简体   繁体   English

从文本文件读取输入并将其提取-C ++

[英]Reading input from a text file and extracting it - C++

I'm trying to read a text file line by line and extract the data into my program. 我正在尝试逐行读取文本文件并将数据提取到程序中。 Essentially, the C++ program will be reading an input file already made that is in this format: 本质上,C ++程序将读取以下格式的输入文件:

     Republican Senator John McMahon
     Democrat Mayor Steven Markel
     Republican Judge Matt Stevens
     Democrat Senator Anthony Quizitano
     S R
     M D
     J R
     ...
     ..
     ..

The format is basically first 3 lines include the Party, Position, and Name and the next following lines indicate the "results" which are in the form of: 格式基本上是前三行,包括“聚会”,“职位”和“姓名”,接下来的几行表示“结果”,其形式为:

[first letter of position] [party voted for] [第一份立场书] [当事方投票]

So, for example if you see SR, that means 1 vote for a Senator, and it goes to the Republican candidate. 因此,例如,如果您看到SR,则表示参议员需要1票,而该票将由共和党候选人获得。

Here's what I have so far: 这是我到目前为止的内容:

            #include<fstream>
            int main()
            {
            std::ifstream input("file.txt");
            }

It's my understanding that this will allow me to input the file, and go through it line by line, but I'm not sure how I should proceed to achieve this from here...any help? 我的理解是,这将允许我输入文件并逐行处理它,但是我不确定如何从此处实现此目标……有什么帮助吗?

Thanks! 谢谢!

Here is, for fun and glory, an implementation based on Boost Spirit. 出于乐趣和荣耀,这是基于Boost Spirit的实现。 I added more fake vote input, just so there could be something to display. 我添加了更多的虚假投票输入,只是为了显示某些内容。

  • I wasn't sure whether there is an 1:1 relation between candidates and votes (I'm not a US citizen, and I don't know whether the candidates listed would be voting or being voted for). 我不确定候选人与选票之间是否存在1:1的关系(我不是美国公民,也不知道所列候选人是会投票还是会被投票)。 So I decided to just use fake data. 所以我决定只使用假数据。

     const std::string input = "Republican Senator John McMahon\\n" "Democrat Senator Anthony Quizitano\\n" "SR\\n" "SR\\n" "SR\\n" "Democrat Mayor Steven Markel\\n" "Republican Judge Matt Stevens\\n" "MD\\n" "JR\\n" "SR\\n" "SR\\n"; 

    The code however can be used for both purposes. 但是,该代码可同时用于两种目的。

  • I made it unimportant in what order the input appears. 我以输入显示的顺序使其无关紧要。
  • Optionally you can assert, though, that the single letters (S,M,J) actually correspond to positions listed before that point. 但是,您可以选择断言单个字母(S,M,J)实际上与该点之前列出的位置相对应。 Enable this by uncommenting the check with posletter_check 通过使用posletter_check取消对支票的注释来启用此posletter_check

See the demo live on http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38 http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38上观看现场演示

#define BOOST_SPIRIT_USE_PHOENIX_V3
#define BOOST_RESULT_OF_USE_DECLTYPE
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iomanip>

namespace qi    = boost::spirit::qi;
namespace phx   = boost::phoenix;

struct Candidate { std::string party, position, name; };

BOOST_FUSION_ADAPT_STRUCT(Candidate, (std::string, party)(std::string, position)(std::string, name))

typedef std::map<std::pair<char, char>, size_t> Votes;
typedef std::vector<Candidate> Candidates;

template <typename It>
    struct parser : qi::grammar<It>
{
    mutable Votes _votes;
    mutable Candidates _candidates;

    parser() : parser::base_type(start)
    {
        using namespace qi;
        using phx::bind; using phx::ref; using phx::val;

        start = (line % eol) >> *eol >> eoi;

        line = 
              vote      [ phx::bind(&parser::register_vote, phx::ref(*this), _1) ]
            | candidate [ phx::push_back(phx::ref(_candidates), _1) ]
            ;

        vote %= graph   
                        // Comment the following line to accept any single
                        // letter, even if a matching position wasn't seen
                        // before:
                        [ _pass = phx::bind(&parser::posletter_check, phx::ref(*this), _1) ]  
            >> ' ' 
            >> char_("RD")
            ;

        candidate = (string("Republican") | string("Democrat"))
            >> ' ' 
            >> as_string [ +graph ]
            >> ' ' 
            >> as_string [ +(char_ - eol) ]
            ;
    }

  private:
    bool posletter_check(char posletter) const
    {
        for (auto& c : _candidates)
            if (posletter == c.position[0])
                return true;
        return false;
    }
    void register_vote(Votes::key_type const& key) const
    {
        auto it = _votes.find(key);
        if (_votes.end()==it)
            _votes[key] = 1;
        else
            it->second++;
    }

    qi::rule<It, Votes::key_type()> vote;
    qi::rule<It, Candidate()> candidate;
    qi::rule<It> start, line;
};

int main()
{
    const std::string input = 
        "Republican Senator John McMahon\n"
        "Democrat Senator Anthony Quizitano\n"
        "S R\n"
        "S R\n"
        "S R\n"
        "Democrat Mayor Steven Markel\n"
        "Republican Judge Matt Stevens\n"
        "M D\n"
        "J R\n"
        "S R\n"
        "S R\n";

    std::string::const_iterator f(std::begin(input)), l(std::end(input));

    parser<std::string::const_iterator> p;

    try
    {
        bool ok = qi::parse(f,l,p);
        if (ok)   
        {
            std::cout << "\ncandidate list\n";
            std::cout << "------------------------------------------------\n";
            for (auto& c : p._candidates)
                std::cout << std::setw(20) << c.name << " (" << c.position << " for the " << c.party << "s)\n";

            std::cout << "\nVote distribution:\n";
            std::cout << "------------------------------------------------\n";
            for (auto& v : p._votes)
                std::cout << '(' << v.first.first << "," << v.first.second << "): " << v.second << " votes " << std::string(v.second, '*') << "\n";
        }
        else std::cerr << "parse failed: '" << std::string(f,l) << "'\n";

        if (f!=l) std::cerr << "trailing unparsed: '" << std::string(f,l) << "'\n";
    } catch(const qi::expectation_failure<std::string::const_iterator>& e)
    {
        std::string frag(e.first, e.last);
        std::cerr << e.what() << "'" << frag << "'\n";
    }
}

Output: 输出:

candidate list
------------------------------------------------
        John McMahon (Senator for the Republicans)
   Anthony Quizitano (Senator for the Democrats)
       Steven Markel (Mayor for the Democrats)
        Matt Stevens (Judge for the Republicans)

Vote distribution:
------------------------------------------------
(J,R): 1 votes *
(M,D): 1 votes *
(S,R): 5 votes *****

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM