Python字典：将满足给定条件的值加到键的总和中

Question

I've following nested dictionary, where the first number is resource ID (the total number of IDs is greater than 100 000):

dict = {1: {'age':1,'cost':14,'score':0.3},
        2: {'age':1,'cost':9,'score':0.5},
        ...}

I want to add to each resource a sum of costs of resources with lower score than given resource. I can add 'sum_cost' key which is equal to 0 by following code:

for id in adic:
    dict[id]['sum_cost'] = 0

It gives me following:

dict = {1: {'age':1,'cost':14,'score':0.3, 'sum_cost':0},
        2: {'age':1,'cost':9,'score':0.5,'sum_cost':0},
        ...}

Now I would like to use ideally for loop (to make the code easily readable) to assign to each sum_cost a value equal of sum of cost of IDs with lower score than the given ID.

Ideal output looks like dictionary where 'sum_cost' of each ID is equal to the cost of IDs with lower score than given ID:

dict = {1: {'age':1,'cost':14,'score':0.3, 'sum_cost':0},
        2: {'age':1,'cost':9,'score':0.5,'sum_cost':21},
        3: {'age':13,'cost':7,'score':0.4,'sum_cost':14}}

Is there any way how to do it?

Answer 1

Notes:

Using sorted method for sorting the dictionary output corresponding to the key score

dictionary get method to get dictionary values

and using a temporary variable for cumulative addition os sum_cost

Code:

dicts = {1: {'age': 1, 'cost': 14, 'score': 0.3, 'sum_cost': 0},
         2: {'age': 1, 'cost': 9, 'score': 0.5, 'sum_cost': 0},
         3: {'age': 13, 'cost': 7, 'score': 0.4, 'sum_cost': 0}}

sum_addition = 0

for key, values in sorted(dicts.items(), key=lambda x: x[1].get('score', None)):
    if dicts[key].get('score') is not None: #By default gives None when key is not available
        dicts[key]['sum_cost'] = sum_addition
        sum_addition += dicts[key]['cost']
        print key, dicts[key]

A even more simplified method by @BernarditoLuis and @Kevin Guan advise

Code2:

dicts = {1: {'age': 1, 'cost': 14, 'score': 0.3, 'sum_cost': 0},
         2: {'age': 1, 'cost': 9, 'score': 0.5, 'sum_cost': 0},
         3: {'age': 13, 'cost': 7, 'score': 0.4, 'sum_cost': 0}}

sum_addition = 0

for key, values in sorted(dicts.items(), key=lambda x: x[1].get('score', None)):
    if dicts[key].get('score'): #By default gives None when key is not available
        dicts[key]['sum_cost'] = sum_addition
        sum_addition += dicts[key]['cost']
        print key, dicts[key]

Output:

1 {'sum_cost': 0, 'age': 1, 'cost': 14, 'score': 0.3}
3 {'sum_cost': 14, 'age': 13, 'cost': 7, 'score': 0.4}
2 {'sum_cost': 21, 'age': 1, 'cost': 9, 'score': 0.5}

Answer 2

What about using OrderedDict?

from collections import OrderedDict

origin_dict = {
    1: {'age':1,'cost':14,'score':0.3}, 
    2: {'age':1,'cost':9,'score':0.5}, 
    3: {'age':1,'cost':8,'score':0.45}
}
# sort by score
sorted_dict = OrderedDict(sorted(origin_dict.items(), key=lambda x: x[1]['score']))
# now all you have to do is to count sum_cost successively starting from 0
sum_cost = 0
for key, value in sorted_dict.items():
    value['sum_cost'] = sum_cost
    sum_cost += value['cost']

print sorted_dict