如何用Python实现字符串单词按出现次数分组存入字典

现有代码问题说明 #

• 直接split()拆分得到的单词包含标点（如day.末尾的句号），会导致同一个单词被识别为不同值
• 计数逻辑错误：count = +1是将变量赋值为1，没有实现累加效果，也没有单独统计每个单词的出现次数
• 缺少「单词计数反向映射为次数对应单词列表」的逻辑

完整实现代码 #

基础写法（无需引入第三方库） #

String = "Today was a very a good day. Tomorrow might be a better day"
# 1、清洗单词：去除每个单词附带的标点符号
clean_words = []
for raw_word in String.split():
    # 只保留单词中的字母字符，过滤标点
    clean_word = ''.join([char for char in raw_word if char.isalpha()])
    clean_words.append(clean_word)

# 2、统计每个单词的出现次数
word_counter = {}
for word in clean_words:
    if word in word_counter:
        word_counter[word] += 1
    else:
        word_counter[word] = 1

# 3、按出现次数分组生成目标字典
res = {}
for word, count in word_counter.items():
    if count not in res:
        res[count] = []
    res[count].append(word)

print(res)

运行后输出结果和你给出的预期完全一致。

简化写法（用Python内置工具） #

借助collections.Counter可以简化计数逻辑：

from collections import Counter

String = "Today was a very a good day. Tomorrow might be a better day"
# 清洗+计数一步完成
clean_words = [''.join([c for c in w if c.isalpha()]) for w in String.split()]
word_counter = Counter(clean_words)

res = {}
for word, cnt in word_counter.items():
    # setdefault自动处理键不存在的情况，无需单独判断
    res.setdefault(cnt, []).append(word)

print(res)

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