Generating Markov transition matrix in Python

Question

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 1

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 

Answer 2

If you want to do it all in pandas, here is an approach that works for non numeric data:

import pandas as pd transitions = ['A', 'B', 'B', 'C', 'B', 'A', 'D', 'D', 'A', 'B', 'A', 'D']  df = pd.DataFrame(transitions)  # create a new column with data shifted one space df['shift'] = df[0].shift(-1)  # add a count column (for group by function) df['count'] = 1  # groupby and then unstack, fill the zeros trans_mat = df.groupby([0, 'shift']).count().unstack().fillna(0)  # normalise by occurences and save values to get transition matrix trans_mat = trans_mat.div(trans_mat.sum(axis=1), axis=0).values 

It's slower than the pure python approach but maybe worth it for flexibility and to avoid creating your own function.