将字典的长度修改为确定的值并保留具有最高值的键

Question

大家好，我正在编写这个脚本，我需要更新一个字典words，其中最常用的单词仅限于limit.

from typing import List, Dict, TextIO, Tuple

def most_frequent(words: Dict[str, int], limit: int) -> None:

new_dict = {}

new_list = []

#I decided to create a list for easier sort

for w in words:

    new_list.append((keys, words.get(w)))

    new_list.sort(key=sorting, reverse=True)

    #key=sorting: used to sort by the value of the key from big to small 

for n_w in new_list:

    if len(new_dict) < limit:

        new_dict[n_w[0]] = n_w[1]

#this part add the words to a new dictionary up to the value of limit

words = new_dict

print(words)

#print just to check my result, I know it's supposed to return None

这是问题所在，我需要实现以下测试用例，其中：len(words) <= limit，如果添加了最常用的单词并导致结果，len(words) > limit则不添加任何单词；如果最后一个单词不是唯一的并且与下一个单词具有相同的值，则也不会添加任何一个。

>>> most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 3, 'rat': 1}, 4)

{'cat': 3, 'dog': 3, 'pig': 3, 'bee': 3}

#This one passes

>>> most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 2, 'rat': 2}, 4)

{'cat': 3, 'dog': 3, 'pig': 3}

#what I get {'cat': 3, 'dog': 3, 'pig': 3, 'bee': 2},  'bee' doesn't get added because is tied with 'rat'

>>> most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 3, 'rat': 1}, 3)  

{}

#what I get {'cat': 3, 'dog': 3, 'pig': 3}, none of them are added because there are 4 with high frequency but if they get added words > limit and it can't be

我觉得我现在使用的方法对于我需要的东西效率不高，而且我陷入了最后两种情况。我不允许使用模块，我应该使用什么方法？或者至少我可以在这里改进什么以获得我需要的东西？

Answer 1

我会做这样的事情：def most_frequent(words, limit):&nbsp; &nbsp; frequencies = words.items()&nbsp; &nbsp; inverse = {}&nbsp; &nbsp; for word, frequency in frequencies:&nbsp; &nbsp; &nbsp; &nbsp; inverse.setdefault(frequency, []).append(word)&nbsp; &nbsp; result = {}&nbsp; &nbsp; remaining = limit&nbsp; &nbsp; for frequency in sorted(inverse.keys(), reverse=True):&nbsp; &nbsp; &nbsp; &nbsp; if len(inverse[frequency]) <= remaining:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; result.update({word: frequency for word in inverse[frequency]})&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; remaining -= len(inverse[frequency])&nbsp; &nbsp; &nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break&nbsp; &nbsp; return resultprint(most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 3, 'rat': 1}, 4))print(most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 2, 'rat': 2}, 4))print(most_frequent({'cat': 3, 'dog': 3, 'pig': 3, 'bee': 3, 'rat': 1}, 3))输出{'bee': 3, 'dog': 3, 'pig': 3, 'cat': 3}{'dog': 3, 'pig': 3, 'cat': 3}{}这个想法是创建一个倒排字典（inverse），其中键是频率，值是具有该频率的单词列表，然后您可以按非升序迭代频率，并将单词列表添加到最终结果中如果剩余预算允许。