当字典值是列表时，我如何从int中获得字典键?

Question

我有python字典:

my_dict = {'C': [1,2,3,4,5,6,7,8,9,10,11,12,15,18,46,64,67,73,78,83],
'B': [13,14,22,32,38,39,42,59,68,74,79,84],
'A': [16,17,19,23,31,37,40,50,51,60,70,75,80,85],
'S': [20,25,33,36,43,44,48,49,57,61,62,63,71,76,81,86,88,89,92,94],
'SS': [21,24,26,47,52,53,54,56,66,69,72,77,82,87,91,93],
'SS+': [27,28,29,30,34,35,41,45,55,58,65,90,95,96,97,98],
'SSS': [99,100]}

当我给它整数值78作为代码的输入时，我如何得到键'C'作为输出:

var_playerXp = 0
my_dict_key = int(input("Give an integer between 1 and 100: "))
if my_dict_key.lower() == 'c':
var_playerXp += 5
elif my_dict_key.lower() == 'b':
var_playerXp += 10
elif my_dict_key.lower() == 'a':
var_playerXp += 25

等。通过"SSS"。

我试图检查与整数值字典相同的方法:

my_dict_key = list(my_dict.keys())[list(my_dict.values()).index(my_dict_key)]

然而，我只得到错误:ValueError: 78不在列表中

Answer 1

my_dict_key = 78
out = [k for k, v in my_dict.items() if my_dict_key in v]
print(*out)

C