【Python】DataFrame差集/交集/并集求解

公众号：尤而小屋
作者：Peter
编辑：Peter

本文讲解的是如何利用Pandas函数求解两个DataFrame的差集、交集、并集。

模拟数据

模拟一份简单的数据：

In [1]:

import pandas as pd

In [2]:

df1 = pd.DataFrame({"col1":[1,2,3,4,5],
                    "col2":[6,7,8,9,10]
                   })

df2 = pd.DataFrame({"col1":[1,3,7],
                    "col2":[6,8,10]
                   })

In [3]:

df1

Out[3]:

In [4]:

df2

Out[4]:

两个DataFrame的相同部分：

差集

方法1：concat + drop_duplicates

In [5]:

df3 = pd.concat([df1,df2])
df3

Out[5]:

In [6]:

# 结果1

df3.drop_duplicates(["col1","col2"],keep=False)

Out[6]:

方法2：append + drop_duplicates

In [7]:

df4 = df1.append(df2)
df4

Out[7]:

In [8]:

# 结果2

df4.drop_duplicates(["col1","col2"],keep=False)

Out[8]:

交集

方法1：merge

In [9]:

# 结果

# 等效：df5 = pd.merge(df1, df2, how="inner")
df5 = pd.merge(df1,df2)

df5

Out[9]:

方法2：concat + duplicated + loc

In [10]:

df6 = pd.concat([df1,df2])
df6

Out[10]:

In [11]:

s = df6.duplicated(subset=['col1','col2'], keep='first')
s

Out[11]:

0    False
1    False
2    False
3    False
4    False
0     True
1     True
2    False
dtype: bool

In [12]:

# 结果
df8 = df6.loc[s == True]
df8

Out[12]:

方法3：concat + groupby + query

In [13]:

# df6 = pd.concat([df1,df2])

df6

Out[13]:

In [14]:

df9 = df6.groupby(["col1", "col2"]).size().reset_index()
df9.columns = ["col1", "col2", "count"]

df9

Out[14]:

In [15]:

df10 = df9.query("count > 1")[["col1", "col2"]]
df10

Out[15]:

并集

方法1：concat + drop_duplicates

In [16]:

df11 = pd.concat([df1,df2])
df11

Out[16]:

In [17]:

# 结果

# df12 = df11.drop_duplicates(subset=["col1","col2"],keep="last")
df12 = df11.drop_duplicates(subset=["col1","col2"],keep="first")
df12

Out[17]:

方法2：append + drop_duplicates

In [18]:

df13 = df1.append(df2)

# df13.drop_duplicates(subset=["col1","col2"],keep="last")
df13.drop_duplicates(subset=["col1","col2"],keep="first")

Out[18]:

方法3：merge

In [19]:

pd.merge(df1,df2,how="outer")

Out[19]: