Question

我有一个带有多索引列的pandas数据框：

arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
      ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
df = pd.DataFrame(np.random.randn(3, 8), index=['A', 'B', 'C'], columns=index)

现在我需要将df [＆＃34; bar＆＃34;]中的值除以df [＆＃34; baz＆＃34;]中的值，并将其保存在名称为＆＃34; new＆＃34;的数据框中。（第二级索引为1和2）。

df [＆＃34; bar＆＃34;] / df [＆＃34; baz＆＃34;]为我提供了正确的值，但我不了解如何将其存储在数据框中。

我试过了：
df["new"] = df["bar"]/df["baz"]和df.loc[:, ("new", ["one", "two"])] = df["bar"]/df["baz"]，但都会出错。有关如何在数据框中以新名称存储数据的任何想法？

Answer 1

您可以按MultiIndex.from_product添加级别，然后使用concat：

a = df["bar"] / df["baz"]
a.columns = pd.MultiIndex.from_product([['new'], a.columns])
print (a)
        new          
        one       two
A -1.080108 -0.876062
B  0.171536  0.278908
C  2.045792  0.795082

df1 = pd.concat([df, a], axis=1)
print (df1)
first        bar                 baz                 foo                 qux  \
second       one       two       one       two       one       two       one   
A      -0.668129 -0.498210  0.618576  0.568692  1.350509  1.629589  0.301966   
B      -0.345811 -0.315231 -2.015971 -1.130231 -1.111846  0.237851 -0.325130   
C       1.915676  0.920348  0.936398  1.157552 -0.106208 -0.088752 -0.971485   

first                  new            
second       two       one       two  
A       0.449483 -1.080108 -0.876062  
B       1.944702  0.171536  0.278908  
C      -0.384060  2.045792  0.795082

另一种解决方案，xs选择并重命名，最后join为原始：

a = (df.xs("bar", axis=1, level=0, drop_level=False) / df["baz"])
       .rename(columns={'bar':'new'})

df1 = df.join(a)
print (df1)
first        bar                 baz                 foo                 qux  \
second       one       two       one       two       one       two       one   
A      -0.668129 -0.498210  0.618576  0.568692  1.350509  1.629589  0.301966   
B      -0.345811 -0.315231 -2.015971 -1.130231 -1.111846  0.237851 -0.325130   
C       1.915676  0.920348  0.936398  1.157552 -0.106208 -0.088752 -0.971485   

first                  new            
second       two       one       two  
A       0.449483 -1.080108 -0.876062  
B       1.944702  0.171536  0.278908  
C      -0.384060  2.045792  0.795082

通过stack和unstack进行重塑的解决方案在大型df中应该更慢：

df1 = df.stack()
df1['new'] = df1["bar"] / df1["baz"]
df1 = df1.unstack()
print (df1)
first        bar                 baz                 foo                 qux  \
second       one       two       one       two       one       two       one   
A      -0.668129 -0.498210  0.618576  0.568692  1.350509  1.629589  0.301966   
B      -0.345811 -0.315231 -2.015971 -1.130231 -1.111846  0.237851 -0.325130   
C       1.915676  0.920348  0.936398  1.157552 -0.106208 -0.088752 -0.971485   

first                  new            
second       two       one       two  
A       0.449483 -1.080108 -0.876062  
B       1.944702  0.171536  0.278908  
C      -0.384060  2.045792  0.795082

loc的解决方案：

a = (df.loc(axis=1)['bar', :] / df["baz"]).rename(columns={'bar':'new'})
print (a)
first        new          
second       one       two
A      -1.080108 -0.876062
B       0.171536  0.278908
C       2.045792  0.795082

df1 = df.join(a)
print (df1)
first        bar                 baz                 foo                 qux  \
second       one       two       one       two       one       two       one   
A      -0.668129 -0.498210  0.618576  0.568692  1.350509  1.629589  0.301966   
B      -0.345811 -0.315231 -2.015971 -1.130231 -1.111846  0.237851 -0.325130   
C       1.915676  0.920348  0.936398  1.157552 -0.106208 -0.088752 -0.971485   

first                  new            
second       two       one       two  
A       0.449483 -1.080108 -0.876062  
B       1.944702  0.171536  0.278908  
C      -0.384060  2.045792  0.795082

<强>设置：

np.random.seed(456)
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
      ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
df = pd.DataFrame(np.random.randn(3, 8), index=['A', 'B', 'C'], columns=index)
print (df)
first        bar                 baz                 foo                 qux  \
second       one       two       one       two       one       two       one   
A      -0.668129 -0.498210  0.618576  0.568692  1.350509  1.629589  0.301966   
B      -0.345811 -0.315231 -2.015971 -1.130231 -1.111846  0.237851 -0.325130   
C       1.915676  0.920348  0.936398  1.157552 -0.106208 -0.088752 -0.971485   

first             
second       two  
A       0.449483  
B       1.944702  
C      -0.384060

Answer 2

选项1：

In [200]: df.join((df[['bar']]/df['baz']).rename(columns={'bar':'new'}))
Out[200]:
first        bar                 baz                 foo                 qux                 new
second       one       two       one       two       one       two       one       two       one       two
A      -1.089798  2.053026  0.470218  1.440740 -0.536965 -0.667857  0.717725 -1.202051 -2.317647  1.424980
B       0.488875  0.428836  1.413451 -0.683677 -1.293274  0.374481  0.074252 -1.195414  0.345873 -0.627250
C      -0.243064 -0.069446 -0.911166  0.478370 -0.948390 -0.366823 -1.499948  1.513508  0.266761 -0.145172

说明：

In [201]: df[['bar']]/df['baz']
Out[201]:
first        bar
second       one       two
A      -2.317647  1.424980
B       0.345873 -0.627250
C       0.266761 -0.145172

In [202]: (df[['bar']]/df['baz']).rename(columns={'bar':'new'})
Out[202]:
first        new
second       one       two
A      -2.317647  1.424980
B       0.345873 -0.627250
C       0.266761 -0.145172

将计算数据存储在pandas数据帧的新多列中

2 个答案: