矩陣的基本運算 —— C 實現

最近寫了一個C程式 ------------- 矩陣的基本運算。

包括四則運算， 求逆， rref（行最簡型）， 轉置。

並有幾個實際應用例子， 求解方程組， 線性迴歸， 線性規劃進行中中。 

貼出來請大牛們指點指點。就貼最難的一個， rref吧。

/************************************ * pTag is the pivots tag * if 1, the column is pivot * and 0, the column is not pivot * Return: *return elimination matrix*************************************/matrixrref(matrix A, int *pTag)         // reduced row echelon form{    // initialize Elimination matrix    matrix E = Imat(A->rows);    double **MA = A->data;    double **ME = E->data;    double e[A->rows];// factor of reduced    // echelon loop    int k;    for (int pivots = 0, j = 0; j < A->cols && pivots < A->rows; j++)    {if (MA[j][pivots] == 0.0){    // if this element is 0.0, find a nonzero element behind in j's column    for (k = pivots + 1; k < A->rows && MA[j][k] == 0.0; k++);    // row swap if find else operate next line    if (k != A->rows)    {RowSwap(A, pivots, k);RowSwap(E, pivots, k);    }    else    {pTag[j] = 0;continue;    }}// eliminatepTag[j] = 1;for (int i = 0; i < A->rows; i++)    e[i] = (i == pivots ? 0 : MA[j][i] / MA[j][pivots]);for (int col = 0; col < A->cols; col++)    for (int row = 0; row < A->rows; row++)MA[col][row] -= e[row] * MA[col][pivots];for (int col = 0; col < E->cols; col++)    for (int row = 0; row < E->rows; row++)ME[col][row] -= e[row] * ME[col][pivots];pivots++;    }        // normalise after echelon, maybe can layout it to     // echelon loop, but it will reduce the precision    double fac;    for (int i = 0, k = 0; i < A->rows; i++)    {while (pTag[k++] == 0);fac = MA[k - 1][i];for (int j = 0; j < E->cols; j++)    ME[j][i] /= fac;for (int j = 0; j < A->cols; j++)    MA[j][i] /= fac;    }    return E;}

上傳半天不見有反應, CSDN, 有點怪. 完整曆程及樣本請到這裡下載

這裡