1,1,2,6,8,5,5,7,8,8,1,1,4,5,5,0,0,0,1,1,4,4,5,1,3,3,4,5,4,1,1]
m = transition_matrix(t)
for row in m: print(' '.join('{0:.2f}'.format(x) for x in row))
输出:
0.67 0.33 0.00 0.00 0.00 0.00 0.00 0.00 0.00
0.00 0.50 0.12 0.12 0.25 0.00 0.00 0.00 0.00
0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00
0.00 0.00 0.00 0.50 0.50 0.00 0.00 0.00 0.00
0.00 0.20 0.00 0.00 0.20 0.60 0.00 0.00 0.00
0.17 0.17 0.00 0.00 0.17 0.33 0.00 0.17 0.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00
0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00
0.00 0.33 0.00 0.00 0.00 0.33 0.00 0.00 0.33
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