如何统计文本文件中包含单引号的高频词？

要让代码能统计包含单引号的单词，核心是修改单词的提取逻辑和有效单词的判断条件——原来的isalnum()方法会直接排除带单引号的单词，具体修改如下：

修改后的完整代码 #

import re

# 读取文件并提取包含字母、数字和单引号的单词
file_name = input('Enter the name of the file: ')
with open(file_name) as f:
    content = f.read().lower()
    # 用正则匹配所有由字母、数字、单引号组成的单词
    words = re.findall(r"[a-zA-Z0-9']+", content)

number_of_words = int(input('Enter how many top words you want to see: '))
uniques = []
stop_words = ["a", "an", "and", "in", "is", "the"]

for word in words:
    # 确保单词不是纯单引号，且只包含允许的字符
    if re.match(r"^[a-zA-Z0-9']*[a-zA-Z0-9][a-zA-Z0-9']*$", word):
        check_special = True
    else:
        check_special = False
    
    if word not in uniques and word not in stop_words and check_special:
        uniques.append(word)

counts = []
for unique in uniques:
    count = 0
    for word in words:
        if word == unique:
            count += 1
    counts.append((count, unique))

counts.sort(reverse=True)

counts_dict = {}
for count, word in counts:
    if count not in counts_dict:
        counts_dict[count] = []
    counts_dict[count].append(word)

count_num_word = 0
for count in sorted(counts_dict.keys(), reverse=True):
    if count_num_word >= number_of_words:
        break
    print('The following words appeared %d times each: %s' % (count, ', '.join(sorted(counts_dict[count]))))
    count_num_word += 1

关键修改点 #

1. 单词提取逻辑：
   原来直接用split()分割会把带标点的单词（比如don't.）当成一个整体，现在用正则表达式re.findall(r"[a-zA-Z0-9']+", content)，可以精准提取包含单引号的有效单词，同时自动去掉单词前后的无关标点（比如句号、逗号）。

2. 有效单词判断：
   替换原来的word.isalnum()判断，改用正则re.match(r"^[a-zA-Z0-9']*[a-zA-Z0-9][a-zA-Z0-9']*$", word)，既允许单词中包含单引号，又能排除纯单引号的无效字符串（比如'''）。

3. 排序逻辑优化：
   把原来的counts.sort()+counts.reverse()改成counts.sort(reverse=True)更简洁；遍历counts_dict时按次数从高到低排序，确保输出顺序符合预期。

内容的提问来源于stack exchange，提问作者thenorthape