规则化
规则化器缩放单个样本让其拥有单位$L^{p}$范数。这是文本分类和聚类常用的操作。例如,两个$L^{2}$规则化的TFIDF
向量的点乘就是两个向量的cosine
相似度。
Normalizer
实现VectorTransformer
,将一个向量规则化为转换的向量,或者将一个RDD
规则化为另一个RDD
。下面是一个规则化的例子。
import org.apache.spark.SparkContext._
import org.apache.spark.mllib.feature.Normalizer
import org.apache.spark.mllib.linalg.Vectors
import org.apache.spark.mllib.util.MLUtils
val data = MLUtils.loadLibSVMFile(sc, "data/mllib/sample_libsvm_data.txt")
//默认情况下,p=2。计算2阶范数
val normalizer1 = new Normalizer()
val normalizer2 = new Normalizer(p = Double.PositiveInfinity)
// Each sample in data1 will be normalized using $L^2$ norm.
val data1 = data.map(x => (x.label, normalizer1.transform(x.features)))
// Each sample in data2 will be normalized using $L^\infty$ norm.
val data2 = data.map(x => (x.label, normalizer2.transform(x.features)))
规则化的实现很简单,我们看它的transform
方法。
override def transform(vector: Vector): Vector = {
//求范数
val norm = Vectors.norm(vector, p)
if (norm != 0.0) {
//稀疏向量可以重用index
vector match {
case DenseVector(vs) =>
val values = vs.clone()
val size = values.size
var i = 0
while (i < size) {
values(i) /= norm
i += 1
}
Vectors.dense(values)
case SparseVector(size, ids, vs) =>
val values = vs.clone()
val nnz = values.size
var i = 0
while (i < nnz) {
values(i) /= norm
i += 1
}
Vectors.sparse(size, ids, values)
case v => throw new IllegalArgumentException("Do not support vector type " + v.getClass)
}
} else {
vector
}
}
求范数调用了Vectors.norm
方法,我们可以看看该方法的实现。
def norm(vector: Vector, p: Double): Double = {
val values = vector match {
case DenseVector(vs) => vs
case SparseVector(n, ids, vs) => vs
case v => throw new IllegalArgumentException("Do not support vector type " + v.getClass)
}
val size = values.length
if (p == 1) {
var sum = 0.0
var i = 0
while (i < size) {
sum += math.abs(values(i))
i += 1
}
sum
} else if (p == 2) {
var sum = 0.0
var i = 0
while (i < size) {
sum += values(i) * values(i)
i += 1
}
math.sqrt(sum)
} else if (p == Double.PositiveInfinity) {
var max = 0.0
var i = 0
while (i < size) {
val value = math.abs(values(i))
if (value > max) max = value
i += 1
}
max
} else {
var sum = 0.0
var i = 0
while (i < size) {
sum += math.pow(math.abs(values(i)), p)
i += 1
}
math.pow(sum, 1.0 / p)
}
}
这里分四种情况。当p=1
时,即计算一阶范数,它的值为所有元素绝对值之和。当p=2
时,它的值为所有元素的平方和。当p == Double.PositiveInfinity
时,返回所有元素绝对值的最大值。
如果以上三种情况都不满足,那么按照下面的公式计算。