Python 如何在熊猫中拆开（或旋转？）
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how to unstack (or pivot?) in pandas
提问by codingknob
I have a dataframe that looks like the following:
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)
This gives me:
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
I want the final dataframe to look something like:
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
I've done the following but I'm not quite there:
我已经完成了以下操作，但还没有完成：
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
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
I thought I could unstack df2 to get something that resembles my final dataframe but I get all sorts of errors. I have also tried to pivot this dataframe but can't quite get what I want.
我以为我可以解开 df2 以获得类似于我的最终数据帧的东西，但是我遇到了各种各样的错误。我也尝试过旋转这个数据框，但不能完全得到我想要的。
采纳答案by unutbu
We want values (e.g. 'GERMANY') to become column names, and column names (e.g. 'HOUR1') to become values -- a swap of sorts.
我们希望值（例如'GERMANY'）成为列名，而列名（例如'HOUR1'）成为值——排序的交换。
The stackmethod turns column names into index values, and
the unstackmethod turns index values into column names.
该stack方法将列名转换为索引值，该unstack方法将索引值转换为列名。
So by shifting the values into the index, we can use stackand unstackto perform the swap.
因此，通过将值转移到索引中，我们可以使用stack和unstack来执行交换。
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)
yields
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
回答by BrenBarn
Using your df2:
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]
There is still a bit of cleanup to do if you want to convert the index level "variable" into a column called "HOUR" and strip out the text "HOUR" from the values, but I think that is the basic format you want.
如果您想将索引级别“变量”转换为名为“HOUR”的列并从值中去除文本“HOUR”，仍有一些清理工作要做，但我认为这是您想要的基本格式。
回答by Rafaell
Try using pivot. You can make it in one line. Eg.
尝试使用枢轴。您可以在一行中完成。例如。
df.pivot(index='start_time', columns='venue_name', values='ocupation')