Python 按两列分组，然后得到最早和最晚日期
[英]Python Group by two columns and then get the earliest and latest date
问题描述
When I was trying to get the earliest and latest date after groupby, I found that max results will be attached after min:
当我试图获取 groupby 之后的最早和最晚日期时，我发现 max 结果将在 min 之后附加：
ATR_table.groupby(['USAGEID', 'STAT']).agg({'DATADTTM':'min','DATADTTM':'max'})
| USAGEID使用 ID |
STAT统计数据 |
DATADTTM数据DTTM |
|---|---|---|
| 10140 10140 |
0 0 |
2020-01-01 2020-01-01 |
| 10140 10140 |
1 1 |
2020-01-01 2020-01-01 |
| 10141 10141 |
0 0 |
2020-01-01 2020-01-01 |
| 10141 10141 |
1 1 |
2020-01-01 2020-01-01 |
| 10140 10140 |
0 0 |
2020-07-18 2020-07-18 |
| 10140 10140 |
1 1 |
2020-07-18 2020-07-18 |
| 10141 10141 |
0 0 |
2020-07-18 2020-07-18 |
| 10141 10141 |
1 1 |
2020-07-18 2020-07-18 |
Is there a way that I can have the following result by using groupby?有没有办法通过使用 groupby 来获得以下结果？
| USAGEID使用 ID |
STAT统计数据 |
DATADTTM Min DATADTTM 最小值 |
DATADTTM Max DATADTTM 最大值 |
|---|---|---|---|
| 10140 10140 |
0 0 |
2020-01-01 2020-01-01 |
2020-07-18 2020-07-18 |
| 10140 10140 |
1 1 |
2020-01-01 2020-01-01 |
2020-07-18 2020-07-18 |
| 10141 10141 |
0 0 |
2020-01-01 2020-01-01 |
2020-07-18 2020-07-18 |
| 10141 10141 |
1 1 |
2020-01-01 2020-01-01 |
2020-07-18 2020-07-18 |
1 个解决方案
解决方案1
1 已采纳 2021-04-02 15:21:54
If you have no other columns, you could simply pass a plain list:如果你没有其他列，你可以简单地传递一个简单的列表：
ATR_table.groupby(['USAGEID', 'STAT']).agg(['min', 'max'])
If you want to be able to use other functions on other columns, you should include a list in the dictionary:如果您希望能够在其他列上使用其他功能，您应该在字典中包含一个列表：
ATR_table.groupby(['USAGEID', 'STAT']).agg({'DATADTTM':['min', 'max']})
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