2 回答

TA贡献1852条经验 获得超7个赞
假设你的数据是连续的,那么你可以做通常的移位:
df1['t'] = pd.to_timedelta(df1['t'].add(':00'))
df = pd.DataFrame({i:df1.A - df1.A.shift(i) >= 4 for i in range(1,5)})
df1['t'] - pd.to_timedelta('1min') * df.idxmax(axis=1).where(df.any(1))
输出:
0 NaT
1 NaT
2 NaT
3 NaT
4 23:03:00
5 23:01:00
6 NaT
7 23:06:00
8 NaT
9 23:06:00
10 23:06:00
dtype: timedelta64[ns]

TA贡献1826条经验 获得超6个赞
我添加了一个索引并使用 ,它现在包括了简单索引窗口之外的时间窗口功能。datetimerolling()
import pandas as pd
import numpy as np
import datetime
df1 = pd.DataFrame({'t' : [
datetime.datetime(2020, 5, 17, 23, 0, 0),
datetime.datetime(2020, 5, 17, 23, 0, 1),
datetime.datetime(2020, 5, 17, 23, 0, 2),
datetime.datetime(2020, 5, 17, 23, 0, 3),
datetime.datetime(2020, 5, 17, 23, 0, 4),
datetime.datetime(2020, 5, 17, 23, 0, 5),
datetime.datetime(2020, 5, 17, 23, 0, 6),
datetime.datetime(2020, 5, 17, 23, 0, 7),
datetime.datetime(2020, 5, 17, 23, 0, 8),
datetime.datetime(2020, 5, 17, 23, 0, 9),
datetime.datetime(2020, 5, 17, 23, 0, 10)
], 'A' : [2,1,2,2,6,5,4,9,7,10,8]}, columns=['t', 'A'])
df1.index = df1['t']
df2 = df1
cond = df1['A'] >= df1.rolling('5s')['A'].apply(lambda x: x[0] + 4)
result = df1[cond]
给
t A
2020-05-17 23:00:04 6
2020-05-17 23:00:05 5
2020-05-17 23:00:07 9
2020-05-17 23:00:09 10
2020-05-17 23:00:10 8
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