我有一个带有两个索引的数据框,如下所示:
Index1 Index2 200701 200702 200703
alphas Fourth Quartile 41.7421 41.1807 39.071
Third Quartile 74.1573 95.0195 90.6572
Second Quartile -34.2001 -42.0068 -21.6236
First Quartile 39.293 37.3475 34.1704
All_Quartiles 37.6624 38.5957 38.0504
betas Fourth Quartile 18.1041 23.0865 33.7109
Third Quartile -51.9743 -93.1191 -87.1772
Second Quartile 121.262 131.556 103.549
First Quartile 26.1859 28.5129 31.8663
All_Quartiles 24.511 23.1601 0.159067
我需要新的索引,像这样:
New_index Index1 Index 2 200701 200702 200703
Sector alphas Fourth Quartile 41.7421 41.1807 39.071
Third Quartile 74.1573 95.0195 90.6572
Second Quartile -34.2001 -42.0068 -21.6236
First Quartile 39.293 37.3475 34.1704
All_Quartiles 37.6624 38.5957 38.0504
betas Fourth Quartile 18.1041 23.0865 33.7109
Third Quartile -51.9743 -93.1191 -87.1772
Second Quartile 121.262 131.556 103.549
First Quartile 26.1859 28.5129 31.8663
All_Quartiles 24.511 23.1601 0.159067
我有许多属于不同部门的数据框架,因此我需要将每个框架与一个循环合并。
答案 0 :(得分:2)
您可以手动重新创建整个MultiIndex,但这很繁琐。我更喜欢将concat
与keys
参数一起添加附加级别。 names
参数使我们可以为其命名。
pd.concat([df], keys=['Sector'], names=['New_index']+df.index.names)
200701 200702 200703
New_index Index1 Index2
Sector alphas Fourth Quartile 41.7421 41.1807 39.071000
Third Quartile 74.1573 95.0195 90.657200
Second Quartile -34.2001 -42.0068 -21.623600
First Quartile 39.2930 37.3475 34.170400
All_Quartiles 37.6624 38.5957 38.050400
betas Fourth Quartile 18.1041 23.0865 33.710900
Third Quartile -51.9743 -93.1191 -87.177200
Second Quartile 121.2620 131.5560 103.549000
First Quartile 26.1859 28.5129 31.866300
All_Quartiles 24.5110 23.1601 0.159067
此处与手动重新创建MultiIndex相同。
arrays = []
arrays.append(pd.Index(['Sector']*len(df), name='New_Index')) # 0th level sector
# Add all existing levels
for i in range(df.index.nlevels):
arrays.append(df.index.get_level_values(i))
new_idx = pd.MultiIndex.from_arrays(arrays)
df.index = new_idx
上面的内容基本上是DataFrame.set_index(append=True)
的内部内容,因此您可以用它清理一点。
df['New_index'] = 'Sector' # New column
df = df.set_index('New_index', append=True) # Bring it to index
df = df.reorder_levels([2, 0, 1]) # Move it to the front