如何更改字典的键和值

Question

所以我得到了下面的代码，我想做的是获取输入元组的每一个秒值，并将其保存到字典中作为dict.key，并将dict.value首先设置为0。之后，我想检查保存的字典，每次这个dict.key出现在元组中，我想把它的计数放在dict.value中。代码当前给我一个错误。

def movie_awards(results):
"""
I want to ignore every first value of the tuple
"""
results = list(results)

answer = {x[1]:0 for x in results}

for x in results:
answer[x[1]] = answer[x] + 1

print(answer)
result = {("Best Picture", "Parasite"),
("Best Actor", "Joker"),
("Best Actress", "Judy"),
("Best Sound Editing", "Ford v Ferrari"),
("Best Sound Editing", "1917"),
("Best Original Score", "Joker"),
("Best Costume Design", "Little Women"),
("Best Visual Effects", "Little Women"),
("Best Director", "Parasite")}

Answer 1

使用+=:

将数字相加

for x in results:
answer[x[1]] += 1

print(answer)

做:

answer[x]

python正在尝试获取密钥为('Best Original Score', 'Joker')的key:value对。因为它不存在，所以会引发一个键错误。

Answer 2

您也可以直接使用collections.Counter:

from collections import Counter
def movie_awards(results):
return Counter(x[1] for x in results)

Answer 3

你可以使用Counter:

from collections import Counter
c = Counter([t[1] for t in result])
>>> Counter({'Joker': 2, 'Parasite': 2, 'Little Women': 2, '1917': 1, 'Ford v Ferrari': 1, 'Judy': 1})