Python 如何在不同的子图中绘制对，侧面有差异

我想在seaborn拍一个情节，但我遇到了一些困难.数据有两个变量:时间(2级)和状态(2级).我想在x轴上绘制时间，并表示为不同的子图，显示各个数据线.最后，在这些的右边，我想展示一个时间2和时间1之间的差异图，对于每个状态级别.我不能做得很好，因为我不能让第二个情节显示在右边.以下是我的try :

import numpy as np
import pandas as pd
import seaborn as sns

# Just making some fake data
ids = [1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5]
times = [1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2,1,1,2,2]
states = ['A', 'B', 'A', 'B'] * 5

np.random.seed(121)
resps = [(i*t) + np.random.normal() for i, t in zip(ids, times)]

DATA = {
    'identity': ids,
    'time': times,
    'state': states,
    'resps': resps
}
df = pd.DataFrame(DATA)
# Done with data

g = sns.relplot(
    data=df, kind='line',
    col='state', x='time', y='resps', units='identity',
    estimator=None, alpha=.5, height=5, aspect=.7)

# # Draw a line onto each Axes
g.map(sns.lineplot,"time", "resps", lw=5, ci=None)

# Make a wide data to make the difference
wide = df.set_index(['identity', 'state', 'time']).unstack().reset_index()

A = wide['state']=='A'
B = wide['state']=='B'

wide['diffA'] = wide[A][('resps', 2)] - wide[A][('resps', 1)]
wide['diffB'] = wide[B][('resps', 2)] - wide[B][('resps', 1)]

wide['difference'] = wide[['diffA', 'diffB']].sum(axis=1)
wide = wide.drop(columns=[('diffA', ''), ('diffB', '')])

sns.pointplot(x='state', y='difference', data=wide, join=False)

Output from the first

And output from the second:

Is there no way to put them together? Even though they are different data? I did try to use matplotlib. And then achieved slightly better results but this still had a problem because I wanted the two left plots to have a shared y axis but not the difference. This created lots of work as well, because I want to be flexible for different numbers of the state variable, but only kept to 2 for simplicity. Here is a paint version of what I want to do (sorry for the poor quality), hopefully with some more control over appearance but this is secondary:

有没有可靠的方法以更简单的方式实现这一点？谢谢