数据挖掘wine数据集分类实验报告及代码

package classification; import java.text.DecimalFormat; import com.sun.org.apache.regexp.internal.recompile; import Jama.Matrix; //矩阵操作类，包含对矩阵的各种操作 public class MOs { ///// 全零矩阵 public static double[][] zeros(int m, int n) { double[][] x = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[i][j] = 0; return x; } //////// 全1矩阵 public static double[][] ones(int m, int n) { double[][] x = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[i][j] = 1; return x; } // 矩阵+矩阵 public static double[][] mPlus(double a[][], double b[][]) { int m1 = a.length, m2 = b.length; int n1 = a[0].length, n2 = b[0].length; double[][] mpm = new double[m1][n1]; if (m1 != m2 || n1 != n2) System.out.println("mPius出错：矩阵不匹配！"); else for (int i = 0; i < m1; i++) for (int j = 0; j < n1; j++) mpm[i][j] = a[i][j] + b[i][j]; return mpm; } // 矩阵+数值 public static double[][] mPlus(double a[][], double b) { int m = a.length; int n = a[0].length; double[][] c = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) c[i][j] = a[i][j] + b; return c; } // 矩阵-矩阵 public static double[][] mMinus(double a[][], double b[][]) { int m1 = a.length, m2 = b.length; int n1 = a[0].length, n2 = b[0].length; double[][] mMm = new double[m1][n1]; if (m1 != m2 || n1 != n2) System.out.println("mMinus出错：矩阵不匹配！"); else for (int i = 0; i < m1; i++) for (int j = 0; j < n1; j++) mMm[i][j] = a[i][j] - b[i][j]; return mMm; } // 数值-矩阵 public static double[][] mMinus(double b, double a[][]) { int m = a.length; int n = a[0].length; double[][] c = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) c[i][j] = b - a[i][j]; return c; } /////// 矩阵*数值 public static double[][] mMulti(double a[][], double b) { int m = a.length; int n = a[0].length; double[][] c = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) c[i][j] = a[i][j] * b; return c; } //// 矩阵*矩阵 public static double[][] mMulti(double a[][], double b[][]) { int m1 = a.length, m2 = b.length; int n1 = a[0].length, n2 = b[0].length; double[][] mmm = new double[m1][n2]; if (n1 != m2) System.out.println("mMuti出错：矩阵不匹配！"); else for (int i = 0; i < m1; i++) for (int k = 0; k < n2; k++) { mmm[i][k] = 0; for (int j = 0; j < n1; j++) mmm[i][k] += a[i][j] * b[j][k]; } return mmm; } ///// 矩阵对应乘-点乘 public static void dotProduct() { } ///// 矩阵各元素幂次方 public static double[][] pow(double a[][], double b) { int m = a.length; int n = a[0].length; double[][] x = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[i][j] = Math.pow(a[i][j], b); return x; } ////// e的指数 public static double[][] e(double a[][]) { int m = a.length; int n = a[0].length; double[][] x = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[i][j] = Math.exp(a[i][j]); return x; } ///// 对数以e为底 public static double[][] log(double a[][]) { int m = a.length; int n = a[0].length; double[][] x = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[i][j] = Math.log(a[i][j]); return x; } //// 矩阵转置 public static double[][] transpose(double a[][]) { int m = a.length; int n = a[0].length; double[][] x = new double[n][m]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) x[j][i] = a[i][j]; return x; } ///// 对矩阵求和 public static double sum(double a[][]) { int m = a.length; int n = a[0].length; double sum = 0; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) sum += a[i][j]; return sum; } ///// 对矩阵行、列求和 public static double[][] sum(double a[][], int b) { int m = a.length; int n = a[0].length; if (b == 1) {// 对行求和 double[][] sum = new double[m][1]; for (int i = 0; i < m; i++) { sum[i][0] = 0; for (int j = 0; j < n; j++) sum[i][0] = sum[i][0] + a[i][j]; } return sum; } else if (b == 2) {// 对列求和 double[][] sum = new double[1][n]; for (int i = 0; i < n; i++) { sum[0][i] = 0; for (int j = 0; j < m; j++) sum[0][i] = sum[0][i] + a[j][i]; } return sum; } return new double[][] {}; } /// 矩阵行连接 public static double[][] rowCombination(double a[][], double b[][]) { int m1 = a.length; int n1 = a[0].length; int m2 = b.length; int n2 = b[0].length; double[][] rowCombination = new double[m1][n1 + n2]; if (m1 != m2) System.out.println("rowCombination出错：矩阵行数不匹配！"); else { for (int j = 0; j < n1; j++)// 一列一列地存入 for (int i = 0; i < m1; i++) rowCombination[i][j] = a[i][j]; for (int j = n1, k = 0; j < n1 + n2; j++, k++) for (int i = 0; i < m1; i++) rowCombination[i][j] = b[i][k]; } return rowCombination; } /// 矩阵列连接 public static double[][] colCombination(double a[][], double b[][]) { int m1 = a.length; int n1 = a[0].length; int m2 = b.length; int n2 = b[0].length; double[][] colCombination = new double[m1 + m2][n1]; if (n1 != n2) System.out.println("colCombination出错：矩阵列数不匹配！"); else { for (int i = 0; i < m1; i++) for (int j = 0; j < n1; j++) colCombination[i][j] = a[i][j]; for (int i = 0, j = m1; i < m2; i++, j++) for (int k = 0; k < n2; k++) colCombination[j][k] = b[i][k]; } return colCombination; } /////// 被一除 public static double[][] dividedByOne(double a[][]) { int m = a.length; int n = a[0].length; double[][] c = new double[m][n]; for (int i = 0; i < m; i++) for (int j = 0; j < n; j++) c[i][j] = 1 / a[i][j]; return c; } //// 获取子矩阵,行为m1:m2，列为n1:n2 public static double[][] getMatrix(double a[][], int m1, int m2, int n1, int n2) { double[][] x = new double[m2 - m1 + 1][n2 - n1 + 1]; for (int i = 0; i < m2 - m1 + 1; i++) for (int j = 0; j < n2 - n1 + 1; j++) x[i][j] = a[m1 - 1 + i][n1 - 1 + j]; return x; } //// 输出矩阵 public static void putMatrix(double a[][]) { int m = a.length; int n = a[0].length; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) System.out.print(a[i][j] + " "); System.out.println(); } } ///// 对矩阵元素统一形式-此处小数点后保留三位（四舍五入） public static double[][] mFormat(double a[][]) { int m = a.length; int n = a[0].length; DecimalFormat format = new DecimalFormat("0.000"); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { String result = format.format(a[i][j]); a[i][j] = Double.parseDouble(result); } } return a; } ////// 计算两个矩阵间相同元素的个数 public static int numSame(double a[][], double b[][]) { int num = 0; int m1 = a.length, m2 = b.length; int n1 = a[0].length, n2 = b[0].length; if (m1 != m2 || n1 != n2) System.out.println("numSame出错：矩阵不匹配"); else for (int i = 0; i < m1; i++) for (int j = 0; j < n1; j++) if (a[i][j] == b[i][j]) num++; return num; } //// 计算一个矩阵中含有某个数的个数 public static int numSame(double a[][], int x) { int num = 0; int m1 = a.length; int n1 = a[0].length; for (int i = 0; i < m1; i++) for (int j = 0; j < n1; j++) if (a[i][j] == x) num++; return num; } ///// 寻找矩阵每行中最大元素的索引（返回实际的序数，即以1开始） public static double[][] indexMax(double a[][]) { int m = a.length; int n = a[0].length; double[][] index = new double[m][1];