Question

我有一个DF，其中包含每月的数据A和B，以获取完整的姓名。如何使用相同的数据将此DF扩展为每周DF（一周结束）？

<?php
    $datetime1 = new DateTime('2009-10-11');
    $datetime2 = new DateTime('2009-10-13');
    $interval = $datetime1->diff($datetime2);
    echo $interval->format('%R%a days');
?>

Answer 1

首先将“日期”转换为日期时间对象：

date=pd.to_datetime(df['Date'])
df['Date']=date

然后使用“日期”作为索引设置DF：

 df=df.set_index(['Date'])

然后将groupby（）和resample（）与pad（）配合使用以进行重新采样：

df.groupby('Name').resample('W-FRI').pad()

这有效！

Answer 2

尝试：

# align Date to weekends
# if Date is already index, reset index first
df.Date = df.Date + pd.to_timedelta(6-df.Date.dt.dayofweek, unit='d')

# set_index
df.set_index(['Date', 'Name'], inplace=True)

# groupby and interpolate
# drop `interpolate()` if you want to fill the new dates with something else
df.groupby(level=1).apply(lambda x: x.reset_index(level=1)
                          .drop('Name', axis=1).asfreq('W')
                          .interpolate())

输出：

+-------+------------+-----------+-----------+
|       |            |     A     |     B     |
+-------+------------+-----------+-----------+
| name  | date       |           |           |
+-------+------------+-----------+-----------+
| henry | 2019-03-03 | -0.775991 | -0.370060 |
|       | 2019-03-10 | -0.583608 | 0.001735  |
|       | 2019-03-17 | -0.391225 | 0.373531  |
|       | 2019-03-24 | -0.198842 | 0.745326  |
|       | 2019-03-31 | -0.006460 | 1.117122  |
| jack  | 2019-03-03 | 0.529327  | -0.317801 |
|       | 2019-03-10 | -0.011261 | 0.336492  |
|       | 2019-03-17 | -0.551850 | 0.990785  |
|       | 2019-03-24 | -1.092438 | 1.645078  |
|       | 2019-03-31 | -1.633027 | 2.299371  |
| phil  | 2019-03-03 | 0.319571  | 0.282131  |
|       | 2019-03-10 | 0.190124  | 0.027607  |
|       | 2019-03-17 | 0.060676  | -0.226918 |
|       | 2019-03-24 | -0.068771 | -0.481442 |
|       | 2019-03-31 | -0.198218 | -0.735966 |
+-------+------------+-----------+-----------+

使用多索引进行DF重采样

2 个答案: