org.apache.mahout.math.jet.random.Gamma.nextDouble()方法的使用及代码示例

x33g5p2x  于2022-01-20 转载在 其他  
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本文整理了Java中org.apache.mahout.math.jet.random.Gamma.nextDouble()方法的一些代码示例，展示了Gamma.nextDouble()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。Gamma.nextDouble()方法的具体详情如下：
包路径：org.apache.mahout.math.jet.random.Gamma
类名称：Gamma
方法名：nextDouble

Gamma.nextDouble介绍

[英]Returns a random number from the distribution.
[中]从分布中返回一个随机数。

代码示例

代码示例来源：origin: apache/mahout

  1. /** Returns a random number from the distribution. */
  2. @Override
  3. public double nextDouble() {
  4.  return nextDouble(alpha, rate);
  5. }

代码示例来源：origin: apache/mahout

  1. /**
  2.  * Returns a sample from this distribution.  The value returned will
  3.  * be the number of negative samples required before achieving r
  4.  * positive samples.  Each successive sample is taken independently
  5.  * from a Bernouli process with probability p of success.
  6.  *
  7.  * The algorithm used is taken from J.H. Ahrens, U. Dieter (1974):
  8.  * Computer methods for sampling from gamma, beta, Poisson and
  9.  * binomial distributions, Computing 12, 223--246.
  10.  *
  11.  * This algorithm is essentially the same as described at
  12.  * http://en.wikipedia.org/wiki/Negative_binomial_distribution#Gamma.E2.80.93Poisson_mixture
  13.  * except that the notion of positive and negative outcomes is uniformly
  14.  * inverted.  Because the inversion is complete and consistent, this
  15.  * definition is effectively identical to that defined on wikipedia.
  16.  */
  17. public int nextInt(int r, double p) {
  18.  return this.poisson.nextInt(gamma.nextDouble(r, p / (1.0 - p)));
  19. }

代码示例来源：origin: apache/mahout

  1. int[] values = new int[1000];
  2. for (int k = 0; k < 1000; k++) {
  3.  int j = (int) Math.min(1000, gen.nextDouble());
  4.  values[j]++;
  5. int[] values = new int[1000];
  6. for (int k = 0; k < 1000; k++) {
  7.  int j = (int) Math.min(1000, gen.nextDouble());
  8.  values[j]++;

代码示例来源：origin: apache/mahout

  1. @Test
  2. public void testNextDouble() {
  3.  double[] z = new double[100000];
  4.  Random gen = RandomUtils.getRandom();
  5.  for (double alpha : new double[]{1, 2, 10, 0.1, 0.01, 100}) {
  6.   Gamma g = new Gamma(alpha, 1, gen);
  7.   for (int i = 0; i < z.length; i++) {
  8.    z[i] = g.nextDouble();
  9.   }
  10.   Arrays.sort(z);
  11.   // verify that empirical CDF matches theoretical one pretty closely
  12.   for (double q : seq(0.01, 1, 0.01)) {
  13.    double p = z[(int) (q * z.length)];
  14.    assertEquals(q, g.cdf(p), 0.01);
  15.   }
  16.  }
  17. }

代码示例来源：origin: apache/mahout

  1. int[] values = new int[1000];
  2. for (int k = 0; k < 1000; k++) {
  3.  int j = (int) Math.min(1000, gen.nextDouble());
  4.  values[j]++;

代码示例来源：origin: apache/mahout

  1. @Test(timeout=50000)
  2. public void testTimesOtherSparseEfficiency() {
  3.  Random raw = RandomUtils.getRandom();
  4.  Gamma gen = new Gamma(0.1, 0.1, raw);
  5.  // build a sequential sparse matrix and a diagonal matrix and multiply them
  6.  Matrix x = new SparseRowMatrix(1000, 2000, false);
  7.  for (int i = 0; i < 1000; i++) {
  8.   int[] values = new int[1000];
  9.   for (int k = 0; k < 1000; k++) {
  10.    int j = (int) Math.min(1000, gen.nextDouble());
  11.    values[j]++;
  12.   }
  13.   for (int j = 0; j < 1000; j++) {
  14.    if (values[j] > 0) {
  15.     x.set(i, j, values[j]);
  16.    }
  17.   }
  18.  }
  19.  Vector d = new DenseVector(2000).assign(Functions.random());
  20.  Matrix y = new DiagonalMatrix(d);
  21.  long t0 = System.nanoTime();
  22.  Matrix z = x.times(y);
  23.  double elapsedTime = (System.nanoTime() - t0) * 1e-6;
  24.  System.out.printf("done in %.1f ms\n", elapsedTime);
  25.  for (MatrixSlice row : z) {
  26.   for (Vector.Element element : row.nonZeroes()) {
  27.    assertEquals(x.get(row.index(), element.index()) * d.get(element.index()), element.get(), 1e-12);
  28.   }
  29.  }
  30. }

