singularvaluedecomposition.java

代码是一个分类器的实现,其中使用了部分weka的源代码。可以将项目导入eclipse运行

/*
 * This software is a cooperative product of The MathWorks and the National
 * Institute of Standards and Technology (NIST) which has been released to the
 * public domain. Neither The MathWorks nor NIST assumes any responsibility
 * whatsoever for its use by other parties, and makes no guarantees, expressed
 * or implied, about its quality, reliability, or any other characteristic.
 */

/*
 * SingularValueDecomposition.java
 * Copyright (C) 1999 The Mathworks and NIST
 *
 */

package weka.core.matrix;

import java.io.Serializable;

/** 
 * Singular Value Decomposition.
 * <P>
 * For an m-by-n matrix A with m &gt;= n, the singular value decomposition is an
 * m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and an n-by-n
 * orthogonal matrix V so that A = U*S*V'.
 * <P>
 * The singular values, sigma[k] = S[k][k], are ordered so that sigma[0] &gt;=
 * sigma[1] &gt;= ... &gt;= sigma[n-1].
 * <P>
 * The singular value decompostion always exists, so the constructor will never
 * fail.  The matrix condition number and the effective numerical rank can be
 * computed from this decomposition.
 * <p/>
 * Adapted from the <a href="http://math.nist.gov/javanumerics/jama/" target="_blank">JAMA</a> package.
 *
 * @author The Mathworks and NIST 
 * @author Fracpete (fracpete at waikato dot ac dot nz)
 * @version $Revision: 1.2 $
 */
public class SingularValueDecomposition 
  implements Serializable {

  /** 
   * Arrays for internal storage of U and V.
   * @serial internal storage of U.
   * @serial internal storage of V.
   */
  private double[][] U, V;

  /** 
   * Array for internal storage of singular values.
   * @serial internal storage of singular values.
   */
  private double[] s;

  /** 
   * Row and column dimensions.
   * @serial row dimension.
   * @serial column dimension.
   */
  private int m, n;

  /** 
   * Construct the singular value decomposition
   * @param Arg    Rectangular matrix
   */
  public SingularValueDecomposition(Matrix Arg) {

    // Derived from LINPACK code.
    // Initialize.
    double[][] A = Arg.getArrayCopy();
    m = Arg.getRowDimension();
    n = Arg.getColumnDimension();
    int nu = Math.min(m,n);
    s = new double [Math.min(m+1,n)];
    U = new double [m][nu];
    V = new double [n][n];
    double[] e = new double [n];
    double[] work = new double [m];
    boolean wantu = true;
    boolean wantv = true;

    // Reduce A to bidiagonal form, storing the diagonal elements
    // in s and the super-diagonal elements in e.

    int nct = Math.min(m-1,n);
    int nrt = Math.max(0,Math.min(n-2,m));
    for (int k = 0; k < Math.max(nct,nrt); k++) {
      if (k < nct) {

        // Compute the transformation for the k-th column and
        // place the k-th diagonal in s[k].
        // Compute 2-norm of k-th column without under/overflow.
        s[k] = 0;
        for (int i = k; i < m; i++) {
          s[k] = Maths.hypot(s[k],A[i][k]);
        }
        if (s[k] != 0.0) {
          if (A[k][k] < 0.0) {
            s[k] = -s[k];
          }
          for (int i = k; i < m; i++) {
            A[i][k] /= s[k];
          }
          A[k][k] += 1.0;
        }
        s[k] = -s[k];
      }
      for (int j = k+1; j < n; j++) {
        if ((k < nct) & (s[k] != 0.0))  {

          // Apply the transformation.

          double t = 0;
          for (int i = k; i < m; i++) {
            t += A[i][k]*A[i][j];
          }
          t = -t/A[k][k];
          for (int i = k; i < m; i++) {
            A[i][j] += t*A[i][k];
          }
        }

        // Place the k-th row of A into e for the
        // subsequent calculation of the row transformation.

        e[j] = A[k][j];
      }
      if (wantu & (k < nct)) {

        // Place the transformation in U for subsequent back
        // multiplication.

        for (int i = k; i < m; i++) {
          U[i][k] = A[i][k];
        }
      }
      if (k < nrt) {

        // Compute the k-th row transformation and place the
        // k-th super-diagonal in e[k].
        // Compute 2-norm without under/overflow.
        e[k] = 0;
        for (int i = k+1; i < n; i++) {
          e[k] = Maths.hypot(e[k],e[i]);
        }
        if (e[k] != 0.0) {
          if (e[k+1] < 0.0) {
            e[k] = -e[k];
          }
          for (int i = k+1; i < n; i++) {
            e[i] /= e[k];
          }
          e[k+1] += 1.0;
        }
        e[k] = -e[k];
        if ((k+1 < m) & (e[k] != 0.0)) {

          // Apply the transformation.

          for (int i = k+1; i < m; i++) {
            work[i] = 0.0;
          }
          for (int j = k+1; j < n; j++) {
            for (int i = k+1; i < m; i++) {
              work[i] += e[j]*A[i][j];
            }
          }
          for (int j = k+1; j < n; j++) {
            double t = -e[j]/e[k+1];
            for (int i = k+1; i < m; i++) {
              A[i][j] += t*work[i];
            }
          }
        }
        if (wantv) {

