Bog*_*mir 3 python datetime dataframe pandas
我有这个数据框 df
id date time record
0 1 2021-07-08 3:00:00 8
1 1 2021-07-08 5:30:00 7
2 1 2021-07-08 9:00:00 10
3 1 2021-01-08 6:30:00 5
4 1 2021-01-08 9:30:00 7
5 2 2021-07-08 3:00:00 7
6 2 2021-07-08 9:00:00 14
7 2 2021-07-08 5:30:00 10
8 2 2021-01-08 3:00:00 11
9 2 2021-01-08 3:00:00 13
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我需要创建一个max等于最大分组的新列id和date。但我也只需要time在早于 7:00:00 时获得最大值。即 forid=1和date=2021-07-08 maxcolumns 应该等于 8,而不是 10,因为我们在time9:00:00时得到了 10 。
这df是更易于访问的数据框
import io
import pandas as pd
data1_txt = """
id,date,time,record
1,2021-07-08,3:00:00,8
1,2021-07-08,5:30:00,7
1,2021-07-08,9:00:00,10
1,2021-01-08,6:30:00,5
1,2021-01-08,9:30:00,7
2,2021-07-08,3:00:00,7
2,2021-07-08,9:00:00,14
2,2021-07-08,5:30:00,10
2,2021-01-08,3:00:00,11
2,2021-01-08,3:00:00,13
"""
df = pd.read_csv(io.StringIO(data1_txt))
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想要的结果是
id date time record max
0 1 2021-07-08 3:00:00 8 8
1 1 2021-07-08 5:30:00 7 8
2 1 2021-07-08 9:00:00 10 8
3 1 2021-01-08 6:30:00 5 5
4 1 2021-01-08 9:30:00 7 5
5 2 2021-07-08 3:00:00 7 10
6 2 2021-07-08 9:00:00 14 10
7 2 2021-07-08 5:30:00 10 10
8 2 2021-01-08 3:00:00 11 13
9 2 2021-01-08 3:00:00 13 13
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让我们分步做
7创建布尔掩码进行比较record小时大于的列中的值7Group由掩蔽柱id和date并transform使用max计算每个组的最大值m = pd.to_timedelta(df['time']).dt.components['hours'].ge(7)
df['max'] = df['record'].mask(m).groupby([df['id'], df['date']]).transform('max')
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id date time record max
0 1 2021-07-08 3:00:00 8 8.0
1 1 2021-07-08 5:30:00 7 8.0
2 1 2021-07-08 9:00:00 10 8.0
3 1 2021-01-08 6:30:00 5 5.0
4 1 2021-01-08 9:30:00 7 5.0
5 2 2021-07-08 3:00:00 7 10.0
6 2 2021-07-08 9:00:00 14 10.0
7 2 2021-07-08 5:30:00 10 10.0
8 2 2021-01-08 3:00:00 11 13.0
9 2 2021-01-08 3:00:00 13 13.0
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