C++ - 矩阵减法

Question

我正在做一些非常基础的线性代数，我可能完全忽略了这里的要点。

假设我有以下矩阵:

v1 = [5, 8]
v2 = [3, 4]
v3 = [4, 4]
v4 = [2, 1]

预期输出:

M1 = [5 - 3, 8 - 4] = [2, 4] 
M2 = [4 - 2, 4 - 1] = [2, 3]

实际输出:

0 0 
0 0 
2 4 
-1 0 
2 3 

代码如下:

std::vector<double> calculate(std::vector<double> values, std::vector<double> values2)
{
    std::vector<double> vel(2, 0);

    for(unsigned i=0; (i < values.size()); i++)
    {
        vel[i] = values[i] - values2[i];
    }

    return vel;
}



  std::vector<std::vector<double> > values = { {5,8}, {3, 4}, {4, 4}, {2, 1}};

    std::vector<std::vector<double> > v;

    v.resize(2);

    for(unsigned i=0; (i < values.size()-1); i++)
    {   
        v[i].resize(2);
        v.push_back(calculate(values[i], values[i + 1]));
        //v[i] = calculate(values[i], values[i + 1]);
    }

    for(unsigned i=0; (i < v.size()); i++)
    {
        for(unsigned j=0; (j < v[i].size()); j++)
        {
            std::cout << v[i][j] << " ";
        }
        std::cout << std::endl;
    }

问题是，下面应该迭代 4 次以上，计算 4 个矩阵，最终生成的 2D vector 应该只包含 2 个值。

我可能遗漏了一些愚蠢的东西。

Answer 1

v.resize(2); // now it contains `{{} {}}`.
for(unsigned i=0; (i < values.size()-1); i++) //for each input except the last (3 iterations)
                    //starting with {5,8} and {3,4}
    v[i].resize(2); //resize one of the vectors already in v to 2 
                    //now v contains {{0,0}, {}
    v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
                    //now v contains {{0,0}, {}, {2,4}}
for(............. (i < values.size()-1); i++)
                    //next the middle pair of inputs  {3,4} and {4,4}
    v[i].resize(2); //resize one of the vectors already in v to 2 
                    //now v contains {{0,0}, {0,0}, {2,4}}
    v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
                    //now v contains {{0,0}, {0,0}, {2,4}, {-1,0}}
for(............. (i < values.size()-1); i++)
                    //finally the last pair of inputs  {4,4} and {2,1}
    v[i].resize(2); //resize the {2,4} to 2, but it was already two
                    //now v contains {{0,0}, {0,0}, {2,4}, {-1,0}}
    v.push_back(calculate(values[i], values[i + 1])); //push back the calculations
                    //now v contains {{0,0}, {0,0}, {2,4}, {-1,0}, {2,3}}
for(............. (i < values.size()-1); ....) //done iterating

你有调试器吗？学习如何像这样单步执行代码、请人向您展示或查找教程非常重要。在调试器中逐步执行此代码会使您清楚地知道发生了什么。

幸运的是，代码很容易修复:

std::vector<std::vector<double> > v;

for(unsigned i=0; i<values.size()-1; i+=2) //NOTE: i+=2!!!
{   
    v.push_back(calculate(values[i], values[i + 1]));
}

证明:http://coliru.stacked-crooked.com/a/827a0183f1e7c582