以指定的方式从字典中导出值和键

Question

Python新手，遇到这个问题。

场景:我有一个包含多个未知整数的列表。我需要把它们拿出来，进行分类，提取出最频繁出现的。如果一个项目有多个实例，则应该首先选择较高的值。

到目前为止，我已经制作了一个字典来处理示例请求列表，但我不确定如何提取上述指定的键和值。

def frequency(requests):
freq = {}
for x in requests:
if (x in freq):
freq[x] += 1
else:
freq[x] = 1
print(freq)   # provides expected result

#my attempt to sort dictionary and extract required values  
sorted_freq = dict(sorted(freq.items(), key=lambda x:x[1], reverse=True))
print(sorted_freq) #printing the keys at this stage doesn't factor in if the key is bigger/smaller for items with same frequency
print(sorted_freq.keys())
return

requests = [2,3,6,5,2,7,2,3,6,5,2,7,11,2,77] #example of request
frequency(requests)

#Output for freq = {2: 5, 3: 2, 6: 2, 5: 2, 7: 2, 11: 1, 77: 1}
#Output for sorted_freq = {2: 5, 3: 2, 6: 2, 5: 2, 7: 2, 11: 1, 77: 1}
#Output for sorted_freq.keys = [2, 3, 6, 5, 7, 11, 77]

在上面，3,6,5 &7都有两次出现，类似地11 &都是同一件事。我要找的输出是[2,7,6,5,3,77,11]

我已经在上面添加了额外的打印来可视化问题，只需要在实际代码中最终打印。

不知道解决这个问题的最佳方法是什么，任何帮助都会很感激。由于

Answer 1

from collections import Counter
from itertools import groupby
requests = [2,3,6,5,2,7,2,3,6,5,2,7,11,2,77]
c = Counter(requests)
freq = list()
for i,g in groupby(c.items(), key=lambda t:t[1]):
freq.extend(sorted([j for j,k in g],reverse=True))
print(freq)

尽量使用内置插件，因为它们真的很有用，不要重新发明轮子:)

输出:

[2, 7, 6, 5, 3, 77, 11]