Python字典：添加到满足给定条件的值的键和

我有以下嵌套字典，其中第一个数字是资源ID（ID总数大于100000）：

我想给每个资源增加一个资源成本的总和，其得分低于给定的资源。我可以通过以下代码添加等于0的'sum_cost'键：

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

它给了我以下信息：

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

现在我想使用理想的for循环（使代码易于阅读）为每个sum_cost分配一个值，该值等于得分低于给定ID的ID的cost之和

理想输出类似于字典，其中每个ID的“总和成本”等于分数低于给定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}}

---

有办法吗？

注意：

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]

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]

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}

用于对与键

得分

获取字典值的步骤

并使用一个临时变量进行累计加法os

sum\u cost

代码：

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]

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]

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}

由@BernarditoLuis和@Kevin Guan提出的更为简化的方法

代码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') 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]

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]

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}

输出：

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]

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]

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}

---

使用OrderedICT怎么样

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

---

如果dicts[key].get（'score'，False）=False:在这里，如果dicts[key].get（'score'，False:？：）@KevinGuan可以，但只是想遵循python zen

隐式比显式好

很好的建议思想简单的

dicts[key]怎么样。get（'score'）

？如果该键不在dict中，它将返回

None

，请参阅。顺便说一句，如果你真的想更含蓄一些，你可以使用

如果dicts[key]。get（'score'，False）不是False：

（我认为它比

！=

。但是我仍然建议

如果dicts[key]。get（'score'，False）：

（或者只要

如果dicts[key]。get（'score'）：

）@KevinGuan我可以把你的评论作为额外的代码添加到我的答案中吗