我有這本詞典串列詞典（我無法更改作品的結構）：
dict_countries = {'gb': [{'datetime': '1955-10-10 17:00:00', 'city': 'chester'},
{'datetime': '1974-10-10 23:00:00', 'city': 'chester'}],
'us': [{'datetime': '1955-10-10 17:00:00', 'city': 'hudson'}]
}
和功能：
def Seen_in_the_city(dict_countries:dict,)-> dict:
city_dict = {}
for each_country in dict_countries.values():
for each_sight in each_country:
citi = each_sight["city"]
if citi in city_dict.keys():
city_dict[each_sight["city"]] = 1
else:
city_dict[citi] = 1
return city_dict
我得到：
{'chester': 1,'hudson': 1}
代替
{'chester': 2,'hudson': 1}
uj5u.com熱心網友回復：
您可以嘗試使用Python 標準庫中模塊中的Counter（的子類dict） ：collections
from collections import Counter
c = Counter()
for key in dict_countries:
for d in dict_countries[key]:
c.update(v for k, v in d.items() if k == 'city')
print(c)
輸出
Counter({'chester': 2, 'hudson': 1})
uj5u.com熱心網友回復：
嘗試：
output = dict()
for country, cities in dict_countries.items():
for city in cities:
if city["city"] not in output:
output[city["city"]] = 0
output[city["city"]] = 1
uj5u.com熱心網友回復：
您無需說 1 即可添加正數。同樣在 if citi 陳述句中， = 1 表示在現有值 (1 1) 上加 1，其中 = 1 基本上是說再次給它一個值 1。
if citi in city_dict.keys():
city_dict[each_sight["city"]] =1
else:
city_dict[citi] = 1
uj5u.com熱心網友回復：
你可以使用groupby從itertools
from itertools import groupby
print({i: len(list(j)[0]) for i,j in groupby(dict_countries.values(), key=lambda x: x[0]["city"])})
uj5u.com熱心網友回復：
如果您不想要額外的匯入（不是說您不應該使用 Counter），還有另一種方法：
dict_countries = {'gb': [{'datetime': '1955-10-10 17:00:00', 'city': 'chester'},
{'datetime': '1974-10-10 23:00:00', 'city': 'chester'}],
'us': [{'datetime': '1955-10-10 17:00:00', 'city': 'hudson'}]
}
def Seen_in_the_city(dict_countries:dict,)-> dict:
city_dict = {}
for each_country in dict_countries.values():
for each_sight in each_country:
citi = each_sight["city"]
city_dict[citi] = city_dict.get(citi, 0) 1
return city_dict
print(Seen_in_the_city(dict_countries))
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