Pandas 中数据集的导数

Question

我有一个循环数据集（图片和下面的数据），我正在尝试获取 dy/dt 的导数。我想要每个周期（上升）的导数，而不仅仅是像我在这里所做的每个点。部分问题是样本的日期时间间隔不均匀。理想情况下，我希望每个周期的斜率从最小值到最大值。这是一个理想化的情况，我的整个数据集都会有噪声，并且每个周期的斜率不一定与这个建模集相同。

数据看起来是这样的，

time,y
12/15/18 01:10 AM,130352.146180556
12/16/18 01:45 AM,130355.219097222
12/17/18 01:47 AM,130358.223263889
12/18/18 02:15 AM,130361.281701389
12/19/18 03:15 AM,130364.406597222
12/20/18 03:25 AM,130352.427430556
12/21/18 03:27 AM,130355.431597222
12/22/18 05:18 AM,130358.663541667
12/23/18 06:44 AM,130361.842430556
12/24/18 07:19 AM,130364.915243056
12/25/18 07:33 AM,130352.944409722
12/26/18 07:50 AM,130355.979826389
12/27/18 09:13 AM,130359.153472222
12/28/18 11:53 AM,130362.4871875
12/29/18 01:23 PM,130365.673263889
12/30/18 02:17 PM,130353.785763889
12/31/18 02:23 PM,130356.798263889
01/01/19 04:41 PM,130360.085763889
01/02/19 05:01 PM,130363.128125

这是我的代码，

import pandas as pd
import numpy as np
import matplotlib.pyplot as plot
from datetime import date, timedelta
import datetime

df=pd.read_csv('saw_data.csv') 
df['time']=pd.to_datetime(df['time'])

将日期时间设置为索引，

df.set_index(df['time'], inplace=True) 

在这里我试图找到每个数据点之间的时间间隔。

df['Time_diff'] = pd.to_timedelta(df['time']-df['time'].shift()).dt.total_seconds().div(60)
#I believe units are minutes. For 'y' row 0 to 1, the diff is ~3 in about a day (86400 sec)
# so 3/86400 x 60 sec/min yields similar result to slope of 0.002 #/min.

'Time_diff' 来自 SO 帖子： calculate the time difference between two consecutive rows in pandas

这应该是导数，

df['slope']=np.gradient(df['y'],1)/df['Time_diff']

这是结果，

print(df.head())

                                   time              y  Time_diff     slope
time                                                                       
2018-12-15 01:10:00 2018-12-15 01:10:00  130352.146181        NaN       NaN
2018-12-16 01:45:00 2018-12-16 01:45:00  130355.219097     1475.0  0.002060
2018-12-17 01:47:00 2018-12-17 01:47:00  130358.223264     1442.0  0.002102
2018-12-18 02:15:00 2018-12-18 02:15:00  130361.281701     1468.0  0.002106
2018-12-19 03:15:00 2018-12-19 03:15:00  130364.406597     1500.0 -0.002951

这是数据集图片，

Answer 1

好的，下面的代码假设您只是使用经典的“尺子和笔”方法来获取渐变（表示为“m”） - 即 y 距离 / x 距离，从每个点之间的坐标差中获得.

下面的代码会给你正梯度（我假设这是你想要的）。如果你想要负梯度（即向相反方向倾斜的线条），你需要稍微调整 shift() 方法。

import pandas as pd
import numpy as np
import matplotlib.pyplot as plot
from datetime import date, timedelta
import datetime, time

df=pd.read_csv('saw_data.csv')
df['time']=pd.to_datetime(df['time'])
df.set_index(df['time'], inplace=True)

df['Time_diff'] = pd.to_timedelta(df['time']-df['time'].shift()).dt.total_seconds().div(60)
df['slope']=np.gradient(df['y'],1)/df['Time_diff']
#df['y_diff'] = pd.to_timedelta(df['y']-df['y'].shift()).dt.total_seconds().div(60)

df['time+1']=df['time'].shift()
df['y+1']=df['y'].shift()
df['y_diff'] = df['y'] - df['y+1']
df['m']= df['y_diff']/df['Time_diff']