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如何进行自定义熊猫数据帧合并?

pandas python

Denis · 技术社区 · 5 年前

假设我有:

data = [['tom', 10, 20], ['nick', 15, 30], ['juli', 14, 40]] 
df = pd.DataFrame(data, columns = ['Name', 'Low-Age', 'High-Age']) 
print(df)
None
   Name  Low-Age  High-Age
0   tom       10        20
1  nick       15        30
2  juli       14        40

然后我有另一张桌子:

data = [[10, 'school'], [30, 'college']] 
edu = pd.DataFrame(data, columns = ['Age', 'Education']) 
print(edu)
None
   Age Education
0   10    school
1   30   college

我如何得到一个表格,将edu[‘年龄’]与df[‘低年龄’]或df[‘高年龄’]进行匹配。如果他们匹配,我想在df后面加上edu[“Education”]。(假设低年龄或高年龄可以匹配,而不是两者都匹配)

所以我希望我的输出是:

  Name  Low-Age  High-Age   Education
0   tom       10        20    school
1  nick       15        30    college
2  juli       14        40     NaN

3 回复 | 直到 5 年前

piRSquared 5 年前

`stack` -&燃气轮机; `map`

edu_dict = dict(zip(edu.Age, edu.Education))

Education = df[['Low-Age', 'High-Age']].stack().map(edu_dict).groupby(level=0).first()
df.assign(Education=Education)

   Name  Low-Age  High-Age Education
0   tom       10        20    school
1  nick       15        30   college
2  juli       14        40       NaN

Vaishali 5 年前

首先将地图与combine\u一起使用

mapper = edu.set_index('Age')['Education']
df['Education'] = df['Low-Age'].map(mapper).combine_first(df['High-Age'].map(mapper))

    Name    Low-Age High-Age    Education
0   tom     10      20          school
1   nick    15      30          college
2   juli    14      40          NaN

ansev 5 年前

使用 Series.map + pd.concat :

edu2=edu.set_index('Age')
s=pd.concat([df['Low-Age'].map(edu2['Education']),df['High-Age'].map(edu2['Education'])])
df['Education']=s[s.notna()].reindex(index=df.index)
print(df)

   Name  Low-Age  High-Age Education
0   tom       10        20    school
1  nick       15        30   college
2  juli       14        40       NaN

也可以求和 pd.concat :

edu2=edu.set_index('Age')
df['Education']= ( df['High-Age'].map(edu2['Education']).fillna('')+
                  df['Low-Age'].map(edu2['Education']).fillna('') )

或

edu2=edu.set_index('Age')
df['Education']= df[['High-Age','Low-Age']].apply(lambda x: x.map(edu2['Education']).fillna('')).sum(axis=1)

print(df)

   Name  Low-Age  High-Age Education
0   tom       10        20    school
1  nick       15        30   college
2  juli       14        40

Sai Kiran 5 年前

这种方法可以在处理大型数据集时节省时间。使用apply()。

low_age_list = df['Low-Age'].tolist()
high_age_list = df['High-Age'].tolist()

def match(row):
   print(row[1])
      if row['Age'] in low_age_list or row['Age'] in high_age_list:
         return row[1]

df['Education'] = edu.apply(match,axis=1)
print(df)