Python 列表：对列表中的值求和并删除重复项

Question

I have the following list in Python:

my_list = [{'code_id': 'A', 'amount': 100.0},{'code_id': 'B', 'amount': 150.0},{'code_id': 'C', 'amount': 200.0},{'code_id': 'A', 'amount': 120.0},{'code_id': 'B', 'amount': 300.0},{'code_id': 'D', 'amount': 180.0}]

From the list above, I need to create a new list with no duplicates; but sum the "amount" for all items in the above list.

I need to achieve a final list like:

final_list = [{'code_id': 'A', 'amount': 220.0},{'code_id': 'B', 'amount': 450.0},{'code_id': 'C', 'amount': 200.0},{'code_id': 'D', 'amount': 180.0}]

I was able to remove duplicates but do not know how to sum values in the process. The sample code I have used:

final_list = []
seen = set()
for dic in my_list:
    key = (dic['code_id'])
    if key in seen:
        continue

    final_list.append(dic)
    seen.add(key)

How can I achieve this in Python?

Answer 1

I would start with a temporary defaultdict that keeps track of the 'amount' for each 'code_id' :

from collections import defaultdict  

my_list = [{'code_id': 'A', 'amount': 100.0},{'code_id': 'B', 'amount': 150.0},{'code_id': 'C', 'amount': 200.0}, {'code_id': 'A', 'amount': 120.0},{'code_id': 'B', 'amount': 300.0},{'code_id': 'D', 'amount': 180.0}]          
tmp = defaultdict(int)                                                                                           
                                                                                                                  
for d in my_list: 
    tmp[d['code_id']] += d['amount']
    # if tmp was a normal dict, you could use
    # tmp[d['code_id']] = tmp.get(d['code_id'], 0) + d['amount']
                                                                                                                
print(tmp)                                                                                                              
# defaultdict(int, {'A': 220.0, 'B': 450.0, 'C': 200.0, 'D': 180.0})

... and then transform the structure of tmp to arrive at the desired result

result = [{'code_id': k, 'amount': v} for k, v in tmp.items()]                                                   
print(result)                                                                                                    
# [{'code_id': 'A', 'amount': 220.0}, {'code_id': 'B', 'amount': 450.0}, {'code_id': 'C', 'amount': 200.0}, {'code_id': 'D', 'amount': 180.0}]

For the pandas users out there:

>>> pd.DataFrame(my_list).groupby('code_id', as_index=False).sum().to_dict(orient='records')                         
[{'code_id': 'A', 'amount': 220.0},
 {'code_id': 'B', 'amount': 450.0},
 {'code_id': 'C', 'amount': 200.0},
 {'code_id': 'D', 'amount': 180.0}]