如何在熊猫中取消堆叠(或转动？)

Question

我有一个如下所示的数据框:

import pandas as pd
datelisttemp = pd.date_range('1/1/2014', periods=3, freq='D')
s = list(datelisttemp)*3
s.sort()
df = pd.DataFrame({'BORDER':['GERMANY','FRANCE','ITALY','GERMANY','FRANCE','ITALY','GERMANY','FRANCE','ITALY' ], 'HOUR1':[2 ,2 ,2 ,4 ,4 ,4 ,6 ,6, 6],'HOUR2':[3 ,3 ,3, 5 ,5 ,5, 7, 7, 7], 'HOUR3':[8 ,8 ,8, 12 ,12 ,12, 99, 99, 99]}, index=s)

这给了我:

Out[458]: df

             BORDER  HOUR1  HOUR2  HOUR3
2014-01-01  GERMANY      2      3      8
2014-01-01   FRANCE      2      3      8
2014-01-01    ITALY      2      3      8
2014-01-02  GERMANY      4      5     12
2014-01-02   FRANCE      4      5     12
2014-01-02    ITALY      4      5     12
2014-01-03  GERMANY      6      7     99
2014-01-03   FRANCE      6      7     99
2014-01-03    ITALY      6      7     99

我希望最终的数据框看起来像:

             HOUR  GERMANY  FRANCE  ITALY
2014-01-01   1     2        2       2     
2014-01-01   2     3        3       3
2014-01-01   3     8        8       8 
2014-01-02   1     4        4       4
2014-01-02   2     5        5       5
2014-01-02   3    12       12      12
2014-01-03   1     6        6       6
2014-01-03   2     7        7       7
2014-01-03   3    99       99      99

我做了以下但是我不在那里:

df['date_col'] = df.index

df2 = melt(df, id_vars=['date_col','BORDER'])  
#Can I keep the same index after melt or do I have to set an index like below?
df2.set_index(['date_col', 'variable'], inplace=True, drop=True)
df2 = df2.sort()

DF

Out[465]: df2

                         BORDER   value
date_col   variable                 
2014-01-01 HOUR1           GERMANY   2
           HOUR1           FRANCE    2
           HOUR1           ITALY     2
           HOUR2           GERMANY   3
           HOUR2           FRANCE    3
           HOUR2           ITALY     3
           HOUR3           GERMANY   8
           HOUR3           FRANCE    8
           HOUR3           ITALY     8
2014-01-02 HOUR1           GERMANY   4
           HOUR1           FRANCE    4
           HOUR1           ITALY     4
           HOUR2           GERMANY   5
           HOUR2           FRANCE    5
           HOUR2           ITALY     5
           HOUR3           GERMANY  12
           HOUR3           FRANCE   12
           HOUR3           ITALY    12
2014-01-03 HOUR1           GERMANY   6
           HOUR1           FRANCE    6
           HOUR1           ITALY     6
           HOUR2           GERMANY   7
           HOUR2           FRANCE    7
           HOUR2           ITALY     7
           HOUR3           GERMANY  99
           HOUR3           FRANCE   99
           HOUR3           ITALY    99

我以为我可以拆开df2来获得类似于我最终数据帧的东西,但是我得到了各种各样的错误.我也尝试过调整这个数据框但是不能完全得到我想要的.

Answer 1

我们希望值(例如'GERMANY')成为列名,列名(例如'HOUR1')成为值 - 排序的交换.

该stack方法将列名称转换为索引值,该unstack方法将索引值转换为列名称.

因此,通过将值移动到索引中,我们可以使用stack和unstack执行交换.

import pandas as pd

datelisttemp = pd.date_range('1/1/2014', periods=3, freq='D')
s = list(datelisttemp)*3
s.sort()
df = pd.DataFrame({'BORDER':['GERMANY','FRANCE','ITALY','GERMANY','FRANCE','ITALY','GERMANY','FRANCE','ITALY' ], 'HOUR1':[2 ,2 ,2 ,4 ,4 ,4 ,6 ,6, 6],'HOUR2':[3 ,3 ,3, 5 ,5 ,5, 7, 7, 7], 'HOUR3':[8 ,8 ,8, 12 ,12 ,12, 99, 99, 99]}, index=s)

df = df.set_index(['BORDER'], append=True)
df.columns.name = 'HOUR'
df = df.unstack('BORDER')
df = df.stack('HOUR')
df = df.reset_index('HOUR')
df['HOUR'] = df['HOUR'].str.replace('HOUR', '').astype('int')
print(df)

产量

BORDER      HOUR  FRANCE  GERMANY  ITALY
2014-01-01     1       2        2      2
2014-01-01     2       3        3      3
2014-01-01     3       8        8      8
2014-01-02     1       4        4      4
2014-01-02     2       5        5      5
2014-01-02     3      12       12     12
2014-01-03     1       6        6      6
2014-01-03     2       7        7      7
2014-01-03     3      99       99     99

Answer 2

使用您的df2：

>>> df2.pivot_table(values='value', index=['DATE', 'variable'], columns="BORDER")
BORDER               FRANCE  GERMANY  ITALY
DATE       variable                        
2014-01-01 HOUR1          2        2      2
           HOUR2          3        3      3
           HOUR3          8        8      8
2014-01-02 HOUR1          4        4      4
           HOUR2          5        5      5
           HOUR3         12       12     12
2014-01-03 HOUR1          6        6      6
           HOUR2          7        7      7
           HOUR3         99       99     99

[9 rows x 3 columns]

如果您想将索引级别“变量”转换为名为“HOUR”的列并从值中删除文本“HOUR”，则仍然需要进行一些清理工作，但我认为这是您想要的基本格式。