          // Place the transformation in V for subsequent
          // back multiplication.

          for (int i = k+1; i < n; i++) {
            V[i][k] = e[i];
          }
        }
      }
    }

    // Set up the final bidiagonal matrix or order p.

    int p = Math.min(n,m+1);
    if (nct < n) {
      s[nct] = A[nct][nct];
    }
    if (m < p) {
      s[p-1] = 0.0;
    }
    if (nrt+1 < p) {
      e[nrt] = A[nrt][p-1];
    }
    e[p-1] = 0.0;

    // If required, generate U.

    if (wantu) {
      for (int j = nct; j < nu; j++) {
        for (int i = 0; i < m; i++) {
          U[i][j] = 0.0;
        }
        U[j][j] = 1.0;
      }
      for (int k = nct-1; k >= 0; k--) {
        if (s[k] != 0.0) {
          for (int j = k+1; j < nu; j++) {
            double t = 0;
            for (int i = k; i < m; i++) {
              t += U[i][k]*U[i][j];
            }
            t = -t/U[k][k];
            for (int i = k; i < m; i++) {
              U[i][j] += t*U[i][k];
            }
          }
          for (int i = k; i < m; i++ ) {
            U[i][k] = -U[i][k];
          }
          U[k][k] = 1.0 + U[k][k];
          for (int i = 0; i < k-1; i++) {
            U[i][k] = 0.0;
          }
        } else {
          for (int i = 0; i < m; i++) {
            U[i][k] = 0.0;
          }
          U[k][k] = 1.0;
        }
      }
    }

    // If required, generate V.

    if (wantv) {
      for (int k = n-1; k >= 0; k--) {
        if ((k < nrt) & (e[k] != 0.0)) {
          for (int j = k+1; j < nu; j++) {
            double t = 0;
            for (int i = k+1; i < n; i++) {
              t += V[i][k]*V[i][j];
            }
            t = -t/V[k+1][k];
            for (int i = k+1; i < n; i++) {
              V[i][j] += t*V[i][k];
            }
          }
        }
        for (int i = 0; i < n; i++) {
          V[i][k] = 0.0;
        }
        V[k][k] = 1.0;
      }
    }

    // Main iteration loop for the singular values.

    int pp = p-1;
    int iter = 0;
    double eps = Math.pow(2.0,-52.0);
    while (p > 0) {
      int k,kase;

      // Here is where a test for too many iterations would go.

      // This section of the program inspects for
      // negligible elements in the s and e arrays.  On
      // completion the variables kase and k are set as follows.

      // kase = 1     if s(p) and e[k-1] are negligible and k<p
      // kase = 2     if s(k) is negligible and k<p
      // kase = 3     if e[k-1] is negligible, k<p, and
      //              s(k), ..., s(p) are not negligible (qr step).
      // kase = 4     if e(p-1) is negligible (convergence).

      for (k = p-2; k >= -1; k--) {
        if (k == -1) {
          break;
        }
        if (Math.abs(e[k]) <= eps*(Math.abs(s[k]) + Math.abs(s[k+1]))) {
          e[k] = 0.0;
          break;
        }
      }
      if (k == p-2) {
        kase = 4;
      } else {
        int ks;
        for (ks = p-1; ks >= k; ks--) {
          if (ks == k) {
            break;
          }
          double t = (ks != p ? Math.abs(e[ks]) : 0.) + 
                                                  (ks != k+1 ? Math.abs(e[ks-1]) : 0.);
          if (Math.abs(s[ks]) <= eps*t)  {
            s[ks] = 0.0;
            break;
          }
        }
        if (ks == k) {
          kase = 3;
        } else if (ks == p-1) {
          kase = 1;
        } else {
          kase = 2;
          k = ks;
        }
      }
      k++;

      // Perform the task indicated by kase.

      switch (kase) {

        // Deflate negligible s(p).

        case 1: {
                  double f = e[p-2];
                  e[p-2] = 0.0;
                  for (int j = p-2; j >= k; j--) {
                    double t = Maths.hypot(s[j],f);
                    double cs = s[j]/t;
                    double sn = f/t;
                    s[j] = t;
                    if (j != k) {
                      f = -sn*e[j-1];
                      e[j-1] = cs*e[j-1];
                    }
                    if (wantv) {
                      for (int i = 0; i < n; i++) {
                        t = cs*V[i][j] + sn*V[i][p-1];
                        V[i][p-1] = -sn*V[i][j] + cs*V[i][p-1];
                        V[i][j] = t;
                      }
                    }
                  }
        }
        break;

        // Split at negligible s(k).

        case 2: {
                  double f = e[k-1];
                  e[k-1] = 0.0;
                  for (int j = k; j < p; j++) {
                    double t = Maths.hypot(s[j],f);
                    double cs = s[j]/t;
                    double sn = f/t;
                    s[j] = t;
                    f = -sn*e[j];
                    e[j] = cs*e[j];
                    if (wantu) {
                      for (int i = 0; i < m; i++) {
                        t = cs*U[i][j] + sn*U[i][k-1];
                        U[i][k-1] = -sn*U[i][j] + cs*U[i][k-1];
                        U[i][j] = t;
                      }
                    }
                  }
        }
        break;

