Question

我使用的是dict-comprehension，将国家/地区名称返回为唯一键，并希望值成为该国家/地区中的城市数。如何从元组列表中计算城市数量？

country_city_tuples= [('Netherlands', 'Alkmaar'),
                      ('Netherlands', 'Tilburg'),
                      ('Netherlands', 'Den Bosch'),
                      ('Netherlands', 'Eindhoven'),
                      ('Spain', 'Madrid'),
                      ('Spain', 'Barcelona'),
                      ('Spain', 'Cordoba'),
                      ('Spain', 'Toledo'),
                      ('Italy', 'Milano'),
                      ('Italy', 'Roma')]

country_names = { 

}

预期结果为：{'Italy': 2 , 'Netherlands': 4, 'Spain': 4}

Answer 1

您可以使用zip从元组列表中提取国家名称，然后使用collections.Counter来计算国家名称的出现频率

from collections import Counter

country_city_tuples= [('Netherlands', 'Alkmaar'),
                      ('Netherlands', 'Tilburg'),
                      ('Netherlands', 'Den Bosch'),
                      ('Netherlands', 'Eindhoven'),
                      ('Spain', 'Madrid'),
                      ('Spain', 'Barcelona'),
                      ('Spain', 'Cordoba'),
                      ('Spain', 'Toledo'),
                      ('Italy', 'Milano'),
                      ('Italy', 'Roma')]

#Extract out country names using zip and list unpacking
country_names, _ = zip(*country_city_tuples)

#Count the number of countries using Counter
print(dict(Counter(country_names)))

要在不使用collections的情况下进行操作，我们可以使用字典来收集频率

country_city_tuples= [('Netherlands', 'Alkmaar'),
                      ('Netherlands', 'Tilburg'),
                      ('Netherlands', 'Den Bosch'),
                      ('Netherlands', 'Eindhoven'),
                      ('Spain', 'Madrid'),
                      ('Spain', 'Barcelona'),
                      ('Spain', 'Cordoba'),
                      ('Spain', 'Toledo'),
                      ('Italy', 'Milano'),
                      ('Italy', 'Roma')]

#Extract out country names using zip and list unpacking
country_names, _ = zip(*country_city_tuples)

result = {}

#Count each country
for name in country_names:
    result.setdefault(name,0)
    result[name] += 1

print(result)

两种情况下的输出将相同

{'Netherlands': 4, 'Spain': 4, 'Italy': 2}

Answer 2

使用defaultdict：

from collections import defaultdict

country_names  = defaultdict(int)
for i in country_city_tuples:
    country_names[i[0]]+=1

country_names
defaultdict(int, {'Netherlands': 4, 'Spain': 4, 'Italy': 2})

Answer 3

尝试一下：

l = set(i[0] for i in country_city_tuples)
d = {}
for i in l:
   d[i] = sum([1 for j in country_city_tuples if j[0]==i])

输出：

{'Italy': 2, 'Netherlands': 4, 'Spain': 4}

Answer 4

将sum与生成器一起使用，如果国家/地区名称与被检查的国家/地区匹配，则生成器返回1，否则返回0：

{name: sum(1 if c[0] == name else 0
           for c in country_city_tuples)
 for name in set(c[0] for c in country_city_tuples)}

您也可以使用dict.get：

r = {}
for name, city in country_city_tuples:
    r.get(name, 0) += 1

Answer 5

不使用defaultdict和其他模块

country_city_tuples= [('Netherlands', 'Alkmaar'),
                      ('Netherlands', 'Tilburg'),
                      ('Netherlands', 'Den Bosch'),
                      ('Netherlands', 'Eindhoven'),
                      ('Spain', 'Madrid'),
                      ('Spain', 'Barcelona'),
                      ('Spain', 'Cordoba'),
                      ('Spain', 'Toledo'),
                      ('Italy', 'Milano'),
                      ('Italy', 'Roma')]

country_names ={}
for i in country_city_tuples:
    try:
        if country_names[i[0]]:
            country_names[i[0]]+=1
    except:
        country_names[i[0]]=1
print(country_names)

#output {'Netherlands': 4, 'Spain': 4, 'Italy': 2}

使用字典理解将国家作为键返回，将城市总数作为值返回

5 个答案: