打开包装列表为大熊猫

Question

鉴于我有一个名单，我怎么能解开它变成一个熊猫的数据帧，例如：

data = {u'2344': ["id", "value1", "value2", "01", "Addf112", "Addf113", "02", " ", "Addf213"]} 

>> id value1 value2 
    01 Ad112 Ad113 
    02   Ad213 

Answer 1

你不得不提取列名的单个元素，然后构建一个列表包括2只列出了您的2行数据：

In [23]: 

data = {u'2344': ["id", "value1", "value2", "01", "Addf112", "Addf113", "02", " ", "Addf213"]} 

pd.DataFrame(columns = data['2344'][:3], data=[data['2344'][3:6], data['2344'][6:]]) 
Out[23]: 
    id value1 value2 
0 01 Addf112 Addf113 
1 02   Addf213 

的动态方法将是使用一个组块（从答案之一修改，以此question）来构建字典并使用此构造DF ：

In [59]: 

def chunker(seq, stride): 
    cols = seq[:stride] 
    data = [seq[stride:][pos::stride] for pos in range(0, stride)] 
    return dict(zip(cols,data)) 

pd.DataFrame(chunker(data['2344'],3)) 

Out[59]: 
    id value1 value2 
0 01 Addf112 Addf113 
1 02   Addf213