List of solveFToD() Examples

Examples of solveFToD()

com.cloudera.oryx.common.math.Solver.solveFToD()
net.myrrix.common.math.Solver.solveFToD()
Like {@link #solveDToF(double[])} but input is a {@code float} array and output is a {@code double} array.

Examples of com.cloudera.oryx.common.math.Solver.solveFToD()

          // Qu' is 0 except for one value at position i, so it's really (Qui')*Yi
          float[] QuiYi = Yi.clone();
          for (int i = 0; i < QuiYi.length; i++) {
            QuiYi[i] *= targetQui;
          }
          newXu = YTYsolver.solveFToD(QuiYi);
        }
      }


      // Similarly for Y vs X
      double[] newYi = null;

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Examples of net.myrrix.common.math.Solver.solveFToD()

      }
      if (itemFeatures == null) {
        continue;
      }
      anyItemIDFound = true;
      double[] userFoldIn = ytySolver.solveFToD(itemFeatures);
      if (anonymousUserFeatures == null) {
        anonymousUserFeatures = new float[userFoldIn.length];
      }
      double signedFoldInWeight = foldInWeight(0.0, values == null ? 1.0f : values[j]);
      if (signedFoldInWeight != 0.0) {

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Examples of net.myrrix.common.math.Solver.solveFToD()

    // Here, we are using userFeatures, which is a row of X, as if it were a column of X'.
    // This is multiplied on the left by (X'*X)^-1. That's our left-inverse of X or at least the one
    // column we need. Which is what the new data point is multiplied on the left by. The result is a column;
    // we scale to complete the multiplication of the fold-in and add it in.
    Solver xtxSolver = generation.getXTXSolver();
    double[] itemFoldIn = xtxSolver == null ? null : xtxSolver.solveFToD(userFeatures);


    // Same, but reversed. Multiply itemFeatures, which is a row of Y, on the right by (Y'*Y)^-1.
    // This is the right-inverse for Y', or at least the row we need. Because of the symmetries we can use
    // the same method above to carry out the multiply; the result is conceptually a row vector.
    // The result is scaled and added in.

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Examples of net.myrrix.common.math.Solver.solveFToD()

    // Same, but reversed. Multiply itemFeatures, which is a row of Y, on the right by (Y'*Y)^-1.
    // This is the right-inverse for Y', or at least the row we need. Because of the symmetries we can use
    // the same method above to carry out the multiply; the result is conceptually a row vector.
    // The result is scaled and added in.
    Solver ytySolver = generation.getYTYSolver();
    double[] userFoldIn = ytySolver == null ? null : ytySolver.solveFToD(itemFeatures);


    if (itemFoldIn != null) {
      if (SimpleVectorMath.norm(userFoldIn) > BIG_FOLDIN_THRESHOLD) {
        log.warn("Item fold in vector is large; reduce -Dmodel.features?");
      }

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