Question

我有一个类似清单的字典-

from collections import defaultdict
defaultdict(list,
            {'row1': ['Affinity'],
             'row2': ['Ahmc',
              'Garfield',
              'Medical Center'],
             'row3': ['Alamance','Macbeth'],
             'row4': [],
             'row5': ['Mayday']})

我想将其转换为数据帧。输出应类似于-

ID  SYN1    SYN2    SYN3    SYN4    SYN5
row1    Affinity                
row2    Ahmc    Garfield    Medical Center      
row3    Alamance    Macbeth         
row4                    
row5    Mayday

Answer 1

collections.defaultdict是dict的子类。

因此您可以只使用pd.DataFrame.from_dict：

res = pd.DataFrame.from_dict(dd, orient='index')
res.columns = [f'SYN{i+1}' for i in res]

print(res)

          SYN1      SYN2            SYN3
row1  Affinity      None            None
row2      Ahmc  Garfield  Medical Center
row3  Alamance   Macbeth            None
row4      None      None            None
row5    Mayday      None            None

Answer 2

是的，您可以使用Series

df=pd.Series(d).apply(pd.Series).fillna('')
Out[55]: 
             0         1               2
row1  Affinity                          
row2      Ahmc  Garfield  Medical Center
row3  Alamance   Macbeth                
row4                                    
row5    Mayday

或者通过数据框构造函数

df=pd.DataFrame(data=list(d.values()),index=d.keys())
Out[64]: 
             0         1               2
row1  Affinity      None            None
row2      Ahmc  Garfield  Medical Center
row3  Alamance   Macbeth            None
row4      None      None            None
row5    Mayday      None            None

然后我们创建列

df.columns='SYN'+(df.columns+1).astype(str)
df
Out[67]: 
          SYN1      SYN2            SYN3
row1  Affinity      None            None
row2      Ahmc  Garfield  Medical Center
row3  Alamance   Macbeth            None
row4      None      None            None
row5    Mayday      None            None

从具有可变长度的列表字典创建数据框

2 个答案: