Question

我有两个数据帧constructor(props){ super(props); this.state = { loadCard : false; } this.changeCard = this.changeCard.bind(this); } changeCard = (userId) => { this.setState({ loadCard: true}); } render(){ return( console('working state'+ this.state.loadCard); ); }和df1。 df2包含人的年龄信息，而df1包含人的性别信息。并非所有人都在df2或df1

df2

我想在df1 Name Age 0 Tom 34 1 Sara 18 2 Eva 44 3 Jack 27 4 Laura 30 df2 Name Sex 0 Tom M 1 Paul M 2 Eva F 3 Jack M 4 Michelle F中获得有关性别的信息，如果我在df1中没有此信息，请设置NaN。我尝试做df2，但是我把一些我不想要的人的信息保存在df1 = pd.merge(df1, df2, on = 'Name', how = 'outer')中。

df2

Answer 1

Sample：

df1 = pd.DataFrame({'Name': ['Tom', 'Sara', 'Eva', 'Jack', 'Laura'], 
                    'Age': [34, 18, 44, 27, 30]})

#print (df1)
df3 = df1.copy()

df2 = pd.DataFrame({'Name': ['Tom', 'Paul', 'Eva', 'Jack', 'Michelle'], 
                    'Sex': ['M', 'M', 'F', 'M', 'F']})
#print (df2)

由map创建的Series使用set_index：

df1['Sex'] = df1['Name'].map(df2.set_index('Name')['Sex'])
print (df1)
    Name  Age  Sex
0    Tom   34    M
1   Sara   18  NaN
2    Eva   44    F
3   Jack   27    M
4  Laura   30  NaN

使用merge和左联接的替代解决方案：

df = df3.merge(df2[['Name','Sex']], on='Name', how='left')
print (df)
    Name  Age  Sex
0    Tom   34    M
1   Sara   18  NaN
2    Eva   44    F
3   Jack   27    M
4  Laura   30  NaN

如果需要通过多列映射（例如Year和Code），则需要merge并使用左连接：

df1 = pd.DataFrame({'Name': ['Tom', 'Sara', 'Eva', 'Jack', 'Laura'], 
                    'Year':[2000,2003,2003,2004,2007],
                    'Code':[1,2,3,4,4],
                    'Age': [34, 18, 44, 27, 30]})

print (df1)
    Name  Year  Code  Age
0    Tom  2000     1   34
1   Sara  2003     2   18
2    Eva  2003     3   44
3   Jack  2004     4   27
4  Laura  2007     4   30

df2 = pd.DataFrame({'Name': ['Tom', 'Paul', 'Eva', 'Jack', 'Michelle'], 
                    'Sex': ['M', 'M', 'F', 'M', 'F'],
                    'Year':[2001,2003,2003,2004,2007],
                    'Code':[1,2,3,5,3],
                    'Val':[21,34,23,44,67]})
print (df2)
       Name Sex  Year  Code  Val
0       Tom   M  2001     1   21
1      Paul   M  2003     2   34
2       Eva   F  2003     3   23
3      Jack   M  2004     5   44
4  Michelle   F  2007     3   67

#merge by all columns
df = df1.merge(df2, on=['Year','Code'], how='left')
print (df)
  Name_x  Year  Code  Age Name_y  Sex   Val
0    Tom  2000     1   34    NaN  NaN   NaN
1   Sara  2003     2   18   Paul    M  34.0
2    Eva  2003     3   44    Eva    F  23.0
3   Jack  2004     4   27    NaN  NaN   NaN
4  Laura  2007     4   30    NaN  NaN   NaN

#specified columns - columns for join (Year, Code) need always + appended columns (Val)
df = df1.merge(df2[['Year','Code', 'Val']], on=['Year','Code'], how='left')
print (df)
    Name  Year  Code  Age   Val
0    Tom  2000     1   34   NaN
1   Sara  2003     2   18  34.0
2    Eva  2003     3   44  23.0
3   Jack  2004     4   27   NaN
4  Laura  2007     4   30   NaN

如果map错误，则表示按连接列重复，此处为Name：

df1 = pd.DataFrame({'Name': ['Tom', 'Sara', 'Eva', 'Jack', 'Laura'], 
                    'Age': [34, 18, 44, 27, 30]})

print (df1)
    Name  Age
0    Tom   34
1   Sara   18
2    Eva   44
3   Jack   27
4  Laura   30

df3, df4 = df1.copy(), df1.copy()

df2 = pd.DataFrame({'Name': ['Tom', 'Tom', 'Eva', 'Jack', 'Michelle'], 
                    'Val': [1,2,3,4,5]})
print (df2)
       Name  Val
0       Tom    1 <-duplicated name Tom
1       Tom    2 <-duplicated name Tom
2       Eva    3
3      Jack    4
4  Michelle    5

s = df2.set_index('Name')['Val']
df1['New'] = df1['Name'].map(s)
print (df1)

InvalidIndexError：仅对具有唯一值的索引对象有效的索引重新建立索引

通过DataFrame.drop_duplicates删除了重复的解决方案，或者通过dict使用map进行了最后一次重复匹配：

#default keep first value
s = df2.drop_duplicates('Name').set_index('Name')['Val']
print (s)
Name
Tom         1
Eva         3
Jack        4
Michelle    5
Name: Val, dtype: int64

df1['New'] = df1['Name'].map(s)
print (df1)
    Name  Age  New
0    Tom   34  1.0
1   Sara   18  NaN
2    Eva   44  3.0
3   Jack   27  4.0
4  Laura   30  NaN

#add parameter for keep last value 
s = df2.drop_duplicates('Name', keep='last').set_index('Name')['Val']
print (s)
Name
Tom         2
Eva         3
Jack        4
Michelle    5
Name: Val, dtype: int64

df3['New'] = df3['Name'].map(s)
print (df3)
    Name  Age  New
0    Tom   34  2.0
1   Sara   18  NaN
2    Eva   44  3.0
3   Jack   27  4.0
4  Laura   30  NaN

#map by dictionary
d = dict(zip(df2['Name'], df2['Val']))
print (d)
{'Tom': 2, 'Eva': 3, 'Jack': 4, 'Michelle': 5}

df4['New'] = df4['Name'].map(d)
print (df4)
    Name  Age  New
0    Tom   34  2.0
1   Sara   18  NaN
2    Eva   44  3.0
3   Jack   27  4.0
4  Laura   30  NaN

Answer 2

您还可以使用join方法：

df1.set_index("Name").join(df2.set_index("Name"), how="left")

编辑：添加了set_index("Name")

Answer 3

从数据帧创建字典的@jezrael 答案的简单补充。

可能会有帮助..

蟒蛇：

df1 = pd.DataFrame({'Name': ['Tom', 'Sara', 'Eva', 'Jack', 'Laura'],
                    'Age': [34, 18, 44, 27, 30]})


df2 = pd.DataFrame({'Name': ['Tom', 'Paul', 'Eva', 'Paul', 'Jack', 'Michelle', 'Tom'],
                    'Something': ['M', 'M', 'F', 'M', 'A', 'F', 'B']})


df1_dict = pd.Series(df1.Age.values, index=df1.Name).to_dict()

df2['Age'] = df2['Name'].map(df1_dict)

print(df2)

输出：

      Name Something   Age
0       Tom         M  34.0
1      Paul         M   NaN
2       Eva         F  44.0
3      Paul         M   NaN
4      Jack         A  27.0
5  Michelle         F   NaN
6       Tom         B  34.0

Python：如何通过保留第一个数据框的信息来合并两个数据框？

3 个答案: