下面的脚本为给定列表生成概率矩阵：
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]
我现在想做相反的事情，并根据概率矩阵创建一个新的A B C D过渡列表。
我怎样才能做到这一点？
最佳答案
随机库的choices函数可能会有所帮助。由于该问题并不表示如何选择第一个字母，因此在这里选择它的可能性与原始列表的内容相同。
由于Python 3.6 random.choices接受带有权重的参数。严格地规范它们不是严格必要的。
import random
letter = random.choice(transitions) # take a starting letter with the same weights as the original list
new_list = [letter]
for _ in range(len(transitions) - 1):
letter = chr(random.choices(range(4), weights=M[rank(letter)])[0] + ord('A'))
new_list.append(letter)
print(new_list)
可以对整个代码进行某种程度的概括，以使其适用于任何类型的节点，而不仅仅是连续的字母：
from _collections import defaultdict
import random
transitions = ['A', 'B', 'B', 'C', 'B', 'A', 'D', 'D', 'A', 'B', 'A', 'D']
nodes = sorted(set(transitions)) # a list of all letters used
M = defaultdict(int) # dictionary counting the occurrences for each transition i,j)
for (i, j) in zip(transitions, transitions[1:]):
M[(i, j)] += 1
# dictionary with for each node a list of frequencies for the transition to a next node
T = {i: [M[(i, j)] for j in nodes] for i in nodes}
# node = random.choice(transitions) # chose the first node randomly with the same probability as the original list
node = random.choice(nodes) # chose the first node randomly, each node with equal probability
new_list = [node]
for _ in range(9):
node = random.choices(nodes, T[node])[0]
new_list.append(node)
print(new_list)
示例输出：
['D', 'A', 'D', 'A', 'D', 'D', 'A', 'D', 'A', 'B']关于python - 在给定转移概率矩阵的情况下，如何生成随机序列？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59468414/