代码示例来源：origin: tdunning/bandit-ranking

  1. public double nextDouble(double alpha, double beta) {
  2.   double x = gAlpha.nextDouble(alpha, 1);
  3.   double y = gBeta.nextDouble(beta, 1);
  4.   return x / (x + y);
  5. }

代码示例来源：origin: cloudera/mahout

  1. /** Returns a random number from the distribution. */
  2. @Override
  3. public double nextDouble() {
  4.  return nextDouble(alpha, rate);
  5. }

代码示例来源：origin: org.apache.mahout/mahout-math

  1. /** Returns a random number from the distribution. */
  2. @Override
  3. public double nextDouble() {
  4.  return nextDouble(alpha, rate);
  5. }

代码示例来源：origin: tdunning/bandit-ranking

  1. /**
  2.  * Returns a random number from the distribution.
  3.  *
  4.  * @return A new sample from this distribution.
  5.  */
  6. @Override
  7. public double nextDouble() {
  8.   double x = gAlpha.nextDouble(alpha, 1);
  9.   double y = gBeta.nextDouble(beta, 1);
  10.   return x / (x + y);
  11. }

代码示例来源：origin: tdunning/bandit-ranking

  1. private double nextVariance() {
  2.     return 1 / gd.nextDouble(n / 2, ss / 2);
  3.   }
  4. }

代码示例来源：origin: tdunning/log-synth

  1. private int pickField() {
  2.     double xValue = x.nextDouble();
  3.     double yValue = y.nextDouble();
  4.     double beta = xValue / (xValue + yValue);
  5.     return (int) Math.floor(beta * fieldNames.size());
  6.   }
  7. }

代码示例来源：origin: tdunning/log-synth

  1. @Override
  2. public JsonNode sample() {
  3.   return new DoubleNode(rand.nextDouble());
  4. }

代码示例来源：origin: org.apache.mahout/mahout-math

  1. /**
  2.  * Returns a sample from this distribution.  The value returned will
  3.  * be the number of negative samples required before achieving r
  4.  * positive samples.  Each successive sample is taken independently
  5.  * from a Bernouli process with probability p of success.
  6.  *
  7.  * The algorithm used is taken from J.H. Ahrens, U. Dieter (1974):
  8.  * Computer methods for sampling from gamma, beta, Poisson and
  9.  * binomial distributions, Computing 12, 223--246.
  10.  *
  11.  * This algorithm is essentially the same as described at
  12.  * http://en.wikipedia.org/wiki/Negative_binomial_distribution#Gamma.E2.80.93Poisson_mixture
  13.  * except that the notion of positive and negative outcomes is uniformly
  14.  * inverted.  Because the inversion is complete and consistent, this
  15.  * definition is effectively identical to that defined on wikipedia.
  16.  */
  17. public int nextInt(int r, double p) {
  18.  return this.poisson.nextInt(gamma.nextDouble(r, p / (1.0 - p)));
  19. }

代码示例来源：origin: cloudera/mahout

  1. /**
  2.  * Returns a sample from this distribution.  The value returned will
  3.  * be the number of negative samples required before achieving r
  4.  * positive samples.  Each successive sample is taken independently
  5.  * from a Bernouli process with probability p of success.
  6.  *
  7.  * The algorithm used is taken from J.H. Ahrens, U. Dieter (1974):
  8.  * Computer methods for sampling from gamma, beta, Poisson and
  9.  * binomial distributions, Computing 12, 223--246.
  10.  *
  11.  * This algorithm is essentially the same as described at
  12.  * http://en.wikipedia.org/wiki/Negative_binomial_distribution#Gamma.E2.80.93Poisson_mixture
  13.  * except that the notion of positive and negative outcomes is uniformly
  14.  * inverted.  Because the inversion is complete and consistent, this
  15.  * definition is effectively identical to that defined on wikipedia.
  16.  */
  17. public int nextInt(int r, double p) {
  18.  return this.poisson.nextInt(gamma.nextDouble(r, p / (1.0 - p)));
  19. }

代码示例来源：origin: tdunning/log-synth

  1. double averageInterval = TimeUnit.MILLISECONDS.convert(1, TimeUnit.DAYS) / transactionsPerDay.nextDouble();
  2. Exponential interval = new Exponential(1 / averageInterval, gen);

代码示例来源：origin: cloudera/mahout

  1. @Test
  2. public void testNextDouble() {
  3.  double[] z = new double[100000];
  4.  Random gen = RandomUtils.getRandom();
  5.  for (double alpha : new double[]{1, 2, 10, 0.1, 0.01, 100}) {
  6.   Gamma g = new Gamma(alpha, 1, gen);
  7.   for (int i = 0; i < z.length; i++) {
  8.    z[i] = g.nextDouble();
  9.   }
  10.   Arrays.sort(z);
  11.   // verify that empirical CDF matches theoretical one pretty closely
  12.   for (double q : seq(0.01, 1, 0.01)) {
  13.    double p = z[(int) (q * z.length)];
  14.    assertEquals(q, g.cdf(p), 0.01);
  15.   }
  16.  }
  17. }

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