扁平化多索引pandas dataframe，其中列名变为值
扁平化多索引pandas dataframe，其中列名变为值
提问于 2017-09-20 01:44:50
假设我有以下数据帧：
import pandas as pd
df = pd.DataFrame(data={'Status' : ['green','green','red','blue','red','yellow','black'],
'Group' : ['A','A','B','C','A','B','C'],
'City' : ['Toronto','Montreal','Vancouver','Toronto','Edmonton','Winnipeg','Windsor'],
'Sales' : [13,6,16,8,4,3,1]})
df.drop('Status',axis=1,inplace=True)
ndf = pd.pivot_table(df,values=['Sales'],index=['City'],columns=['Group'],fill_value=0,margins=False)结果如下所示：
In [321]: ndf
Out[321]:
Sales
Group A B C
City
Edmonton 4 0 0
Montreal 6 0 0
Toronto 13 0 8
Vancouver 0 16 0
Windsor 0 0 1
Winnipeg 0 3 0我如何将其展平，使其成为单级数据框，但具有指定组的列？
也就是说，结果应该是：
City group sales
Edmonton A 4
Edmonton B 0
Edmonton C 0
Montreal A 6
Montreal B 0
Montreal C 0
Toronto A 13
Toronto B 0
Toronto C 8
Vancouver A 0
Vancouver B 16
Vancouver C 0
Windsor A 0
Windsor B 0
Windsor C 1
Winnipeg A 0
Winnipeg B 3
Winnipeg C 0回答 2
Stack Overflow用户
回答已采纳
发布于 2017-09-20 01:49:01
In [1260]: ndf.stack().reset_index()
Out[1260]:
City Group Sales
0 Edmonton A 4
1 Edmonton B 0
2 Edmonton C 0
3 Montreal A 6
4 Montreal B 0
5 Montreal C 0
6 Toronto A 13
7 Toronto B 0
8 Toronto C 8
9 Vancouver A 0
10 Vancouver B 16
11 Vancouver C 0
12 Windsor A 0
13 Windsor B 0
14 Windsor C 1
15 Winnipeg A 0
16 Winnipeg B 3
17 Winnipeg C 0Stack Overflow用户
发布于 2017-09-20 01:49:16
这很容易做到：
ndf.stack().reset_index()
City Group Sales
0 Edmonton A 4
1 Edmonton B 0
2 Edmonton C 0
3 Montreal A 6
4 Montreal B 0
5 Montreal C 0
6 Toronto A 13
7 Toronto B 0
8 Toronto C 8
9 Vancouver A 0
10 Vancouver B 16
11 Vancouver C 0
12 Windsor A 0
13 Windsor B 0
14 Windsor C 1
15 Winnipeg A 0
16 Winnipeg B 3
17 Winnipeg C 0页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/46306786
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