如何消除温度对裂缝计传感器数据的影响？

背景与数据说明 #

• 持有裂缝计（用于监测地面裂缝宽度随时间变化）的时间索引Pandas DataFrame数据，以及邻近传感器采集的温度时间索引Pandas DataFrame数据
• 可视化结果显示二者存在显著相关性，温度波动会直接影响裂缝宽度测量值，其中2023年数据的相关性表现尤为明显
• 已提供2023年1月的样本数据，可按需补充其他月份数据

核心目标 #

消除温度波动带来的裂缝宽度干扰，提取并分析不受温度影响的裂缝宽度真实变化趋势

数据加载代码 #

import pandas as pd
import numpy as np

df_crack = pd.DataFrame({'date': ['2023-01-01 00:00:00', '2023-01-02 00:00:00', 
                          '2023-01-03 00:00:00', '2023-01-04 00:00:00',
                          '2023-01-05 00:00:00', '2023-01-06 00:00:00',
                          '2023-01-07 00:00:00', '2023-01-08 00:00:00',
                          '2023-01-09 00:00:00', '2023-01-10 00:00:00',
                          '2023-01-11 00:00:00', '2023-01-12 00:00:00',
                          '2023-01-13 00:00:00', '2023-01-14 00:00:00',
                          '2023-01-15 00:00:00', '2023-01-16 00:00:00',
                          '2023-01-17 00:00:00', '2023-01-18 00:00:00',
                          '2023-01-19 00:00:00', '2023-01-20 00:00:00',
                          '2023-01-21 00:00:00', '2023-01-22 00:00:00',
                          '2023-01-23 00:00:00', '2023-01-24 00:00:00',
                          '2023-01-25 00:00:00', '2023-01-26 00:00:00',
                          '2023-01-27 00:00:00', '2023-01-28 00:00:00',
                          '2023-01-29 00:00:00', '2023-01-30 00:00:00',
                          ], 
               'aperture': [0.452762281,0.372262281,0.513928948,0.447762281,
                            0.377095615,0.355095615,0.271428948,0.291762281,
                            0.476762281,0.335928948,0.280428948,0.283762281,
                            0.322928948,0.287262281,0.316928948,0.209262281,
                            0.407928948,0.254262281,0.232095615,0.264262281,
                            0.076095615,-0.025237719,-0.042237719,-0.094904385,
                            0.017428948,-0.036071052,-0.094071052,-0.071404385,
                            0.008095615,-0.141571052]})

df_crack['date'] = pd.to_datetime(df_crack['date'])
df_crack = df_crack.set_index('date')

df_temp = pd.DataFrame({'date': ['2023-01-01 00:00:00', '2023-01-02 00:00:00', 
                          '2023-01-03 00:00:00', '2023-01-04 00:00:00',
                          '2023-01-05 00:00:00', '2023-01-06 00:00:00',
                          '2023-01-07 00:00:00', '2023-01-08 00:00:00',
                          '2023-01-09 00:00:00', '2023-01-10 00:00:00',
                          '2023-01-11 00:00:00', '2023-01-12 00:00:00',
                          '2023-01-13 00:00:00', '2023-01-14 00:00:00',
                          '2023-01-15 00:00:00', '2023-01-16 00:00:00',
                          '2023-01-17 00:00:00', '2023-01-18 00:00:00',
                          '2023-01-19 00:00:00', '2023-01-20 00:00:00',
                          '2023-01-21 00:00:00', '2023-01-22 00:00:00',
                          '2023-01-23 00:00:00', '2023-01-24 00:00:00',
                          '2023-01-25 00:00:00', '2023-01-26 00:00:00',
                          '2023-01-27 00:00:00', '2023-01-28 00:00:00',
                          '2023-01-29 00:00:00', '2023-01-30 00:00:00',
                          ], 
               'temperature': [9.6,8,8.4,6.2,6.2,6,3.9,8.5,8.3,5.3,5.6,5.3,
                               6.2,6.3,6.9,4.8,6.7,3.6,3,4.6,2.3,1.3,1,0.3,
                               1.6,0.4,1.5,1.4,2.2,1.2]})

df_temp['date'] = pd.to_datetime(df_temp['date'])
df_temp = df_temp.set_index('date')

内容的提问来源于stack exchange，提问作者Ferran