计算文件中单词频率的优雅方式

本文介绍了计算文件中单词频率的优雅方式的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在文件中计算每个英语单词的频率的优雅而有效的方法是什么？

What are the elegant and effective ways to count the frequency of each "english" word in a file?

推荐答案

，我定义 letter_only std :: locale ，以便忽略来自流的标点符号，并从输入流中只读取有效的英语字母。这样，流将处理方式，方式。和 ways！作为同一个词ways，因为流将忽略像 / code>和！。

First of all, I define letter_only std::locale so as to ignore punctuations coming from the stream, and to read only valid "english" letters from the input stream. That way, the stream will treat the words "ways", "ways." and "ways!" as just the same word "ways", because the stream will ignore punctuations like "." and "!".

struct letter_only: std::ctype<char> { letter_only(): std::ctype<char>(get_table()) {} static std::ctype_base::mask const* get_table() { static std::vector<std::ctype_base::mask> rc(std::ctype<char>::table_size,std::ctype_base::space); std::fill(&rc['A'], &rc['z'+1], std::ctype_base::alpha); return &rc[0]; } };

解决方案1 ​​

Solution 1

int main() { std::map<std::string, int> wordCount; ifstream input; input.imbue(std::locale(std::locale(), new letter_only())); //enable reading only letters! input.open("filename.txt"); std::string word; while(input >> word) { ++wordCount[word]; } for (std::map<std::string, int>::iterator it = wordCount.begin(); it != wordCount.end(); ++it) { cout << it->first <<" : "<< it->second << endl; } }

解决方案2

Solution 2

struct Counter { std::map<std::string, int> wordCount; void operator()(const std::string & item) { ++wordCount[item]; } operator std::map<std::string, int>() { return wordCount; } }; int main() { ifstream input; input.imbue(std::locale(std::locale(), new letter_only())); //enable reading only letters! input.open("filename.txt"); istream_iterator<string> start(input); istream_iterator<string> end; std::map<std::string, int> wordCount = std::for_each(start, end, Counter()); for (std::map<std::string, int>::iterator it = wordCount.begin(); it != wordCount.end(); ++it) { cout << it->first <<" : "<< it->second << endl; } }