Plotly绘制折线图时实现线条交叉处自动更改颜色

现有包含日期列、S1与S2两个数值数据序列的pandas DataFrame，需使用Plotly绘制折线图，实现两条数据序列交叉时，自动更改对应位置的线条颜色。

最初的实现逻辑为逐行判断S1与S2的大小关系生成颜色列表，传入go.Scatter的line.color参数，代码如下：

import plotly.graph_objects as go
from plotly.subplots import make_subplots

data = {'Date': ['1.1.2020', '2.1.2020', '3.1.2020', '4.1.2020', '5.1.2020'],
    'S1': [20, 21, 19, 18, 20],
    'S2': [10, 11, 30, 18, 10]}
df = pd.DataFrame(data)

df['color']= df.apply(lambda row: 'black' if row['S1'] >= row['S2'] else 'red', axis=1)

fig = make_subplots(rows=1, cols=1, 
                vertical_spacing=0.05,
                shared_xaxes=True)

fig.add_trace(go.Scatter(
     x=df['Date'],
     y=df['S1'],
     line=dict(color='blue')),
     row=1, col=1)

fig.add_trace(go.Scatter(
     x=df['Date'],
     y=df['S2'],
     line=dict(color=list(df['color']))),
     row=1, col=1)

fig.show()

该代码运行后未达到预期效果，无法在交叉处正确切换线条颜色。后续将测试数据集扩大到10条记录验证其他方案时，又出现交叉点位之间存在多余红色连线的问题，测试数据集如下：

data = {'Date': ['1.1.2020', '2.1.2020', '3.1.2020', '4.1.2020', '5.1.2020', '6.1.2020', '7.1.2020', '8.1.2020', '9.1.2020', '10.1.2020'],
'S1': [20, 21, 19, 18, 20, 19, 18, 20, 20, 20],
'S2': [10, 11, 30, 18, 10, 10, 17, 30, 20, 10]}

采用分层绘制轨迹的思路实现效果：

• 第一层绘制固定为蓝色的S1完整折线
• 第二层绘制固定为黑色的S2基础折线
• 第三层提取S1小于S2的交叉区间及前后相邻点位的S2数值，绘制红色线段覆盖对应交叉段

完整可运行代码如下：

import plotly.graph_objects as go
from plotly.subplots import make_subplots
import pandas as pd

data = {'Date': ['1.1.2020', '2.1.2020', '3.1.2020', '4.1.2020', 
'5.1.2020', '6.1.2020', '7.1.2020', '8.1.2020', '9.1.2020', 
'10.1.2020'],
'S1': [20, 21, 19, 18, 20, 19, 18, 20, 20, 20],
'S2': [10, 11, 30, 18, 10, 10, 17, 30, 20, 10]}
df = pd.DataFrame(data)

df['crossing']= df.loc[(df.S1 < df.S2) | (df.S1.shift(1) < 
df.S2.shift(1)) | (df.S1.shift(-1) < df.S2.shift(-1)), 'S2']

fig = go.Figure()
fig.add_trace(go.Scatter(
  x=df['Date'],
  y=df['S1'],
  line=dict(color='blue')))

fig.add_trace(go.Scatter(
  x=df['Date'],
  y=df['S2'],
  line=dict(color='black')))

fig.add_trace(go.Scatter(
  x=df['Date'],
  y=df['crossing'],
  line=dict(color='red')))

fig.show()

内容的提问来源于stack exchange，提问作者khrizz