包含键中项目的字典中所有值的总和

Question

给定一个字典和一个字符串作为参数，返回一个新字典，其中包含指定为类别的项目（第二个参数，'city'、'sport'、'name' 之一）作为键和它的关联值。如果该项目不止一次出现，则取值的总和。

例如。

>>> get_wins_by_category(d, 'city')
{'Toronto': 34, 'Ottawa': 45}
>>> get_wins_by_category(d, 'sport')
{'basketball': 31, 'hockey': 48}
>>> get_wins_by_category(d, 'name')
{'Raptors': 10, 'Blues': 21, 'Senators': 45, 'Leafs': 3}

到目前为止我得到了什么：

d = {('Raptors', 'Toronto', 'basketball'): 10,
     ('Blues', 'Toronto', 'basketball'): 21,
     ('Senators', 'Ottawa', 'hockey'): 45,
     ('Leafs', 'Toronto', 'hockey'): 3}

def get_wins_by_category(dct, category):
    new_dict = {}
    if category == 'city':
        for key in dct.keys():
            new_dict[key[1]] = #todo
    elif category == 'sport':
        for key in dct.keys():
            new_dict[key[2]] = #todo
    elif category == 'name':
        for key in dct.keys():
            new_dict[key[0]] = #todo
    return new_dict

我的问题是等号后写什么。我知道，如果该项目不止一次出现，它采用包含该项目的所有值的总和，但我不知道如何将其编写为代码。另请注意，三元组将始终按以下顺序排列：姓名、城市、运动。

Answer 1

使用collections.defaultdict，不需要if-else：

from collections import defaultdict
def get_wins_by_category(team_to_win, category):

    d = {'name':0, 'city':1, 'sport':2}
    dic = defaultdict(int)
    for k, v in team_to_win.items():
        dic[k[d[category]]] += v
    return dic
... 
>>> get_wins_by_category(d, 'city')
defaultdict(<type 'int'>, {'Toronto': 34, 'Ottawa': 45})
>>> get_wins_by_category(d, 'sport')
defaultdict(<type 'int'>, {'basketball': 31, 'hockey': 48})
>>> get_wins_by_category(d, 'name')
defaultdict(<type 'int'>, {'Senators': 45, 'Blues': 21, 'Raptors': 10, 'Leafs': 3})

另一种选择是collections.Counter：

from collections import Counter
def get_wins_by_category(team_to_win, category):
    #index each category points to
    d = {'name':0, 'city':1, 'sport':2}
    dic = Counter()       
    for k, v in team_to_win.items():
        dic[k[d[category]]] += v
    return dic
... 
>>> get_wins_by_category(d, 'city')
Counter({'Ottawa': 45, 'Toronto': 34})
>>> get_wins_by_category(d, 'sport')
Counter({'hockey': 48, 'basketball': 31})
>>> get_wins_by_category(d, 'name')
Counter({'Senators': 45, 'Blues': 21, 'Raptors': 10, 'Leafs': 3})

Answer 2

使用错误的数据结构总是会让你的代码更难写、更难读，而且通常也会更高效地运行。

如果您想按值查找某些内容，请使用以该值作为键的 dict（或命名元组），不要遍历整个列表并逐个搜索。如果您需要创建多个 dicts，请执行此操作。

例如：

from collections import Counter
teams, cities, sports = Counter(), Counter(), Counter()
for keys, score in d.items():
    team, city, sport = keys
    teams[team] += score
    cities[city] += score
    sports[sport] += score
categories = {'team': teams, 'city': cities, 'sport': sports}

现在你的代码很简单：

def get_wins_by_category(category):
    return categories[category]

或者，或者，保留每个分数的所有分数，这样除了对分数求和（例如，平均分数）之外，您还可以做其他事情：

from collections import Counter
teams, cities, sports = defaultdict(list), defaultdict(list), defaultdict(list)
for keys, score in d.items():
    team, city, sport = keys
    teams[team].append(score)
    cities[city].append(score)
    sports[sport].append(score)
categories = {'team': teams, 'city': cities, 'sport': sports}

def get_wins_by_category(category):
    return {key: sum(scores) for key, scores in categories[category].items()}

def get_avg_wins_by_category(category):
    return {key: sum(scores)/len(scores) 
            for key, scores in categories[category].items()}

Answer 3

if category == 'city':
        for key, value in dct.items():
            new_dict[key[1]] += value

你明白了..

Answer 4

以下通过在cats 中定义输入来消除对if: 块的需求

d = {('Raptors', 'Toronto', 'basketball'): 10,
     ('Blues', 'Toronto', 'basketball'): 21,
     ('Senators', 'Ottawa', 'hockey'): 45,
     ('Leafs', 'Toronto', 'hockey'): 3}

cats = ("team", "city", "sport")

def get_wins_by_category(d, cats, cat):
    if cat in cats:    
        return {t: sum(v for k, v in d.items() if t in k) 
                for t in set(key[cats.index(cat)] for key in d)}