将我的字典变成 pandas 数据框 [英] Turning my dictionary into a pandas dataframe

问题描述

我有一个函数，可以根据某些条件创建多个命令.

I have a function which create several dicts of dicts, based on some conditions.

但是，我真的很想在收集字典后将其转换为数据框. 但是我找不到一种简单的方法...现在，我在想解决方案是将字典中的每个键乘以最内部字典中键的数量，但是希望有更好的方法

However, I would really like to turn the dict into a dataframe after collecting it. But I can't find an easy way to do so... Right now I'm thinking the solution is to multiply each key in the dict by the number of keys in the most inner dict, but hopefully there is a better way

由于我的函数创建了字典，因此，如果有更好的方法可以更改它.

Since my function creates the dict I can alter it in any way, if there is a better way to do this.

这是我的字典

{'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}}}}

我想要的输出将是这样的，当然带有值:

My desired output would be something like this, of course with the values:

                    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

推荐答案

您可以将嵌套字典平整两次以获取键的元组，然后传递给

You can flatten nested dictionaries 2 times for tuples for keys and pass to 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

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