有没有办法在 python 或 R 中的字符串上应用转换矩阵？答案

【问题标题】：Is there any way to apply transition matrix on strings in either python or R?有没有办法在 python 或 R 中的字符串上应用转换矩阵？
【发布时间】：2019-05-18 23:37:28
【问题描述】：

我有以下几行：

johnsonsu(a=0.35, b=0.76, loc=973796.40, scale=134834.36)
johnsonsu(a=0.35, b=0.76, loc=973796.40, scale=134834.36)
gausshyper(a=1.50, b=0.67, c=2.50, z=3.68, loc=77873.97, scale=2249451.03)
gausshyper(a=1.50, b=0.67, c=2.50, z=3.68, loc=77873.97, scale=2249451.03)
gausshyper(a=1.50, b=0.67, c=2.50, z=3.68, loc=77873.97, scale=2249451.03)
johnsonsu(a=0.35, b=0.76, loc=973796.40, scale=134834.36)

它们是一些数据的分布和参数。我们想对它们应用一个转移矩阵来获得它们的概率。我们尝试了许多不同的代码，但由于数据类型不同，我们总是会出错。

我们在这些帖子中尝试了这些解决方案：

Generating Markov transition matrix in Python

Building a Transition Matrix using words in Python/Numpy

Calculate transition matrix of letters

迄今为止我们尝试过的最佳解决方案：

import pandas as pd
transitions #Larger instances than the ones above in the post
df = pd.DataFrame(columns = ['state', 'next_state'])
for i, val in enumerate(transitions[:-1]): # We don't care about last state
    df_stg = pd.DataFrame(index=[0])
    df_stg['state'], df_stg['next_state'] = transitions[i], transitions[i+1]
    df = pd.concat([df, df_stg], axis = 0)
cross_tab = pd.crosstab(df['state'], df['next_state'])
cross_tab.div(cross_tab.sum(axis=1), axis=0)

