pandas 第三节 (merge用法)_df.merge()-优快云博客

本文链接：https://blog.youkuaiyun.com/learningcoder/article/details/103657107

本文介绍了pandas库中用于数据融合的两个关键方法——merge和join。join操作基于行索引进行，要求列名不重复，返回结果包含df1的所有行，df2的列如果不存在则用NaN填充。而merge则更接近关系型数据库的操作，通过列索引进行关联。此外，还提到了pandas的分组功能。

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df1

df2

join是按照行索引来关联操作,列名不能相同

df1.join(df2) #返回df1的行数，df1和df2的所有列名值没有的话用NaN

merge是按照列索引来关联操作，和关系型数据库很相近

1. df1.merge(df2,on=0) #等价于df1.0=df2.0
    0  1_x   2   3   4  1_y
0  18   15  19  15  12   10
1  18   15  19  15  12   11
2  16   13  11  15  18   13
SELECT * FROM df1
JOIN df2 ON df1.[0]=df2.[0]
2. df1.merge(df2,left_on=3,right_on=0) #默认是inner
   0_x  1_x   2   3   4  0_y  1_y
0   11   18  12  18  14   18   10
1   11   18  12  18  14   18   11
SELECT * FROM df1
JOIN df2 ON df1.[3]=df2.[0]

3. df1.merge(df2,left_on=3,right_on=0,how="outer")
    0_x   1_x     2   3     4   0_y   1_y
0  18.0  15.0  19.0  15  12.0   NaN   NaN
1  16.0  13.0  11.0  15  18.0   NaN   NaN
2  10.0  16.0  15.0  14  16.0   NaN   NaN
3  11.0  18.0  12.0  18  14.0  18.0  10.0
4  11.0  18.0  12.0  18  14.0  18.0  11.0
5   NaN   NaN   NaN  19   NaN  19.0  10.0
6   NaN   NaN   NaN  16   NaN  16.0  13.0
SELECT * FROM df1 
FULL OUTER JOIN df2 on df1.[3]=df2.[0]

4. df1.merge(df2,left_on=3,right_on=0,how="left")
   0_x  1_x   2   3   4   0_y   1_y
0   18   15  19  15  12   NaN   NaN
1   10   16  15  14  16   NaN   NaN
2   11   18  12  18  14  18.0  10.0
3   11   18  12  18  14  18.0  11.0
4   16   13  11  15  18   NaN   NaN
SELECT * FROM df1 
LEFT JOIN df2 on df1.[3]=df2.[0]

5. df1.merge(df2,left_on=3,right_on=0,how="right")
    0_x   1_x     2    3     4   0_y   1_y
0   11.0  18.0   12.0   18   14.0   18   10
1   11.0  18.0   12.0   18   14.0   18   11
2   NaN   NaN   NaN  19   NaN   19   10
3   NaN   NaN   NaN  16   NaN   16   13
SELECT * FROM df1 
RIGHT JOIN df2 on df1.[3]=df2.[0]

分组 group

grouped=df2.groupby([0]) #pandas.core.groupby.generic.DataFrameGroupBy

遍历分组
for i,j in grouped:
    print(type(i),type(j))
    print("$"*50)
    print(i)
    print("-"*50)
    print(j)
    print("*"*50

c=grouped.count()

select [0],count([1]),count([2]) 
from df2
group by [0]