我有一个数据框如下:
df=pd.DataFrame({ 'family' : ["A","A","B","B"],
'V1' : [5,5,40,10,],
'V2' :[50,10,180,20],
'gr_0' :["all","all","all","all"],
'gr_1' :["m1","m1","m2","m3"],
'gr_2' :["m12","m12","m12","m9"],
'gr_3' :["NO","m14","m15","NO"]
})
我想用以下方式来改变它:
df_new=pd.DataFrame({ 'family' : ["A","A","A","A","B","B","B","B","B","B"],
'gr' : ["all","m1","m12","m14","all","m2","m3","m12","m9","m15"],
"calc(sumV2/sumV1)":[6,6,6,2,4,4.5,2,4.5,2,4.5]
})
family gr calc(sumV2/sumV1)
0 A all 6.0
1 A m1 6.0
2 A m12 6.0
3 A m14 2.0
4 B all 4.0
5 B m2 4.5
6 B m3 2.0
7 B m12 4.5
8 B m9 2.0
9 B m15 4.5
为了达到新的目的:
我对Python很陌生。软编码对我来说似乎相当复杂。 最好,我不希望“No”记录被列在这个df\u new中,但它也可以保留在输出中。你知道吗
melt
+groupby
:一种类似于this answer的方法,但其优点是完全矢量化,避免了
apply
性能:
您可以这样做:
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