Python字典，查找重复的键值的总和

Question

I have a list of dictionaries that look like the following:

[{
    "id": "42f409d0-2cef-49b0-a027-59ed571cc2a9",
    "cost": 0.868422,
    "environment": "nonprod"
},
{
    "id": "42f409d0-2cef-49b0-a027-59ed571cc2a9",
    "cost": 0.017017,
    "environment": "prod"
},
{
    "id": "aa385029-afa6-4f1a-a1d9-d88b7d934699",
    "cost": 0.010304,
    "environment": "prod"
},
{
    "id": "b13a0676-6926-49db-808c-3c968a9278eb",
    "cost": 2.870336,
    "environment": "nonprod"
},
{
    "id": "b13a0676-6926-49db-808c-3c968a9278eb",
    "cost": 0.00455,
    "environment": "prod"
},
{
    "id": "b13a0676-6926-49db-808c-3c968a9278eb",
    "cost": 0.032458,
    "environment": "prod"
}]

The part im having a hard time understanding is how can I group these by id and environment and add up there costs for the given environment.

End result should be, only having a single pair of prod and nonprod for a given ID and have all costs for prod or nonprod added up under a single id in prod or a single id in nonprod.

I hope this is enough detail, thank you!

Answer 1

d = {}
for el in data:
    d[(el["id"], el["environment"])] = d.get((el["id"], el["environment"]), 0) + el["cost"]
d
# {('42f409d0-2cef-49b0-a027-59ed571cc2a9', 'nonprod'): 0.868422,
#  ('42f409d0-2cef-49b0-a027-59ed571cc2a9', 'prod'): 0.017017,
#  ('aa385029-afa6-4f1a-a1d9-d88b7d934699', 'prod'): 0.010304,
#  ('b13a0676-6926-49db-808c-3c968a9278eb', 'nonprod'): 2.870336,
#  ('b13a0676-6926-49db-808c-3c968a9278eb', 'prod'): 0.037008}

Answer 2

Try placing the dict values into a pandas dataframe, then use pandas's groupby function (I set the variable dicts to equal your above list of dictionaries):

import pandas as pd
df = pd.DataFrame(dicts)
df.groupby(["id", "environment"], as_index=False).sum()

Output:

                                      id    environment  cost
0   42f409d0-2cef-49b0-a027-59ed571cc2a9    nonprod      0.868422
1   42f409d0-2cef-49b0-a027-59ed571cc2a9    prod         0.017017
2   aa385029-afa6-4f1a-a1d9-d88b7d934699    prod         0.010304
3   b13a0676-6926-49db-808c-3c968a9278eb    nonprod      2.870336
4   b13a0676-6926-49db-808c-3c968a9278eb    prod         0.037008

Answer 3

In addition of above answers, a solution of python by creating a unique key of each transaction id|type will also work. This piece of code does exactly that and can even be made easier to read with defaultdict .

dictCounter = dict()
#assuming test is the list of dicts
for eachEntry in test:
    newUniqueKey = eachEntry["id"]+"|"+eachEntry["environment"]
    if newUniqueKey not in dictCounter.keys():
        dictCounter[newUniqueKey]=eachEntry["cost"]
    else:
        dictCounter[newUniqueKey]+=eachEntry["cost"]