Python字典：將滿足給定條件的值加到鍵的總和中

Question

我遵循的是嵌套字典，其中第一個數字是資源ID（ID的總數大於100 000）：

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

我想向每種資源添加總成本低於給定資源的資源成本總和。 我可以通過以下代碼添加等於0的'sum_cost'鍵：

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

它給我以下內容：

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

現在，我想理想地使用for循環（以使代碼易於閱讀）為每個sum_cost分配一個值，該值等於比給定ID得分低的ID的成本之和。

理想的輸出看起來像字典，其中每個ID的“ sum_cost”等於得分低於給定ID的ID的成本：

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}}

有什么辦法嗎？

Answer 1

筆記：

使用排序方法對與關鍵score對應的字典輸出進行排序

dictionary get方法獲取字典值

並使用臨時變量進行累加os sum_cost

碼：

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]

@BernarditoLuis和@Kevin Guan提出的甚至更簡化的方法建議

代碼2：

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]

輸出：

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

怎樣使用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