熊猫:根据栏目中的独特价值创建栏目

Question

我有一个有三列的Pandas数据框架，其中两列有标识符(日期和id)，另一列有我真正关心的值(值)。看起来是这样的：

,date,id,value
0,20210801,269277473,-1114389.6
1,20210802,269277473,-1658061.0
2,20210803,269277473,-1338010.2
3,20210804,269277473,-475779.6
4,20210805,269277473,-1417980.0
5,20210806,269277473,-1673400.6
6,20210807,269277473,-1438969.8
12,20210801,269277476,504300.0
13,20210802,269277476,519889.8
14,20210803,269277476,513899.4
15,20210804,269277476,526258.8
16,20210805,269277476,524730.0
17,20210806,269277476,548010.6
18,20210807,269277476,539031.0
24,20210801,269277480,477399.0
25,20210802,269277480,443499.0
26,20210803,269277480,394801.2
27,20210804,269277480,440100.0
28,20210805,269277480,455499.6
29,20210806,269277480,441100.2
30,20210807,269277480,438899.4

我希望将这些值滚动到一个表中，其中索引中的日期、列是is，而内容是值，如下所示：

date,269277473,269277476,269277480
20210801,-1114389.6,504300.0,477399.0
20210802,-1658061.0,519889.8,443499.0
20210803,-1338010.2,513899.4,394801.2
20210804,-475779.6,526258.8,440100.0
20210805,-1417980.0,524730.0,455499.6
20210806,-1673400.6,548010.6,441100.2
20210807,-1438969.8,539031.0,438899.4

鉴于我的表我们巨大(数亿的价值)，什么是最有效的方式来实现这一点？

Answer 1

您需要应用pivot

df.pivot(*df)

在您的情况下(根据列的顺序)，它等价于：

df.pivot(index='date', columns='id', values='value')

产出：

id        269277473  269277476  269277480
date                                     
20210801 -1114389.6   504300.0   477399.0
20210802 -1658061.0   519889.8   443499.0
20210803 -1338010.2   513899.4   394801.2
20210804  -475779.6   526258.8   440100.0
20210805 -1417980.0   524730.0   455499.6
20210806 -1673400.6   548010.6   441100.2
20210807 -1438969.8   539031.0   438899.4

Answer 2

使用pivot

>>> df.pivot(index='date', columns='id', values='value')
id        269277473  269277476  269277480
date
20210801 -1114389.6   504300.0   477399.0
20210802 -1658061.0   519889.8   443499.0
20210803 -1338010.2   513899.4   394801.2
20210804  -475779.6   526258.8   440100.0
20210805 -1417980.0   524730.0   455499.6
20210806 -1673400.6   548010.6   441100.2
20210807 -1438969.8   539031.0   438899.4

如果有其他列，则可以使用pivot_table对每个列的值(平均值、和、.)应用函数。