        // Perform one qr step.

        case 3: {

                  // Calculate the shift.

                  double scale = Math.max(Math.max(Math.max(Math.max(
                            Math.abs(s[p-1]),Math.abs(s[p-2])),Math.abs(e[p-2])), 
                        Math.abs(s[k])),Math.abs(e[k]));
                  double sp = s[p-1]/scale;
                  double spm1 = s[p-2]/scale;
                  double epm1 = e[p-2]/scale;
                  double sk = s[k]/scale;
                  double ek = e[k]/scale;
                  double b = ((spm1 + sp)*(spm1 - sp) + epm1*epm1)/2.0;
                  double c = (sp*epm1)*(sp*epm1);
                  double shift = 0.0;
                  if ((b != 0.0) | (c != 0.0)) {
                    shift = Math.sqrt(b*b + c);
                    if (b < 0.0) {
                      shift = -shift;
                    }
                    shift = c/(b + shift);
                  }
                  double f = (sk + sp)*(sk - sp) + shift;
                  double g = sk*ek;

                  // Chase zeros.

                  for (int j = k; j < p-1; j++) {
                    double t = Maths.hypot(f,g);
                    double cs = f/t;
                    double sn = g/t;
                    if (j != k) {
                      e[j-1] = t;
                    }
                    f = cs*s[j] + sn*e[j];
                    e[j] = cs*e[j] - sn*s[j];
                    g = sn*s[j+1];
                    s[j+1] = cs*s[j+1];
                    if (wantv) {
                      for (int i = 0; i < n; i++) {
                        t = cs*V[i][j] + sn*V[i][j+1];
                        V[i][j+1] = -sn*V[i][j] + cs*V[i][j+1];
                        V[i][j] = t;
                      }
                    }
                    t = Maths.hypot(f,g);
                    cs = f/t;
                    sn = g/t;
                    s[j] = t;
                    f = cs*e[j] + sn*s[j+1];
                    s[j+1] = -sn*e[j] + cs*s[j+1];
                    g = sn*e[j+1];
                    e[j+1] = cs*e[j+1];
                    if (wantu && (j < m-1)) {
                      for (int i = 0; i < m; i++) {
                        t = cs*U[i][j] + sn*U[i][j+1];
                        U[i][j+1] = -sn*U[i][j] + cs*U[i][j+1];
                        U[i][j] = t;
                      }
                    }
                  }
                  e[p-2] = f;
                  iter = iter + 1;
        }
        break;

        // Convergence.

        case 4: {

                  // Make the singular values positive.

                  if (s[k] <= 0.0) {
                    s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
                    if (wantv) {
                      for (int i = 0; i <= pp; i++) {
                        V[i][k] = -V[i][k];
                      }
                    }
                  }

                  // Order the singular values.

                  while (k < pp) {
                    if (s[k] >= s[k+1]) {
                      break;
                    }
                    double t = s[k];
                    s[k] = s[k+1];
                    s[k+1] = t;
                    if (wantv && (k < n-1)) {
                      for (int i = 0; i < n; i++) {
                        t = V[i][k+1]; V[i][k+1] = V[i][k]; V[i][k] = t;
                      }
                    }
                    if (wantu && (k < m-1)) {
                      for (int i = 0; i < m; i++) {
                        t = U[i][k+1]; U[i][k+1] = U[i][k]; U[i][k] = t;
                      }
                    }
                    k++;
                  }
                  iter = 0;
                  p--;
        }
        break;
      }
    }
  }

  /** 
   * Return the left singular vectors
   * @return     U
   */
  public Matrix getU() {
    return new Matrix(U,m,Math.min(m+1,n));
  }

  /** 
   * Return the right singular vectors
   * @return     V
   */
  public Matrix getV() {
    return new Matrix(V,n,n);
  }

  /** 
   * Return the one-dimensional array of singular values
   * @return     diagonal of S.
   */
  public double[] getSingularValues() {
    return s;
  }

  /** 
   * Return the diagonal matrix of singular values
   * @return     S
   */
  public Matrix getS() {
    Matrix X = new Matrix(n,n);
    double[][] S = X.getArray();
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        S[i][j] = 0.0;
      }
      S[i][i] = this.s[i];
    }
    return X;
  }

  /** 
   * Two norm
   * @return     max(S)
   */
  public double norm2() {
    return s[0];
  }

  /** 
   * Two norm condition number
   * @return     max(S)/min(S)
   */
  public double cond() {
    return s[0]/s[Math.min(m,n)-1];
  }

  /** 
   * Effective numerical matrix rank
   * @return     Number of nonnegligible singular values.
   */
  public int rank() {
    double eps = Math.pow(2.0,-52.0);
    double tol = Math.max(m,n)*s[0]*eps;
    int r = 0;
    for (int i = 0; i < s.length; i++) {
      if (s[i] > tol) {
        r++;
      }
    }
    return r;
  }
}