Pandas学习笔记系列：
- Pandas学习笔记（一）基本介绍
- Pandas学习笔记（二）选择数据
- Pandas学习笔记（三）修改&添加值
- Pandas学习笔记（四）处理丢失值
- Pandas学习笔记（五）合并 concat
- Pandas学习笔记（六）合并 merge
- Pandas学习笔记（七）plot画图
原文：https://morvanzhou.github.io/tutorials/data-manipulation/np-pd/3-7-pd-merge/ 本文有删减
要点
pandas中的merge和concat类似,但主要是用于两组有key column的数据,统一索引的数据. 通常也被用在Database的处理当中.
依据一组key合并
import pandas as pd
#定义资料集并打印出
left = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3']})
right = pd.DataFrame({'key': ['K0', 'K1', 'K2', 'K3'],
'C': ['C0', 'C1', 'C2', 'C3'],
'D': ['D0', 'D1', 'D2', 'D3']})
print(left)
"""
A B key
0 A0 B0 K0
1 A1 B1 K1
2 A2 B2 K2
3 A3 B3 K3
"""
print(right)
"""
C D key
0 C0 D0 K0
1 C1 D1 K1
2 C2 D2 K2
3 C3 D3 K3
"""
#依据key column合并，并打印出
res = pd.merge(left, right, on='key')
print(res)
"""
A B key C D
0 A0 B0 K0 C0 D0
1 A1 B1 K1 C1 D1
2 A2 B2 K2 C2 D2
3 A3 B3 K3 C3 D3
"""
依据两组key合并
合并时有4种方法how = ['left', 'right', 'outer', 'inner']，预设值how='inner'。
import pandas as pd
#定义资料集并打印出
left = pd.DataFrame({'key1': ['K0', 'K0', 'K1', 'K2'],
'key2': ['K0', 'K1', 'K0', 'K1'],
'A': ['A0', 'A1', 'A2', 'A3'],
'B': ['B0', 'B1', 'B2', 'B3']})
right = pd.DataFrame({'key1': ['K0', 'K1', 'K1', 'K2'],
'key2': ['K0', 'K0', 'K0', 'K0'],
'C': ['C0', 'C1', 'C2', 'C3'],
'D': ['D0', 'D1', 'D2', 'D3']})
print(left)
"""
A B key1 key2
0 A0 B0 K0 K0
1 A1 B1 K0 K1
2 A2 B2 K1 K0
3 A3 B3 K2 K1
"""
print(right)
"""
C D key1 key2
0 C0 D0 K0 K0
1 C1 D1 K1 K0
2 C2 D2 K1 K0
3 C3 D3 K2 K0
"""
依据key1与key2 columns进行合并，并打印出四种结果['left', 'right', 'outer', 'inner']
inner表示如果两个keys对应的value值相等，就交叉合并，否则丢弃，也就是求交集∩
例如left和right都有(K0,K0)和(K1,K0)。其中right有两个(K1,K0),left只有一个，从下面代码中的结果可以看到具有相同key的值会交叉合并，所以此时会生成两个新的(K1,K0)值。
res = pd.merge(left, right, on=['key1', 'key2'], how='inner')
print(res)
"""
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A2 B2 K1 K0 C1 D1
2 A2 B2 K1 K0 C2 D2
"""
# outter求并集∪
res = pd.merge(left, right, on=['key1', 'key2'], how='outer')
print(res)
"""
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A2 B2 K1 K0 C2 D2
4 A3 B3 K2 K1 NaN NaN
5 NaN NaN K2 K0 C3 D3
"""
# 以左边的key为准，如果右边和左边的值相等
res = pd.merge(left, right, on=['key1', 'key2'], how='left')
print(res)
"""
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A1 B1 K0 K1 NaN NaN
2 A2 B2 K1 K0 C1 D1
3 A2 B2 K1 K0 C2 D2
4 A3 B3 K2 K1 NaN NaN
"""
res = pd.merge(left, right, on=['key1', 'key2'], how='right')
print(res)
"""
A B key1 key2 C D
0 A0 B0 K0 K0 C0 D0
1 A2 B2 K1 K0 C1 D1
2 A2 B2 K1 K0 C2 D2
3 NaN NaN K2 K0 C3 D3
"""
Indicator
indicator=True会将合并的记录放在新的一列。
import pandas as pd
#定义资料集并打印出
df1 = pd.DataFrame({'col1':[0,1], 'col_left':['a','b']})
df2 = pd.DataFrame({'col1':[1,2,2],'col_right':[2,2,2]})
print(df1)
"""
col1 col_left
0 0 a
1 1 b
"""
print(df2)
"""
col1 col_right
0 1 2
1 2 2
2 2 2
"""
# 依据col1进行合并，并启用indicator=True，最后打印出
res = pd.merge(df1, df2, on='col1', how='outer', indicator=True)
print(res)
"""
col1 col_left col_right _merge
0 0.0 a NaN left_only
1 1.0 b 2.0 both
2 2.0 NaN 2.0 right_only
3 2.0 NaN 2.0 right_only
"""
# 自定indicator column的名称，并打印出
res = pd.merge(df1, df2, on='col1', how='outer', indicator='indicator_column')
print(res)
"""
col1 col_left col_right indicator_column
0 0.0 a NaN left_only
1 1.0 b 2.0 both
2 2.0 NaN 2.0 right_only
3 2.0 NaN 2.0 right_only
"""
依据index合并
import pandas as pd
#定义资料集并打印出
left = pd.DataFrame({'A': ['A0', 'A1', 'A2'],
'B': ['B0', 'B1', 'B2']},
index=['K0', 'K1', 'K2'])
right = pd.DataFrame({'C': ['C0', 'C2', 'C3'],
'D': ['D0', 'D2', 'D3']},
index=['K0', 'K2', 'K3'])
print(left)
"""
A B
K0 A0 B0
K1 A1 B1
K2 A2 B2
"""
print(right)
"""
C D
K0 C0 D0
K2 C2 D2
K3 C3 D3
"""
#依据左右资料集的index进行合并，how='outer',并打印出
res = pd.merge(left, right, left_index=True, right_index=True, how='outer')
print(res)
"""
A B C D
K0 A0 B0 C0 D0
K1 A1 B1 NaN NaN
K2 A2 B2 C2 D2
K3 NaN NaN C3 D3
"""
#依据左右资料集的index进行合并，how='inner',并打印出
res = pd.merge(left, right, left_index=True, right_index=True, how='inner')
print(res)
"""
A B C D
K0 A0 B0 C0 D0
K2 A2 B2 C2 D2
"""
解决overlapping的问题
import pandas as pd
#定义资料集
boys = pd.DataFrame({'k': ['K0', 'K1', 'K2'], 'age': [1, 2, 3]})
girls = pd.DataFrame({'k': ['K0', 'K0', 'K3'], 'age': [4, 5, 6]})
#使用suffixes解决overlapping的问题
res = pd.merge(boys, girls, on='k', suffixes=['_boy', '_girl'], how='inner')
print(res)
"""
age_boy k age_girl
0 1 K0 4
1 1 K0 5
"""