Pandas：如何通过保留第一个的信息来合并列上的两个数据帧？

Question

我有两个数据df1和df2 。 df1包含人的年龄信息，而df2包含人的性别信息。 并非所有人都在df1或df2

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

如果我在df2中没有此信息，我想在df1中获取人们的性别信息并设置NaN 。 我试图做df1 = pd.merge(df1, df2, on = 'Name', how = 'outer')但我在df2中保留了一些我不想要的人的信息。

df1
     Name   Age     Sex
0     Tom    34      M
1     Sara   18     NaN
2     Eva    44      F
3     Jack   27      M
4     Laura  30     NaN

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)

使用由set_index创建的Series map ：

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

与左连接merge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

---

如果需要多列映射（例如Year和Code ）需要与左连接merge ：

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出错，则表示连接列重复，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：重新索引仅对具有唯一值的索引对象有效

解决方案通过DataFrame.drop_duplicates删除重复项，或使用 map by dict进行最后一次重复匹配：

#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set_index("Name")

Answer 3

从数据框创建字典的@jezrael 答案的简单补充。

它可能会有所帮助..

Python：

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

Answer 4

尚未提及重新索引，但它非常快，并且可以根据需要自动填充缺失值。

---

DataFrame.reindex

使用公共键 ( Name ) 作为映射数据帧 ( df2 ) 的索引：

• 如果df2的索引已经是Name ，只需直接reindex ：

   df2['Sex'].reindex(df1['Name'])

• 否则事先set_index ：

   df2.set_index('Name')['Sex'].reindex(df1['Name'])

请注意，当分配到现有数据帧时，重新索引的索引将未对齐，因此仅分配数组值：

df1['Sex'] = df2.set_index('Name')['Sex'].reindex(df1['Name']).array

#     Name  Age  Sex
# 0    Tom   34    M
# 1   Sara   18  NaN
# 2    Eva   44    F
# 3   Jack   27    M
# 4  Laura   30  NaN

我还注意到一个常见的假设，即重新索引很慢，但实际上很快（est）：

---

填充缺失值

reindex支持自动填充缺失值：

• fill_value :静态替换
• method ：给定单调索引的算法替换（ ffill 、 bfill或nearest ）

例如，用不想说(PNS) 填充空的Sex值：

df2.set_index('Name')['Sex'].reindex(df1['Name'], fill_value='PNS')

#     Name  Age  Sex
# 0    Tom   34    M
# 1   Sara   18  PNS
# 2    Eva   44    F
# 3   Jack   27    M
# 4  Laura   30  PNS

使用fill_value重新索引比链接fillna更快：

---

处理重复项

映射数据框 ( df2 ) 不能有重复的键，因此drop_duplicates如果适用：

df2.drop_duplicates('Name').set_index('Name')['Sex'].reindex(df1['Name'])

---

_{时序数据：}

 ''' Note: This is python code in a js snippet, so "run code snippet" will not work. The snippet is just to avoid cluttering the main post with supplemental code. ''' df1 = pd.DataFrame({'Name': np.arange(n), 'Age': np.random.randint(100, size=n)}).sample(frac=1).reset_index(drop=True) df2 = pd.DataFrame({'Name': np.arange(n) + int(n * 0.5), 'Sex': np.random.choice(list('MF'), size=n)}).sample(frac=1).reset_index(drop=True) def reindex_(df1, df2): df1['Sex'] = df2.set_index('Name')['Sex'].reindex(df1['Name']).array return df1 def map_(df1, df2): df1['Sex'] = df1['Name'].map(df2.set_index('Name')['Sex']) return df1 def dict_(df1, df2): df1['Sex'] = df1['Name'].map(dict(zip(df2['Name'], df2['Sex']))) return df1 def merge_(df1, df2): return df1.merge(df2[['Name', 'Sex']], left_on='Name', right_on='Name', how='left') def join_(df1, df2): return df1.set_index('Name').join(df2.set_index('Name'), how='left').reset_index() reindex_fill_value_ = lambda df1, df2: df2.set_index('Name')['Sex'].reindex(df1['Name'], fill_value='PNTS') reindex_fillna_ = lambda df1, df2: df2.set_index('Name')['Sex'].reindex(df1['Name']).fillna('PNTS') map_fillna_ = lambda df1, df2: df1['Name'].map(df2.set_index('Name')['Sex']).fillna('PNTS')