如果我在蟒蛇中有两个列表或数据框架(熊猫),我如何合并/匹配/加入它们?
例如:
清单/ DF 1:
Table_Name Table_Alias
tab_1 t1
tab_2 t2
tab_3 t3清单/ DF 2:
Table_Alias Variable_Name
t1 Owner
t1 Owner_Id
t2 Purchase_date
t3 Maintenance_cost预期结果:
Table_Name Table_Alias Variable_Name
tab_1 t1 Owner
tab_1 t1 Owner_Id
tab_2 t2 Purchase_date
tab_3 t3 Maintenance_cost注意:如果我是在R中这样做的话,我会使用这样的方法:
df3 <- merge(df1, df2, by = 'Table_Alias', all.y = T)在python中做这件事的最好方法是什么?
发布于 2015-07-03 15:16:55
你想要一个“外部”merge
In [9]:
df.merge(df1, how='outer')
Out[9]:
Table_Name Table_Alias Variable_Name
0 tab_1 t1 Owner
1 tab_1 t1 Owner_Id
2 tab_2 t2 Purchase_date
3 tab_3 t3 Maintenance_cost它将在dfs的重叠列上匹配,并返回匹配行的合并。
发布于 2015-07-03 15:36:19
我只会使用pd.merge(df1, df2, how='outer',on='alias')
df1 = pd.DataFrame({ "table_name":['tab1',"tab2","tab3"],"talias ['t1','t2','t3']})
df2 = pd.DataFrame({"talias":['t1',"t1","t2",'t3'], "vname,['Owner','Owner_Id','Purchase_date','Maintenance_cost']})
pd.merge(df1,df2,how='outer', on='talias')
Out:
Table_Alias Table_Name Variable_Name
0 t1 tab1 Owner
1 t1 tab1 Owner_Id
2 t2 tab2 Purchase_date
3 t3 tab3 Maintenance_costhttps://stackoverflow.com/questions/31209908
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