Python的大熊猫转换逗号分隔值的列表，数据帧

Question

我有一个字符串列表，它看起来像这样：

["Name: Alice, Department: HR, Salary: 60000", "Name: Bob, Department: Engineering, Salary: 45000"] 

我想这个列表转换成数据帧，看起来像这样：

Name | Department | Salary 
-------------------------- 
Alice | HR | 60000 

Bob | Engineering | 45000 

最简单的方法是什么？ 我的直觉说丢数据到CSV，并与正则表达式单独标题“^ *：”，但必须有一个更简单的方法

Answer 1

随着一些字符串处理就可以得到类型的字典列表，并传递到数据帧的构造函数：

lst = ["Name: Alice, Department: HR, Salary: 60000", 
     "Name: Bob, Department: Engineering, Salary: 45000"] 
pd.DataFrame([dict([kv.split(': ') for kv in record.split(', ')]) for record in lst]) 
Out: 
    Department Name Salary 
0   HR Alice 60000 
1 Engineering Bob 45000 

Answer 2

你能做到这样：

In [271]: s 
Out[271]: 
['Name: Alice, Department: HR, Salary: 60000', 
'Name: Bob, Department: Engineering, Salary: 45000'] 

In [272]: pd.read_csv(io.StringIO(re.sub(r'\s*(Name|Department|Salary):\s*', r'', '~'.join(s))), 
    ...:    names=['Name','Department','Salary'], 
    ...:    header=None, 
    ...:    lineterminator=r'~' 
    ...:) 
    ...: 
Out[272]: 
    Name Department Salary 
0 Alice   HR 60000 
1 Bob Engineering 45000 

Answer 3

有点创意

s.str.extractall(r'(?P<key>[^,]+)\s*:(?P<value>[^,]+)') \ 
    .reset_index('match', drop=True) \ 
    .set_index('key', append=True).value.unstack() 

设置

l = ["Name: Alice, Department: HR, Salary: 60000", 
    "Name: Bob, Department: Engineering, Salary: 45000"] 
s = pd.Series(l)