计算文件中单词频率的优雅方式 [英] Elegant ways to count the frequency of words in a file
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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;
}
}
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