从文本文件中读取输入并将其解压缩 – C
发布时间:2020-12-16 07:32:43 所属栏目:百科 来源:网络整理
导读:我正在尝试逐行读取文本文件并将数据提取到我的程序中.本质上,C程序将读取已经采用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 ...
我正在尝试逐行读取文本文件并将数据提取到我的程序中.本质上,C程序将读取已经采用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 ... .. .. 格式基本上是前3行包括Party,Position和Name,接下来的行表示“结果”,其形式为: [第一封信] [党投票] 所以,例如,如果你看到S R,这意味着对参议员投了1票,那就是共和党候选人. 这是我到目前为止所拥有的: #include<fstream> int main() { std::ifstream input("file.txt"); } 这是我的理解,这将允许我输入文件,并逐行进行,但我不知道如何从这里开始实现这个…任何帮助? 谢谢! 解决方法
为了娱乐和荣耀,这是一个基于Boost Spirit的实现.我添加了更多假投票输入,只是因此可能会显示一些内容.
>我不确定候选人和投票之间是否存在1:1的关系(我不是美国公民,我不知道列出的候选人是投票还是投票).所以我决定只使用假数据. const std::string input = "Republican Senator John McMahonn" "Democrat Senator Anthony Quizitanon" "S Rn" "S Rn" "S Rn" "Democrat Mayor Steven Markeln" "Republican Judge Matt Stevensn" "M Dn" "J Rn" "S Rn" "S Rn"; 但是,该代码可用于这两个目的. 请参阅http://liveworkspace.org/code/d9e39c19674fbf7b2419ff88a642dc38的现场演示 #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,_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 McMahonn" "Democrat Senator Anthony Quizitanon" "S Rn" "S Rn" "S Rn" "Democrat Mayor Steven Markeln" "Republican Judge Matt Stevensn" "M Dn" "J Rn" "S Rn" "S Rn"; 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 listn"; 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"; } } 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 ***** (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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