在忽略大小写敏感度的列表中查找最频繁的字符串

Question

我有一个名为 的 Twitter 主题标签列表li。我想top_10从中制作一个最常见的标签的新列表。到目前为止我已经完成了（#）：li = ['COVID19', 'Covid19', 'covid19', 'coronavirus', 'Coronavirus',...]tag_counter = dict()for tag in li:    if tag in tag_counter:         tag_counter[tag] += 1    else:         tag_counter[tag] = 1 popular_tags = sorted(tag_counter, key = tag_counter.get, reverse = True)top_10 = popular_tags[:10]print('\nList of the top 10 popular hashtags are :\n',top_10)由于主题标签不区分大小写，我想在创建我的tag_counter.

Answer 1

collections.Counter从标准库中使用

from collections import Counter

list_of_words = ['hello', 'hello', 'world']

lowercase_words = [w.lower() for w in list_of_words]

Counter(lowercase_words).most_common(1)

退货：

[('hello', 2)]

Answer 2

首先规范化数据，使用 lower 或 upper。

li = ['COVID19', 'Covid19', 'covid19', 'coronavirus', 'Coronavirus']

li = [x.upper() for x in li] # OR, li = [x.lower() for x in li]

tag_counter = dict()

for tag in li:

    if tag in tag_counter:

         tag_counter[tag] += 1

    else:

         tag_counter[tag] = 1

 

popular_tags = sorted(tag_counter, key = tag_counter.get, reverse = True)

top_10 = popular_tags[:10]

print('\nList of the top 10 popular hashtags are :\n',top_10)

Answer 3

见下文

from collections import Counter

lst = ['Ab','aa','ab','Aa','Cct','aA']

lower_lst = [x.lower() for x in lst ]

counter = Counter(lower_lst)

print(counter.most_common(1))

Answer 4

您可以使用Counter收藏库

from collections import Counter

li = ['COVID19', 'Covid19', 'covid19', 'coronavirus', 'Coronavirus']

print(Counter([i.lower() for i in li]).most_common(10))

输出：

[('covid19', 3), ('coronavirus', 2)]