如果两个键值相同，如何对词典列表中的元素求和

Question

I have the following list of dictionaries:

dictionary =[{'Flow': 100, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 90, 'Location': 'Europe', 'Name': 'B1'},
            {'Flow': 20, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 70, 'Location': 'Europe', 'Name': 'B1'}]

I want to create a new list of dictionaries, with summed Flow values of all dictionaries where Location and Name are the same. My desired output would be:

new_dictionary =[{'Flow': 120, 'Location': 'USA', 'Name': 'A1'},
            {'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},]

How can I achieve this?

Answer 1

This is possible, but non-trivial to implement in python. Might I suggest using pandas? This is simple with a groupby , sum , and to_dict .

import pandas as pd

(pd.DataFrame(dictionary)
   .groupby(['Location', 'Name'], as_index=False)
   .Flow.sum()
   .to_dict('r'))

[{'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},
 {'Flow': 120, 'Location': 'USA', 'Name': 'A1'}]

To install, use pip install --user pandas .

---

Otherwise, you can apply a pseudo-generic group operation using itertools.groupby .

from itertools import groupby
from operator import itemgetter

grouper = ['Location', 'Name']
key = itemgetter(*grouper)
dictionary.sort(key=key)

[{**dict(zip(grouper, k)), 'Flow': sum(map(itemgetter('Flow'), g))} 
    for k, g in groupby(dictionary, key=key)]

[{'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},
 {'Flow': 120, 'Location': 'USA', 'Name': 'A1'}]

Answer 2

While I would also prefer using Pandas if possible, here is solution using plain python:

In [1]: import itertools

In [2]: dictionary =[{'Flow': 100, 'Location': 'USA', 'Name': 'A1'},
   ...:             {'Flow': 90, 'Location': 'Europe', 'Name': 'B1'},
   ...:             {'Flow': 20, 'Location': 'USA', 'Name': 'A1'},
   ...:             {'Flow': 70, 'Location': 'Europe', 'Name': 'B1'}]
   ...:

In [3]: import operator

In [4]: key = operator.itemgetter('Location', 'Name')

In [5]: [{'Flow': sum(x['Flow'] for x in g),
   ...:   'Location': k[0],
   ...:   'Name': k[1]}
   ...:  for k, g in itertools.groupby(sorted(dictionary, key=key), key=key)]
   ...:
   ...:
Out[5]:
[{'Flow': 160, 'Location': 'Europe', 'Name': 'B1'},
 {'Flow': 120, 'Location': 'USA', 'Name': 'A1'}]

Another way is to use defaultdict, which gives you a slightly different representation (though you can convert it back to list of dicts if you want):

In [11]: import collections

In [12]: cnt = collections.defaultdict(int)

In [13]: for r in dictionary:
    ...:     cnt[(r['Location'], r['Name'])] += r['Flow']
    ...:

In [14]: cnt
Out[14]: defaultdict(int, {('Europe', 'B1'): 160, ('USA', 'A1'): 120})

In [15]: [{'Flow': x, 'Location': k[0], 'Name': k[1]} for k, x in cnt.items()]
Out[15]:
[{'Flow': 120, 'Location': 'USA', 'Name': 'A1'},
 {'Flow': 160, 'Location': 'Europe', 'Name': 'B1'}]

Answer 3

Not exactly the output you expect, but..

Using collections.Counter()

count = Counter()

for i in dictionary:
    count[i['Location'], i['Name']] += i['Flow']

print count

Will give:

Counter({ ('Europe', 'B1'): 160, 
          ('USA', 'A1'): 120 })

I hope this will at least give you some idea.