对字典值求和，使用两个键作为索引：如何实现这一点？

Question

I have the following dictionary:

res = [{'name': 'mfi', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}, 
{'name': 'serv', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}, 
{'name': 'inv', 'percentage': 100.0, 'tax_base': 1200.0, 'tax_amount': 168.0}, 
{'name': 'mfi', 'percentage': 50.0, 'tax_base': 1500.0, 'tax_amount': 210.0}, 
{'name': 'none', 'percentage': 0.0, 'tax_base': 1000.0, 'tax_amount': 0.0}, 
{'name': 'none', 'percentage': 0.0, 'tax_base': 900.0, 'tax_amount': 126.0}, 
{'name': 'mfi', 'percentage': 50.0, 'tax_base': 1000.0, 'tax_amount': 140.0}]

From this dictionary, I need to sum 'tax_base' and 'tax_amount' value keys, and use keys 'name' and 'percentage' as index.

As a result, I need:

res_final = [{'name': 'mfi', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0},
{'name': 'mfi', 'percentage': 50.0, 'tax_base': 2500.0, 'tax_amount': 350.0},  
{'name': 'serv', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}, 
{'name': 'inv', 'percentage': 100.0, 'tax_base': 1200.0, 'tax_amount': 168.0}, 
{'name': 'none', 'percentage': 0.0, 'tax_base': 1900.0, 'tax_amount': 126.0}, 
]

How can I achieve this? Can you provide me a sample please?

Answer 1

Taking your original input, res and your preferred output res_final , the following code would work:

# Creating a temporary dictionary with the aggregation
temp_result = {}
for each in res:
    key = (each['name'], each['percentage'])
    if key not in temp_result:
        temp_result[key] = dict(tax_base=0, tax_amount=0)

    temp_result[key]['tax_base'] += each['tax_base']
    temp_result[key]['tax_amount'] += each['tax_amount']

# Transforming the temporary dictionary to the correct format
final_result = []
for (name, percentage), taxes in temp_result.items():
    final_result.append(dict(
            name=name,
            percentage=percentage,
            tax_base=taxes['tax_base'],
            tax_amount=taxes['tax_amount']
    ))

for each in final_result:
    print(each)

The result will be:

{'name': 'mfi', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}
{'name': 'serv', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}
{'name': 'inv', 'percentage': 100.0, 'tax_base': 1200.0, 'tax_amount': 168.0}
{'name': 'mfi', 'percentage': 50.0, 'tax_base': 2500.0, 'tax_amount': 350.0}
{'name': 'none', 'percentage': 0.0, 'tax_base': 1900.0, 'tax_amount': 126.0}

Explanation

In the first part we create a new dictionary, that has as key the combination of name and percentage as a tuple , and as value a dictionary with the tax_base and tax_amount for that key.

Then we check if the key is already in our dictionary and if it isn't we create the key. The final step is summing the tax_base and tax_amount .

Now we have one dictionary with all the information, but not in the right format. The second part takes care of that. We split the key again into the name and percentage and merge the data with tax_base and tax_amount to one dict.

---

Edit

In case people are wondering how to do it with pd.DataFrame .

import pandas as pd

df = pd.DataFrame(res)
res = df.groupby(['name', 'percentage'], as_index=False).sum()
final_result = res.to_dict('records')

for each in final_result:
    print(each)

Will result in the same output, but it is not guaranteed to be in the same order as the input.

Answer 2

Here's another (very similar) way of doing it but a little bit more concise.

Here we first create a list of unique name:percentage pairs and then loop over those unique keys, filtering out entries from the res list that do not match that unique key.

unique_keys = list(set([f"{d['name']}:{d['percentage']}" for d in res])) # use list(set()) to keep only unique values as keys
output = []

for key in unique_keys:
    name, percentage = key.split(":")
    matching_entries = list(filter(lambda d: d['name'] == name and str(d['percentage']) == percentage, res))
    
    summed = {"name": name, "percentage": float(percentage), "tax_base": 0.0, "tax_amount": 0.0}
    for entry in matching_entries:
        summed["tax_base"] += entry.get("tax_base", 0) # use get in case value is not in dictionary, 0 is default value
        summed["tax_amount"] += entry.get("tax_amount", 0)
    
    output.append(summed)

output.sort(key=lambda d: d['name']) # sorting to organize a bit

Output:

[{'name': 'inv', 'percentage': 100.0, 'tax_base': 1200.0, 'tax_amount': 168.0},
 {'name': 'mfi', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0},
 {'name': 'mfi', 'percentage': 50.0, 'tax_base': 2500.0, 'tax_amount': 350.0},
 {'name': 'none', 'percentage': 0.0, 'tax_base': 1900.0, 'tax_amount': 126.0},
 {'name': 'serv', 'percentage': 100.0, 'tax_base': 1000.0, 'tax_amount': 140.0}]