将我的字典变成熊猫数据框

Question

我有一个函数，它根据某些条件创建多个字典。

但是，我真的很想在收集后将 dict 转换为数据框。但我找不到一个简单的方法来做到这一点......现在我认为解决方案是将字典中的每个键乘以最内部字典中的键数，但希望有更好的方法

由于我的函数创建了 dict，如果有更好的方法，我可以以任何方式更改它。

这是我现在的字典

{'TSLA': {2011: {'negative': {'lowPrice': 185.16,
    'lowDate': '05/27/19',
    'highPrice': 365.71,
    'highDate': '12/10/18',
    'change': -0.49}},
  2012: {'negative': {'lowPrice': 185.16,
    'lowDate': '05/27/19',
    'highPrice': 365.71,
    'highDate': '12/10/18',
    'change': -0.49}},
  2013: {'negative': {'lowPrice': 32.91,
    'lowDate': '01/07/13',
    'highPrice': 37.24,
    'highDate': '03/26/12',
    'change': -0.12},
   'positive': {'lowPrice': 32.91,
    'lowDate': '01/07/13',
    'highPrice': 190.9,
    'highDate': '09/23/13',
    'change': 4.8}}}}

我想要的输出是这样的，当然还有值：

                    lowPrice lowDate highPrice highDate change
ATVI  2012 Negative      NaN     NaN       NaN      NaN  NaN
           Positive      NaN     NaN       NaN      NaN  NaN
      2013 Negative      NaN     NaN       NaN      NaN  NaN
TSLA  2014 Positive      NaN     NaN       NaN      NaN  NaN
      2012 Negative      NaN     NaN       NaN      NaN  NaN
      2013 Positive      NaN     NaN       NaN      NaN  NaN
      2014 Positive      NaN     NaN       NaN      NaN  NaN

Answer 1

对于键的元组，您可以将嵌套字典展平 2 次并传递给DataFrame.from_dict：

d1 = {(k1, k2, k3): v3 
      for k1, v1 in d.items() 
      for k2, v2 in v1.items()
      for k3, v3 in v2.items()}

df = pd.DataFrame.from_dict(d1, orient='index')
#alternative
#df = pd.DataFrame(d1).T

print (df)
                   lowPrice   lowDate highPrice  highDate change
TSLA 2011 negative   185.16  05/27/19    365.71  12/10/18  -0.49
     2012 negative   185.16  05/27/19    365.71  12/10/18  -0.49
     2013 negative    32.91  01/07/13     37.24  03/26/12  -0.12
          positive    32.91  01/07/13     190.9  09/23/13    4.8

Answer 2

类似，但您也可以使用from_dict：

df=pd.DataFrame.from_dict({(i, j, x) : y
                           for i in d.keys()
                           for j in d[i].keys()
                           for x, y in d[i][j].items()},
                           orient='index')

print (df)

                    lowPrice   lowDate  highPrice  highDate  change
TSLA 2011 negative    185.16  05/27/19     365.71  12/10/18   -0.49
     2012 negative    185.16  05/27/19     365.71  12/10/18   -0.49
     2013 negative     32.91  01/07/13      37.24  03/26/12   -0.12
          positive     32.91  01/07/13     190.90  09/23/13    4.80