Python 计数器保持值高于 n 次计数

Question

我有一个单词字典和它们在给定语料库中出现的次数。我怎样才能保留至少出现 n 次的单词（比方说，n=10）？

dictionary = {
  'beryllium': 60,
  'inch': 56,
  'any': 51,
  'such': 31,
  'court': 26,
  'be': 25,
  'by': 23,
  'arsenic': 21,
  'person': 20,
  'land': 20,
  'Lapp': 16,
  'county': 15,
  'Associate_in_Nursing': 15,
  'executor': 15,
  'information_technology': 14,
  'state': 14,
  'angstrom': 14,
  'not': 14,
  'other': 14,
  'boundary': 14,
  'tree': 13,
  'administrator': 12,
  'are': 11,
  'helium': 11,
  'no': 11,
  'action': 10,
  'rich_person': 10,
  'use': 10,
  'astatine': 10,
  'being': 9,
  'tobacco': 9,
  'every': 9,
  'curse': 9,
  'ordain': 8,
  'justice': 8,
  'one': 8,
  'notice': 8,
  'law': 8,
  'pound': 8,
  'debt': 8,
  'creditor': 8
}
}

对于上面的虚拟示例，它应该 return 包含一个停在 'astatine' 的字典。

Answer 1

从现有字典中删除（就地改变原始字典）

thresh = 10

delete = [key for key in dictionary if dictionary[key] < thresh]
 
# delete the key
for key in delete:
    del dictionary[key]

一行：

thresh = 10
for key in [key for key in dictionary if dictionary [key] < thresh]: del dictionary[key]

Answer 2

filtered_dict = {key: value for key, value in count_dict.items() if value >= 10}

Python 计数器保持值高于 n 次计数

Python counter keep values above n counts

python

dictionary

counter