python import docx import nltk from collections import Counter from nltk.tokenize import word_tokenize # 确保下载了nltk的punkt分词器 nltk.download('punkt') def word_frequency_in_docx(file_path): # 创建一个空字典来存储单词频率 word_freq = {} # 读取Word文档 doc = docx.Document(file_path) #...
fromcollectionsimportCounterimportmatplotlib.pyplotaspltdefword_frequency(text):words=text.split()frequency=Counter(words)# 返回最常见的10个词returnfrequency.most_common(10)if__name__=="__main__":frequency=word_frequency(text)words,counts=zip(*frequency)plt.pie(counts,labels=words,autopct='%1.1f%...
doc(text, min_length=): words = jieba.lcut(text) filtered_words = [word word words (word) >= min_length] Counter(filtered_words)(run,
Python dubirajara/go-word-frequency-counter Star4 Golang Word Frequency Counter gogolangstopwordsfrequency-counterword-frequency-count UpdatedFeb 20, 2022 Go techiaith/geiriau-mwyaf-aml Star3 Code Issues Pull requests Rhestrau geiriau mwyaf aml y Gymraeg a Saesneg // Wordlists of the most commo...
接下来使用counter方法来进行词频统计,并取得前n位最频繁的词语: count=Counter(wordDict)rank=count.most_common()[:n] 接下来要使用最开始定义的线性变换方法来计算每个单词的大小了, 因为会有出现数据量小的时候,词频最高的第一位由于是关键词把第二位甩得老远,做出来的词云没有层次不好看,所以在这里,我设...
fromcollectionsimportdefaultdict,Counter importjson # Function to calculate word Frequency and store it into Dictionary defwordListToFreqDict(wordlist): wordfreq=[wordlist.count(p)forpinwordlist] returndict(zip(wordlist,wordfreq)) # Combine all wordslist text files into one and convert to lowercas...
fromcollectionsimportCounter# 定义计算词频的函数defcount_frequency(word_list):returnCounter(word_list)# 返回每个单词及其出现频率# 示例:计算词频frequency=count_frequency(word_list) 1. 2. 3. 4. 5. 6. 7. 8. 4. 应用 WordRank 算法 WordRank 算法的核心是通过词频等指标来评估词的权重。以下是一个...
这是我的代码,它计算文件中某个特定字符串的出现次数。int frequency(std::string s, std::string file_name){ std::stringwordcin >> file_name >>word; std::c 浏览1提问于2015-07-06得票数0 1回答 用Java读取大文件,速度太慢,gc开销超过限制 ...
append(i) if a > prob * n: i += 1 prob += cnt[i] ** 0.75 / z return table if __name__ == '__main__': from collections import Counter # count, aka frequency cnt = [1, 2, 3, 100, 15] # 0.75 power prob = [x ** 0.75 for x in cnt] prob = [x / sum(prob) ...
让我们通过一个例子来进行说明,该例子摘自维基百科关于Python的文章,其中提到“python consistently ranks as one of the most popular programming languages.” 这个句子包含11个单词,那么为什么我们不创建一个长度为11的向量,其中每个索引的值为1表示单词存在,值为0表示单词不存在呢?这通常被称为one-hot编码。