Generating Markov transition matrix in Python

st19297 picture st19297 · Oct 10, 2017 · Viewed 25.7k times · Source

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, ....]

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?

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

Answer

John Coleman picture John Coleman · Oct 10, 2017

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)

output:

[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]

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))

Output:

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