我想通过比较“不”申请两个DF1和DF2摆脱只DF2(所有列)数据。
我的3行代码如下,这个我碰到DF1和DF2无法从DF1修剪领域的所有列。如何实现?
我已经像下面2个熊猫dataframes:
df1:
no,name,salary
1,abc,100
2,def,105
3,abc,110
4,def,115
5,abc,120
df2:
no,name,salary,dept,addr
1,abc,100,IT1,ADDR1
2,abc,101,IT2,ADDR2
3,abc,102,IT3,ADDR3
4,abc,103,IT4,ADDR4
5,abc,104,IT5,ADDR5
6,abc,105,IT6,ADDR6
7,abc,106,IT7,ADDR7
8,abc,107,IT8,ADDR8
df1 = pd.read_csv("D:\\data\\data1.csv")
df2 = pd.read_csv("D:\\data\\data2.csv")
resDF = pd.merge(df1, df2, on='no' , how='inner')
我认为你需要只过滤no
列,然后on
和how
参数是没有必要的:
resDF = pd.merge(df1[['no']], df2)
或与boolean indexing
过滤使用isin
:
resDF = df2[df2['no'].isin(df1['no'])]