我有两个数据帧df1和df2,其中df2是df1的子集。我如何得到一个新的数据帧(df3),这是两个数据帧之间的差异?
换句话说,一个在df1中所有的行/列都不在df2中的数据帧?
我有两个数据帧df1和df2,其中df2是df1的子集。我如何得到一个新的数据帧(df3),这是两个数据帧之间的差异?
换句话说,一个在df1中所有的行/列都不在df2中的数据帧?
当前回答
Pandas现在提供了一个新的API来处理数据帧差异:Pandas . datafframe .compare
df.compare(df2)
col1 col3
self other self other
0 a c NaN NaN
2 NaN NaN 3.0 4.0
其他回答
另一个可能的解决方案是使用numpy广播:
df1[np.all(~np.all(df1.values == df2.values[:, None], axis=2), axis=0)]
输出:
Name Age
1 Mike 45
4 Marry 27
7 Bolt 39
正如这里提到的 那
df1[~df1.apply(tuple,1).isin(df2.apply(tuple,1))]
是正确的解决方案,但它会产生错误的输出如果
df1=pd.DataFrame({'A':[1],'B':[2]})
df2=pd.DataFrame({'A':[1,2,3,3],'B':[2,3,4,4]})
在这种情况下,上面的溶液会给出 空数据帧,相反,你应该使用concat方法后,从每个数据帧删除重复。
使用concate和drop_duplicate
df1=df1.drop_duplicates(keep="first")
df2=df2.drop_duplicates(keep="first")
pd.concat([df1,df2]).drop_duplicates(keep=False)
Pandas现在提供了一个新的API来处理数据帧差异:Pandas . datafframe .compare
df.compare(df2)
col1 col3
self other self other
0 a c NaN NaN
2 NaN NaN 3.0 4.0
pandas DataFrame.compare中有一种新的方法,即比较2个不同的dataframe,并返回数据记录中每列中变化的值。
例子
第一个Dataframe
Id Customer Status Date
1 ABC Good Mar 2023
2 BAC Good Feb 2024
3 CBA Bad Apr 2022
第二个Dataframe
Id Customer Status Date
1 ABC Bad Mar 2023
2 BAC Good Feb 2024
5 CBA Good Apr 2024
比较Dataframes
print("Dataframe difference -- \n")
print(df1.compare(df2))
print("Dataframe difference keeping equal values -- \n")
print(df1.compare(df2, keep_equal=True))
print("Dataframe difference keeping same shape -- \n")
print(df1.compare(df2, keep_shape=True))
print("Dataframe difference keeping same shape and equal values -- \n")
print(df1.compare(df2, keep_shape=True, keep_equal=True))
结果
Dataframe difference --
Id Status Date
self other self other self other
0 NaN NaN Good Bad NaN NaN
2 3.0 5.0 Bad Good Apr 2022 Apr 2024
Dataframe difference keeping equal values --
Id Status Date
self other self other self other
0 1 1 Good Bad Mar 2023 Mar 2023
2 3 5 Bad Good Apr 2022 Apr 2024
Dataframe difference keeping same shape --
Id Customer Status Date
self other self other self other self other
0 NaN NaN NaN NaN Good Bad NaN NaN
1 NaN NaN NaN NaN NaN NaN NaN NaN
2 3.0 5.0 NaN NaN Bad Good Apr 2022 Apr 2024
Dataframe difference keeping same shape and equal values --
Id Customer Status Date
self other self other self other self other
0 1 1 ABC ABC Good Bad Mar 2023 Mar 2023
1 2 2 BAC BAC Good Good Feb 2024 Feb 2024
2 3 5 CBA CBA Bad Good Apr 2022 Apr 2024
使用lambda函数,您可以过滤_merge值为“left_only”的行,以获得df1中df2中缺失的所有行
df3 = df1.merge(df2, how = 'outer' ,indicator=True).loc[lambda x :x['_merge']=='left_only']
df