Cela pourrait vous donner quelques idées:
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)
sortie:
[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]
Sur Editer Voici une fonction qui implémente les idées ci-dessus:
#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))
Sortie:
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
En Python brut, vous devez utiliser une liste de listes. Ce genre de chose est plus naturellement fait dans 'numpy' ou' pandas'. Si vous souhaitez utiliser l'un de ces outils, vous pouvez peut-être ajouter le tag approprié. Dans tous les cas, quelle est l'entrée de votre problème? Une liste finie d'états? –