结果：

state   alpha(a=1.10, loc=-94626.86, scale=1135344.81)  dgamma(a=0.61, loc=820000.00, scale=1885232.33) dgamma(a=0.78, loc=780000.00, scale=349653.54)  dgamma(a=0.81, loc=761200.00, scale=404939.11)  dweibull(c=0.77, loc=730000.00, scale=356863.56)    dweibull(c=0.90, loc=700000.00, scale=375807.48)    foldcauchy(c=2.59, loc=1423.70, scale=313236.41)    gausshyper(a=1.50, b=0.67, c=2.50, z=3.68, loc=77873.97, scale=2249451.03)  gennorm(beta=0.12, loc=725000.01, scale=0.00)   gennorm(beta=0.19, loc=545200.00, scale=38.09)  gennorm(beta=0.33, loc=575900.00, scale=7595.02)    gennorm(beta=0.33, loc=580090.00, scale=9423.99)    gennorm(beta=0.34, loc=532822.50, scale=7547.83)    gennorm(beta=0.42, loc=750000.00, scale=22359.35)   gennorm(beta=0.47, loc=666600.00, scale=42042.13)   johnsonsu(a=-0.02, b=0.50, loc=770186.45, scale=32359.52)   johnsonsu(a=-0.49, b=0.40, loc=561967.63, scale=65812.06)   johnsonsu(a=0.31, b=0.47, loc=835025.10, scale=53272.01)    johnsonsu(a=0.35, b=0.76, loc=973796.40, scale=134834.36)   loglaplace(c=1.63, loc=-927.08, scale=640927.08)    loglaplace(c=2.42, loc=-1009.51, scale=773124.55)   pearson3(skew=2.13, loc=908886.62, scale=577310.56) t(df=0.08, loc=700000.00, scale=1.71)   vonmises_line(kappa=2.01, loc=741142.93, scale=449091.04)
alpha(a=1.10, loc=-94626.86, scale=1135344.81)  19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
dgamma(a=0.61, loc=820000.00, scale=1885232.33) 0   19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0
dgamma(a=0.78, loc=780000.00, scale=349653.54)  0   0   19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0
dgamma(a=0.81, loc=761200.00, scale=404939.11)  0   0   0   19  0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0
dweibull(c=0.77, loc=730000.00, scale=356863.56)    0   0   0   0   19  0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
dweibull(c=0.90, loc=700000.00, scale=375807.48)    0   0   0   0   0   19  0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0
foldcauchy(c=2.59, loc=1423.70, scale=313236.41)    0   0   0   0   1   0   19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
gausshyper(a=1.50, b=0.67, c=2.50, z=3.68, loc=77873.97, scale=2249451.03)  0   0   0   0   0   0   0   19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0
gennorm(beta=0.12, loc=725000.01, scale=0.00)   0   0   0   1   0   0   0   0   19  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
gennorm(beta=0.19, loc=545200.00, scale=38.09)  0   0   0   0   0   0   0   0   0   19  0   0   0   0   0   0   0   0   0   0   0   1   0   0
gennorm(beta=0.33, loc=575900.00, scale=7595.02)    0   0   0   0   0   1   0   0   0   0   19  0   0   0   0   0   0   0   0   0   0   0   0   0
gennorm(beta=0.33, loc=580090.00, scale=9423.99)    0   0   0   0   0   0   0   0   0   0   0   19  1   0   0   0   0   0   0   0   0   0   0   0
gennorm(beta=0.34, loc=532822.50, scale=7547.83)    0   0   0   0   0   0   0   0   0   0   0   0   19  0   0   0   1   0   0   0   0   0   0   0
gennorm(beta=0.42, loc=750000.00, scale=22359.35)   0   0   0   0   0   0   1   0   0   0   0   0   0   19  0   0   0   0   0   0   0   0   0   0
gennorm(beta=0.47, loc=666600.00, scale=42042.13)   0   0   0   0   0   0   0   0   0   0   0   1   0   0   19  0   0   0   0   0   0   0   0   0
johnsonsu(a=-0.02, b=0.50, loc=770186.45, scale=32359.52)   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   19  0   0   0   0   0   0   0   0
johnsonsu(a=-0.49, b=0.40, loc=561967.63, scale=65812.06)   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   19  0   0   0   0   0   0   0
johnsonsu(a=0.31, b=0.47, loc=835025.10, scale=53272.01)    0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   19  0   0   0   0   0   0
johnsonsu(a=0.35, b=0.76, loc=973796.40, scale=134834.36)   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   19  0   0   0   0   0
loglaplace(c=1.63, loc=-927.08, scale=640927.08)    0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   19  0   0   0   0
loglaplace(c=2.42, loc=-1009.51, scale=773124.55)   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   19  0   0   0
pearson3(skew=2.13, loc=908886.62, scale=577310.56) 0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   19  0   0
t(df=0.08, loc=700000.00, scale=1.71)   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   19  1
vonmises_line(kappa=2.01, loc=741142.93, scale=449091.04)   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0   0   19

概率是错误的。最后一个代码为转换矩阵中的大多数值输出 0。但是，如果索引和列彼此相似，则它们的值变为 19

【问题讨论】：

到目前为止你尝试了什么？
@TobiasWilfert 该帖子已在不同链接中更新了解决方案
问题不清楚。请提供您的导入、您的错误和更多详细信息。
@jasonm 帖子已更新
那么错误是什么？

标签： r python-3.x pandas numpy dataframe

【解决方案1】：

我已经解决了这个问题。我刚刚注意到输出给出了 19 和零，因为数据没有被打乱。因此，我打乱了数据，然后运行了代码。数据符合要求。

在此示例中，我将添加字符而不是分布和参数，以便让事情变得更简单。

transitions = ['A', 'B', 'B', 'C', 'A', 'A', 'A', 'Z']
from itertools import islice

def window(seq, n=2):
    "Sliding window width n from seq.  From old itertools recipes."""
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result
    for elem in it:
        result = result[1:] + (elem,)
        yield result

import pandas as pd

pairs = pd.DataFrame(window(transitions), columns=['state1', 'state2'])
counts = pairs.groupby('state1')['state2'].value_counts()
probs = (counts / counts.sum()).unstack()

DF_probs = pd.DataFrame(probs)
df = DF_probs.fillna(0)

结果：

state2         A         B         C         Z
state1                                        
A       0.285714  0.142857  0.000000  0.142857
B       0.000000  0.142857  0.142857  0.000000
C       0.142857  0.000000  0.000000  0.000000

参考：Brad Solomon [Solution]

【讨论】：