熊猫比较多索引数据帧而不循环答案

【问题标题】：Pandas comparing multiindex dataframes without looping熊猫比较多索引数据帧而不循环
【发布时间】：2018-09-27 10:45:56
【问题描述】：

我想比较两个多索引数据帧并添加另一列以显示值的差异（如果第一个数据帧和第二个数据帧之间的所有索引值都匹配）而不使用循环

index_a = [1,2,2,3,3,3]
index_b = [0,0,1,0,1,2]
index_c = [1,2,2,4,4,4]
index = pd.MultiIndex.from_arrays([index_a,index_b], names=('a','b'))
index_1 = pd.MultiIndex.from_arrays([index_c,index_b], names=('a','b'))
df1 = pd.DataFrame(np.random.rand(6,), index=index, columns=['p'])
df2 = pd.DataFrame(np.random.rand(6,), index=index_1, columns=['q'])

df1

df2

生成的矩阵 (df1-df2) 应如下所示

        p  diff
a b 
1 0 .4655  -0.1936 

2 0 .8600   .2916
  1 .9010   .3321   

3 0 .0652    No Match
  1 .5686    No Match
  2 .8965    No Match

【问题讨论】：

标签： python python-3.x pandas

【解决方案1】：

将reindex_like 或reindex 用于索引的交集：

df1['new'] = (df1['p'] - df2['q'].reindex_like(df1)).fillna('No Match')
#alternative
#df1['new'] = (df1['p'] - df2['q'].reindex(df1.index)).fillna('No Match')
print (df1)
            p       new
a b                    
1 0  0.955587  0.924466
2 0  0.312497 -0.310224
  1  0.306256  0.231646
3 0  0.575613  No Match
  1  0.674605  No Match
  2  0.462807  No Match

Index.intersection 和 DataFrame.loc 的另一个想法：

df1['new'] = (df1['p'] - df2.loc[df2.index.intersection(df1.index), 'q']).fillna('No Match')

或merge 左连接：

df = pd.merge(df1, df2, how='left', left_index=True, right_index=True)
df['new'] = (df['p'] - df['q']).fillna('No Match')
print (df)
            p         q       new
a b                              
1 0  0.789693  0.665148  0.124544
2 0  0.082677  0.814190 -0.731513
  1  0.762339  0.235435  0.526905
3 0  0.727695       NaN  No Match
  1  0.903596       NaN  No Match
  2  0.315999       NaN  No Match

【讨论】：

【解决方案2】：

使用以下来获取匹配索引的差异。不匹配索引将为 NaN

diff = df1['p'] - df2['q']

#Output
a  b
1  0   -0.666542
2  0   -0.389033
   1    0.064986
3  0         NaN
   1         NaN
   2         NaN
4  0         NaN
   1         NaN
   2         NaN
dtype: float64

【讨论】：