Pandas差集-交集-并集求解
Pandas求解差集、交集、并集
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 pdIn [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]:
df1Out[3]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
In [4]:
df2Out[4]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
两个DataFrame的相同部分：
差集
方法1：concat + drop_duplicates
In [5]:
df3 = pd.concat([df1,df2])
df3Out[5]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
In [6]:
# 结果1
df3.drop_duplicates(["col1","col2"],keep=False)Out[6]:
col1 | col2 | |
|---|---|---|
1 | 2 | 7 |
3 | 4 | 9 |
4 | 5 | 10 |
2 | 7 | 10 |
方法2：append + drop_duplicates
In [7]:
df4 = df1.append(df2)
df4Out[7]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
In [8]:
# 结果2
df4.drop_duplicates(["col1","col2"],keep=False)Out[8]:
col1 | col2 | |
|---|---|---|
1 | 2 | 7 |
3 | 4 | 9 |
4 | 5 | 10 |
2 | 7 | 10 |
交集
方法1：merge
In [9]:
# 结果
# 等效：df5 = pd.merge(df1, df2, how="inner")
df5 = pd.merge(df1,df2)
df5Out[9]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 3 | 8 |
方法2：concat + duplicated + loc
In [10]:
df6 = pd.concat([df1,df2])
df6Out[10]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
In [11]:
s = df6.duplicated(subset=['col1','col2'], keep='first')
sOut[11]:
0 False
1 False
2 False
3 False
4 False
0 True
1 True
2 False
dtype: boolIn [12]:
# 结果
df8 = df6.loc[s == True]
df8Out[12]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 3 | 8 |
方法3：concat + groupby + query
In [13]:
# df6 = pd.concat([df1,df2])
df6Out[13]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
In [14]:
df9 = df6.groupby(["col1", "col2"]).size().reset_index()
df9.columns = ["col1", "col2", "count"]
df9Out[14]:
col1 | col2 | count | |
|---|---|---|---|
0 | 1 | 6 | 2 |
1 | 2 | 7 | 1 |
2 | 3 | 8 | 2 |
3 | 4 | 9 | 1 |
4 | 5 | 10 | 1 |
5 | 7 | 10 | 1 |
In [15]:
df10 = df9.query("count > 1")[["col1", "col2"]]
df10Out[15]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
2 | 3 | 8 |
并集
方法1：concat + drop_duplicates
In [16]:
df11 = pd.concat([df1,df2])
df11Out[16]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
0 | 1 | 6 |
1 | 3 | 8 |
2 | 7 | 10 |
In [17]:
# 结果
# df12 = df11.drop_duplicates(subset=["col1","col2"],keep="last")
df12 = df11.drop_duplicates(subset=["col1","col2"],keep="first")
df12Out[17]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
2 | 7 | 10 |
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]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
2 | 7 | 10 |
方法3：merge
In [19]:
pd.merge(df1,df2,how="outer")Out[19]:
col1 | col2 | |
|---|---|---|
0 | 1 | 6 |
1 | 2 | 7 |
2 | 3 | 8 |
3 | 4 | 9 |
4 | 5 | 10 |
5 | 7 | 10 |