1

我有多个浮点数行，一行中的数字是空格分隔的，例如

1.2 2.2 3.2
1.1 2.1 3.1

我想将上述数字提取为字符串并解析为二维向量； std::vector< std::vector< std::string > > { {"1.2", "2.2", "3.2"},{"1.1", "2.1", "3.1} }

我的代码如下所示。

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>
#include <string>

namespace client
{
    namespace qi = boost::spirit::qi;
    namespace ascii = boost::spirit::ascii;

    using VecType = std::vector< std::vector< std::string >>; 
    
    struct employee
    {
        VecType name;
    };
}

BOOST_FUSION_ADAPT_STRUCT(
    client::employee,
    (client::VecType, name)
)
//]

namespace client
{
    template <typename Iterator>
    struct employee_parser : qi::grammar<Iterator, VecType(), ascii::space_type>
    {
        employee_parser() : employee_parser::base_type(start)
        {
            using qi::lexeme;
            using ascii::char_;

            number %= lexeme[+char_( "0-9." ) >>  qi::space ];
            start %= +number;
        }

        qi::rule<Iterator, std::string(), ascii::space_type> number;
        qi::rule<Iterator, VecType(), ascii::space_type> start;
    };
}

但这会产生 2D 向量，其外部向量大小为 6，每个内部向量大小为 1。

我不明白如何将字符串从新行中拆分为仅生成 2 个内部向量。

4

1 回答 1

0

你必须拆分规则。让我们从类型开始：

using VecType = std::vector<std::string>;
using VecVecType = std::vector<VecType>;

现在，让我们制定一个规则来解析一个数字、一行数字和多行：

qi::rule<Iterator, std::string()> number;
qi::rule<Iterator, VecType(), qi::blank_type> row;
qi::rule<Iterator, VecVecType()> start;

实现它们（请注意，我将船长移到了语法中，因为将其泄漏到接口中不是一个好主意）：

number = raw [ double_ ]; // raw[] to get string value
row    = +number;
start  = qi::skip(blank) [ row % eol ];

注意：我使用blank而不是space因为我们不想跳过eol对语法很重要的那个。

演示

住在科利鲁

#include <boost/spirit/include/qi.hpp>
#include <string>
#include <iomanip>

namespace client {
    namespace qi = boost::spirit::qi;

    using VecType = std::vector<std::string>;
    using VecVecType = std::vector<VecType>;
} // namespace client

namespace client {
    template <typename Iterator>
    struct my_parser : qi::grammar<Iterator, VecVecType()> {
        my_parser() : my_parser::base_type(start) {
            using namespace qi;

            number = raw [ double_ ]; // raw[] to get string value
            row    = *number;
            start  = qi::skip(blank) [ row % eol ];
        }

        qi::rule<Iterator, std::string()> number;
        qi::rule<Iterator, VecType(), qi::blank_type> row;
        qi::rule<Iterator, VecVecType()> start;
    };
} // namespace client

int main() {
    client::my_parser<std::string::const_iterator> const p;
    for (std::string const& input: {
            "",
            "1.2 2.2 3.2\n1.1 2.1 3.1",
            })
    {
        std::cout << "--- " << std::quoted(input) << " -----\n";
        auto f = begin(input), l = end(input);
        client::VecVecType output;
        if (parse(f, l, p, output)) {
            std::cout << "Parsed:\n";
            for (auto& row : output) {
                for (auto& v : row) {
                    std::cout << "\t" << v;
                }
                std::cout << "\n";
            }
        } else {
            std::cout << "Failed\n";
        }
        if (f!=l) {
            std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
        }
    }
}

印刷

--- "" -----
Parsed:

--- "1.2 2.2 3.2
1.1 2.1 3.1" -----
Parsed:
    1.2 2.2 3.2
    1.1 2.1 3.1

奖金

强类型化让一切变得更有趣：如果可以解析成双精度，为什么还要解析成字符串？

还展示了如何启用规则调试：

住在科利鲁

#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
#include <string>
#include <iomanip>

namespace client {
    namespace qi = boost::spirit::qi;

    using VecType = std::vector<double>;
    using VecVecType = std::vector<VecType>;
} // namespace client

namespace client {
    template <typename Iterator>
    struct my_parser : qi::grammar<Iterator, VecVecType()> {
        my_parser() : my_parser::base_type(start) {
            using namespace qi;

            row    = *double_;
            start  = qi::skip(blank) [ row % eol ];

            BOOST_SPIRIT_DEBUG_NODES((start)(row))
        }

      private:
        qi::rule<Iterator, VecType(), qi::blank_type> row;
        qi::rule<Iterator, VecVecType()> start;
    };
} // namespace client

int main() {
    client::my_parser<std::string::const_iterator> const p;
    for (std::string const& input: {
            "",
            "1.2 2.2 3.2\n1.1 2.1 3.1",
            })
    {
        std::cout << "--- " << std::quoted(input) << " -----\n";
        auto f = begin(input), l = end(input);
        client::VecVecType output;
        if (parse(f, l, p, output)) {
            std::cout << "Parsed:\n";
            for (auto& row : output) {
                for (auto& v : row) {
                    std::cout << "\t" << v;
                }
                std::cout << "\n";
            }
        } else {
            std::cout << "Failed\n";
        }
        if (f!=l) {
            std::cout << "Remaining input: " << std::quoted(std::string(f,l)) << "\n";
        }
    }
}

印刷

--- "" -----
<start>
  <try></try>
  <row>
    <try></try>
    <success></success>
    <attributes>[[]]</attributes>
  </row>
  <success></success>
  <attributes>[[[]]]</attributes>
</start>
Parsed:

--- "1.2 2.2 3.2
1.1 2.1 3.1" -----
<start>
  <try>1.2 2.2 3.2\n1.1 2.1 </try>
  <row>
    <try>1.2 2.2 3.2\n1.1 2.1 </try>
    <success>\n1.1 2.1 3.1</success>
    <attributes>[[1.2, 2.2, 3.2]]</attributes>
  </row>
  <row>
    <try>1.1 2.1 3.1</try>
    <success></success>
    <attributes>[[1.1, 2.1, 3.1]]</attributes>
  </row>
  <success></success>
  <attributes>[[[1.2, 2.2, 3.2], [1.1, 2.1, 3.1]]]</attributes>
</start>
Parsed:
    1.2 2.2 3.2
    1.1 2.1 3.1
于 2020-07-27T14:37:53.013 回答