Question

我关注df1：

         A1       A2        A3        A4       B1        B2        B3      B4  \
3  0.202425  0.13495  0.202425  0.202425  0.94465  0.877175  0.877175  0.8097   

        C1      C2   ...           F3        F4        G1        G2        G3  \
3  1.21455  1.3495   ...     4.925676  4.318401  5.330526  5.600426  5.802851   

         G4        H1        H2       H3       H4  
3  5.398001  0.202425  0.067475  0.13495  0.13495  

[1 rows x 32 columns]

我想创建这样的东西：

       A1       A2        A3        A4
0.202425  0.13495   0.202425  0.202425
0.94465   0.877175  0.877175  0.8097
1.21455   1.3495    1.282025  1.282025
2.429101  2.496576  2.429101  2.429101    
3.441226  3.846076  3.643651  3.643651
4.723251  4.858201  4.925676  4.925676
5.330526  5.600426  5.802851  5.802851
0.202425  0.067475  0.13495   0.13495

这是我的代码：

a_cols = [c for c in df1.columns if c.startswith('A')]
b_cols = [c for c in df1.columns if c.startswith('B')]
c_cols = [c for c in df1.columns if c.startswith('C')]
d_cols = [c for c in df1.columns if c.startswith('D')]
e_cols = [c for c in df1.columns if c.startswith('E')]
f_cols = [c for c in df1.columns if c.startswith('F')]
g_cols = [c for c in df1.columns if c.startswith('G')]
h_cols = [c for c in df1.columns if c.startswith('H')]

col_dict = dict(zip(a_cols, b_cols,c_cols,d_cols,e_cols,f_cols,g_cols,h_cols))
l=pd.concat([df1.loc[:, a_cols], df1.loc[:, b_cols], df1.loc[:, c_cols], df1.loc[:, d_cols], df1.loc[:, e_cols], df1.loc[:, f_cols], df1.loc[:, g_cols].df1.loc[:, h_cols].rename(columns=col_dict)])
print (l)

不知何故，我无法将多个列表压缩...

Answer 1

也许你可以在values和reshape使用numpy数组。

如果列length为4，则可以使用它。

print df
#         A1       A2        A3        A4       B1        B2        B3      B4  \
#0  0.202425  0.13495  0.202425  0.202425  0.94465  0.877175  0.877175  0.8097   
#
#        F1      F2        F3        F4        G1        G2        G3  \
#0  1.21455  1.3495  4.925676  4.318401  5.330526  5.600426  5.802851   
#
#         G4        H1        H2       H3       H4  
#0  5.398001  0.202425  0.067475  0.13495  0.13495  


print (df.values).reshape((5, 4))

#[[ 0.202425  0.13495   0.202425  0.202425]
# [ 0.94465   0.877175  0.877175  0.8097  ]
# [ 1.21455   1.3495    4.925676  4.318401]
# [ 5.330526  5.600426  5.802851  5.398001]
# [ 0.202425  0.067475  0.13495   0.13495 ]]

print (df.values).shape[1]
#20

print pd.DataFrame((df.values).reshape(((df.values).shape[1]/4, 4)), 
                                       columns=['A1','A2','A3','A4'])

#         A1        A2        A3        A4
#0  0.202425  0.134950  0.202425  0.202425
#1  0.944650  0.877175  0.877175  0.809700
#2  1.214550  1.349500  4.925676  4.318401
#3  5.330526  5.600426  5.802851  5.398001
#4  0.202425  0.067475  0.134950  0.134950

Pandas将行与不同的行名称组合到新的df中

1 个答案: