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我正在尝试创建一个对称矩阵矩阵n x n，并n*(n+1)/2使用.boostc++

到目前为止，我能够创建矩阵，并使用以下代码用随机值填充它

#include <iostream>
#include <fstream>    
#include </usr/include/boost/numeric/ublas/matrix.hpp>
#include </usr/include/boost/numeric/ublas/matrix_sparse.hpp>
#include </usr/include/boost/numeric/ublas/symmetric.hpp>
#include </usr/include/boost/numeric/ublas/io.hpp>

using namespace std;



int test_boost () {
    using namespace boost::numeric::ublas;
    symmetric_matrix<double, upper> m_sym (3, 3);

    double filler[6] = {0, 1, 2, 3, 4, 5};        

    for (unsigned i = 0; i < m_sym.size1 (); ++ i)
        for (unsigned j = i; j < m_sym.size2 (); ++ j)
            m_sym (i, j) = filler[i+j*m_sym.size1()];

    std::cout << m_sym << std::endl;
    return 0;
}

我要做的是使用数组中的值填充对称矩阵的上（或下）部分filler。所以输出上对称矩阵应该是

         |      0    |      1    |      2    |
------------------------------------------------
   0 |          0           1           3    
   1 |          1           2           4
   2 |          3           4           5

关于如何做到这一点的任何想法？

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1 回答 1

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我会通过保留一个从头到尾遍历填充物的迭代器来简化这一点：

symmetric_matrix<double, upper> m_sym (3, 3);

double filler[6] = {0, 1, 2, 3, 4, 5};        

assert(m_sym.size1() == m_sym.size2());

double const* in = std::begin(filler);
for (size_t i = 0; i < m_sym.size1(); ++ i)
    for (size_t j = 0; j <= i && in != std::end(filler); ++ j)
        m_sym (i, j) = *in++;

印刷：Live On Coliru

我个人建议创建一个辅助函数，例如：

Live On Wandbox

#include <iostream>
#include <fstream>    
#include <boost/numeric/ublas/matrix.hpp>
#include <boost/numeric/ublas/matrix_sparse.hpp>
#include <boost/numeric/ublas/symmetric.hpp>
#include <boost/numeric/ublas/io.hpp>

namespace bnu = boost::numeric::ublas;

template <typename T = double>
bnu::symmetric_matrix<T, bnu::upper> make_symmetric(std::initializer_list<T> filler) {
    size_t n = (sqrt(8*filler.size() + 1) - 1)/2;
    assert((n*(n+1))/2 == filler.size());

    bnu::symmetric_matrix<T, bnu::upper> result(n, n);

    auto in = std::begin(filler);
    for (size_t i = 0; i < result.size1(); ++ i)
        for (size_t j = 0; j <= i && in != std::end(filler); ++ j)
            result (i, j) = *in++;

    return result;
}

int main() {
    std::cout << make_symmetric({0,1,2}) << "\n";
    std::cout << make_symmetric({0,1,2,3,4,5}) << "\n";
    std::cout << make_symmetric({0,1,2,3,4,5,6,7,8,9}) << "\n";
}

印刷

[2,2]((0,1),(1,2))
[3,3]((0,1,3),(1,2,4),(3,4,5))
[4,4]((0,1,3,6),(1,2,4,7),(3,4,5,8),(6,7,8,9))

注意：尺寸检查使用级数展开1 + ... + n和反之：n = 1/2 (sqrt(8 x + 1) - 1)

于 2018-02-21T19:56:32.267 回答