创建包含字典列表的字典列表

Question

I'm having some trouble trying to bend my brain around this issue I've been working on. I have a list of stock data that I'm pulling from a SQLite database. This is a list of dictionaries and the dictionaries look something like this:

[ {symbol:'AAPL', date:'2021-04-28', close:130},
  {symbol:'AAPL', date:'2021-04-27', close:129}, 
  {symbol:'MSFT', date:'2021-04-28', close:155}, 
  {symbol:'MSFT', date:'2021-04-27', close:156}  ]  

I'm working with open, high, low, and close data and I want to write a script to assign indicator values to each data point/timestamp. In order to assign these indicators, I need to manipulate this data into a new list as follows:

Stock data = [  {'AAPL':[ {date:'2021-04-28', close:130},
                          {date:'2021-04-27', close:129} ] },
                {'MSFT':[ {date:'2021-04-28', close:155}, 
                          {date:'2021-04-27', close:156} ] } ]

I want to create a list of dictionaries with the key being the stock ticker, and the value being a list of all the OHLC data (as dictionary) contained in the database.

It this seems to me to be a list of dictionaries containing lists of dictionaries.

I'm not worried about efficiency as I only want to run this once per day, in the middle of the night.

Answer 1

This can be done with itertools.groupby with a little follow-up to remove group key.

from itertools import groupby

lst = [{'symbol': 'AAPL', 'date': '2021-04-28', 'close': 130},
       {'symbol': 'AAPL', 'date': '2021-04-27', 'close': 129},
       {'symbol': 'MSFT', 'date': '2021-04-28', 'close': 155},
       {'symbol': 'MSFT', 'date': '2021-04-27', 'close': 156}]

group_key = 'symbol'

print([{k: [{sub_k: sub_v for sub_k, sub_v in d.items() if sub_k != group_key}
            for d in v]}
       for k, v in groupby(lst, key=lambda x: x[group_key])])

Output:

[{'AAPL': [{'date': '2021-04-28', 'close': 130}, 
           {'date': '2021-04-27', 'close': 129}]}, 
 {'MSFT': [{'date': '2021-04-28', 'close': 155}, 
           {'date': '2021-04-27', 'close': 156}]}]

---

Edit: Assuming the list is not already ordered by 'symbol' a sorting phase is necessary:

from itertools import groupby

lst = [{'symbol': 'AAPL', 'date': '2021-04-28', 'close': 130},
       {'symbol': 'MSFT', 'date': '2021-04-28', 'close': 155},
       {'symbol': 'AAPL', 'date': '2021-04-27', 'close': 129},
       {'symbol': 'MSFT', 'date': '2021-04-27', 'close': 156}]

group_key = 'symbol'


def get_key(x):
    return x[group_key]


print([{k: [{sub_k: sub_v for sub_k, sub_v in d.items() if sub_k != group_key}
            for d in v]}
       for k, v in groupby(sorted(lst, key=get_key), key=get_key)])

Output:

[{'AAPL': [{'date': '2021-04-28', 'close': 130}, 
           {'date': '2021-04-27', 'close': 129}]}, 
 {'MSFT': [{'date': '2021-04-28', 'close': 155}, 
           {'date': '2021-04-27', 'close': 156}]}]