我有以下df1
数据帧:
t A
0 23:00 2
1 23:01 1
2 23:02 2
3 23:03 2
4 23:04 6
5 23:05 5
6 23:06 4
7 23:07 9
8 23:08 7
9 23:09 10
10 23:10 8
对于每个t
(此处简化的增量,在现实生活中不均匀分布(,我想找到(如果有的话(前 5 分钟内tr
最近的时间,其中A(t)- A(tr) >= 4
。我想得到:
t A tr
0 23:00 2
1 23:01 1
2 23:02 2
3 23:03 2
4 23:04 6 23:03
5 23:05 5 23:01
6 23:06 4
7 23:07 9 23:06
8 23:08 7
9 23:09 10 23:06
10 23:10 8 23:06
目前,我可以使用shift(-1)
将每一行与上一行进行比较,例如cond = df1['A'] >= df1['A'].shift(-1) + 4
.
我怎样才能看得更远?
假设您的数据每分钟都是连续的,那么您可以执行通常的偏移:
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]
我添加了一个datetime
索引并使用了rolling()
,它现在包括了简单索引窗口之外的时间窗口功能。
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