我有一个有两个索引的dataframe:
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,就像这样:
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 我有许多数据,多个索引,属于不同的部门,我需要合并每一个循环为。
发布于 2020-01-31 20:34:11
您可以手动重新创建整个MultiIndex,但这需要大量的编写。我更喜欢带有concat参数的keys来添加额外的级别。names参数允许我们给它命名。
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这里将是相同的手动重新创建MultiIndex。
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以上基本上是DataFrame.set_index(append=True)的内部结构,所以您可以用它来稍微清理一下。
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 fronthttps://stackoverflow.com/questions/60010583
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