# initialize string text = "Python: How to count words in string Python" substring = "Python" ...
words = string.split()#遍历单词列表，更新单词的出现次数 for word in words:if word in word_count:word_count[word] += 1 else:word_count[word] = 1 #找到出现次数最多的单词 max_count = 0 max_word = ""for word, count in word_count.items():if count > max_count:max_count = count ...
Other methods give you information about the string itself. The methodcountreturns how many times a given substring appears within a string. The methodendswithreturns whether the string ends with a certain substring, whereas the methodstartswithreturns whether the string started with a substring: Ano...
clear() # empty the counter | >>> c | Counter() | | Note: If a count is set to zero or reduced to zero, it will remain | in the counter until the entry is deleted or the counter is cleared: | | >>> c = Counter('aaabbc') | >>> c['b'] -= 2 # reduce the count ...
Python之words count 要求: 对文件单词进行统计,不区分大小写,并显示单词重复最多的十个单词 思路: 利用字典key,value的特性存单词及其重复的次数 每行进行特殊字符的处理,分离出被特殊字符包含的单词 defmakekey(s:str)->list: lst=[] s_complex= set(r"""!`#.,-*()\/[]*""") #利用集合装置特殊...
用Python 实现函数count_words(),该函数输入字符串s和数字n,返回s中n个出现频率最高的单词。返回值是一个元组列表,包含出现次数最高的n个单词及其次数,即[(<单词1>, <次数1>), (<单词2>, <次数2>), ... ],按出现次数降序排列。 您可以假设所有输入都是小写形式,并且不含标点符号或其他字符(只包含字...
count() 方法:S.count(sub[, start[, end]]) -> int Return the number of non-overlapping occurrences of substring sub in string S[start:end]. Optional arguments start and end are interpreted as in slice notation. 查找子字符串 sub 在字符串中出现的次数,可选参数,开始 和 结束 可理解为切片...
words=re.findall(r'\b\w+\b',string)# 获取单词列表word_list=Counter(words)# 根据单词出现次数进行排序sorted_word_list=sorted(word_list.items(),key=lambdax:x[1],reverse=True)# 输出结果forword,countinsorted_word_list:print(f"{word}:{count}")# 测试代码string="Thisisa test string.It ...
translator = str.maketrans('', '', string.punctuation) text = text.translate(translator) # 使用空格将文本分割成单词列表 words = text.split() # 创建一个空字典来存储单词计数 word_count = {} # 遍历单词列表,统计每个单词的出现次数 for word in words: ...
for letter1 in range(65,91): #大写字母统计 WD.append(chr(letter1))for letter2 in range(97,123): #小写字母统计 wd.append(chr(letter2))for i in range(26):per = (text.count(wd[i])+text.count(WD[i]))/len(text)per = '{:.2%}'.format(per)sum0 = text.count(wd[i])+...