如何找到两个数据框的补码 [英] how to find the complement of two dataframes
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问题描述
给定两个大数据帧,是否有任何简洁高效的代码(避免直接使用任何for loop
)使我能够获得这两个数据帧的补全?
given two large dataframes, is there any concise and efficient code (avoid using any for loop
directly) that allow me to obtain the complement of these two dataframes?
对我来说,最直接的方法是计算union-intersection
,如下面的朴素示例所示,但是我不知道如何用pandas
或np
the most straight forward way to me is to compute union-intersection
as shown in the naive example below, but I do not know how to implement this in an elegant languages of pandas
or np
df1= pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
'key2': ['K0', 'K1', 'K0', 'K1'],
'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3']})
df2= pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
'key2': ['K0', 'K0', 'K0', 'K0'],
'C': ['C0', 'C1', 'C2', 'C3'],
'D': ['D0', 'D1', 'D2', 'D3']})
intersection= pd.merge(df1, df2, how='inner',on=['key1', 'key2'])
union=pd.merge(df1, df2, how='outer',on=['key1', 'key2'])
complement=union-intersection
感谢您的任何评论和答案
thanks for any comments and answers
推荐答案
从此开始:
df1= pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
'key2': ['K0', 'K1', 'K0', 'K1'],
'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3']})
df2= pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
'key2': ['K0', 'K0', 'K0', 'K0'],
'C': ['C0', 'C1', 'C2', 'C3'],
'D': ['D0', 'D1', 'D2', 'D3']})
intersection = pd.merge(df1, df2, how='inner',on=['key1', 'key2'])
union = pd.merge(df1, df2, how='outer',on=['key1', 'key2'])
打印联盟
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A2 B2 K1 K0 C2 D2
4 A3 B3 K2 K1 NaN NaN
5 NaN NaN K2 K0 C3 D3
打印交叉点
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A2 B2 K1 K0 C1 D1
2 A2 B2 K1 K0 C2 D2
联合路口试试:
union[union.isnull().any(axis=1)]
A B key1 key2 C D
1 A1 B1 K0 K1 NaN NaN
4 A3 B3 K2 K1 NaN NaN
5 NaN NaN K2 K0 C3 D3
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