Pyspark和PCA:如何提取此PCA的特征向量？如何计算他们解释的方差？

Question

我正在使用pyspark(使用库)减少Spark DataFrame带有PCA模型的维度,spark ml如下所示:

pca = PCA(k=3, inputCol="features", outputCol="pca_features")
model = pca.fit(data)

在哪里data是一个Spark DataFrame实验室,其中features一个DenseVector是3维:

data.take(1)
Row(features=DenseVector([0.4536,-0.43218, 0.9876]), label=u'class1')

拟合后,我转换数据:

transformed = model.transform(data)
transformed.first()
Row(features=DenseVector([0.4536,-0.43218, 0.9876]), label=u'class1', pca_features=DenseVector([-0.33256, 0.8668, 0.625]))

我的问题是:如何提取此PCA的特征向量？如何计算他们解释的方差？

Answer 1

[ 更新:从Spark 2.2开始,PCA和SVD都可在PySpark中使用 - 请参阅JIRA票证SPARK-6227以及用于Spark ML 2.2的PCA和PCAModel ; 以下原始答案仍适用于较旧的Spark版本.]

嗯,这似乎令人难以置信,但实际上没有办法从PCA分解中提取这些信息(至少从Spark 1.5开始).但同样,有许多类似的"抱怨" - 例如,请参阅此处,因为无法从a中提取最佳参数CrossValidatorModel.

幸运的是,几个月前,我参加了AMPLab(Berkeley)和Databricks 的"可扩展机器学习" MOOC,即Spark的创建者,我们在那里实施了一个完整的PCA管道"手工"作为家庭作业的一部分.我已经修改了我的函数(请放心,我得到了充分的信任:-),以便将数据框作为输入(而不是RDD),与您的格式相同(即DenseVectors包含数字特征的行).

我们首先需要定义一个中间函数,estimatedCovariance如下所示:

import numpy as np

def estimateCovariance(df):
    """Compute the covariance matrix for a given dataframe.

    Note:
        The multi-dimensional covariance array should be calculated using outer products.  Don't
        forget to normalize the data by first subtracting the mean.

    Args:
        df:  A Spark dataframe with a column named 'features', which (column) consists of DenseVectors.

    Returns:
        np.ndarray: A multi-dimensional array where the number of rows and columns both equal the
            length of the arrays in the input dataframe.
    """
    m = df.select(df['features']).map(lambda x: x[0]).mean()
    dfZeroMean = df.select(df['features']).map(lambda x:   x[0]).map(lambda x: x-m)  # subtract the mean

    return dfZeroMean.map(lambda x: np.outer(x,x)).sum()/df.count()

然后,我们可以编写一个main pca函数如下:

from numpy.linalg import eigh

def pca(df, k=2):
    """Computes the top `k` principal components, corresponding scores, and all eigenvalues.

    Note:
        All eigenvalues should be returned in sorted order (largest to smallest). `eigh` returns
        each eigenvectors as a column.  This function should also return eigenvectors as columns.

    Args:
        df: A Spark dataframe with a 'features' column, which (column) consists of DenseVectors.
        k (int): The number of principal components to return.

    Returns:
        tuple of (np.ndarray, RDD of np.ndarray, np.ndarray): A tuple of (eigenvectors, `RDD` of
        scores, eigenvalues).  Eigenvectors is a multi-dimensional array where the number of
        rows equals the length of the arrays in the input `RDD` and the number of columns equals
        `k`.  The `RDD` of scores has the same number of rows as `data` and consists of arrays
        of length `k`.  Eigenvalues is an array of length d (the number of features).
     """
    cov = estimateCovariance(df)
    col = cov.shape[1]
    eigVals, eigVecs = eigh(cov)
    inds = np.argsort(eigVals)
    eigVecs = eigVecs.T[inds[-1:-(col+1):-1]]  
    components = eigVecs[0:k]
    eigVals = eigVals[inds[-1:-(col+1):-1]]  # sort eigenvals
    score = df.select(df['features']).map(lambda x: x[0]).map(lambda x: np.dot(x, components.T) )
    # Return the `k` principal components, `k` scores, and all eigenvalues

    return components.T, score, eigVals

测试

让我们首先看一下现有方法的结果,使用Spark ML PCA 文档中的示例数据(将它们全部修改DenseVectors):

 from pyspark.ml.feature import *
 from pyspark.mllib.linalg import Vectors
 data = [(Vectors.dense([0.0, 1.0, 0.0, 7.0, 0.0]),),
         (Vectors.dense([2.0, 0.0, 3.0, 4.0, 5.0]),),
         (Vectors.dense([4.0, 0.0, 0.0, 6.0, 7.0]),)]
 df = sqlContext.createDataFrame(data,["features"])
 pca_extracted = PCA(k=2, inputCol="features", outputCol="pca_features")
 model = pca_extracted.fit(df)
 model.transform(df).collect()

 [Row(features=DenseVector([0.0, 1.0, 0.0, 7.0, 0.0]), pca_features=DenseVector([1.6486, -4.0133])),
  Row(features=DenseVector([2.0, 0.0, 3.0, 4.0, 5.0]), pca_features=DenseVector([-4.6451, -1.1168])),
  Row(features=DenseVector([4.0, 0.0, 0.0, 6.0, 7.0]), pca_features=DenseVector([-6.4289, -5.338]))]

然后,用我们的方法:

 comp, score, eigVals = pca(df)
 score.collect()

 [array([ 1.64857282,  4.0132827 ]),
  array([-4.64510433,  1.11679727]),
  array([-6.42888054,  5.33795143])]

让我强调一点,我们在我们定义的函数中没有使用任何collect()方法 - score是RDD应该的.

请注意,我们的第二列的符号与现有方法的符号完全相反; 但这不是问题:根据(免费下载)统计学习简介,由Hastie和Tibshirani共同撰写,p.382

每个主要组件加载向量都是唯一的,直到符号翻转.这意味着两个不同的软件包将产生相同的主成分加载向量,尽管这些加载向量的符号可能不同.符号可能不同,因为每个主成分加载向量指定了p维空间中的方向:由于方向不改变,因此翻转符号无效.[...]类似地,得分向量在符号翻转之前是唯一的,因为Z的方差与-Z的方差相同.

最后,既然我们已经有了特征值,那么编写一个函数来解释方差的百分比是很简单的:

 def varianceExplained(df, k=1):
     """Calculate the fraction of variance explained by the top `k` eigenvectors.

     Args:
         df: A Spark dataframe with a 'features' column, which (column) consists of DenseVectors.
         k: The number of principal components to consider.

     Returns:
         float: A number between 0 and 1 representing the percentage of variance explained
             by the top `k` eigenvectors.
     """
     components, scores, eigenvalues = pca(df, k)  
     return sum(eigenvalues[0:k])/sum(eigenvalues)


 varianceExplained(df,1)
 # 0.79439325322305299

作为测试,我们还检查在我们的示例数据中解释的方差是否为1.0,对于k = 5(因为原始数据是5维的):

 varianceExplained(df,5)
 # 1.0

这应该有效地完成你的工作; 随时提出您可能需要的任何澄清.

[使用Spark 1.5.0和1.5.1开发和测试]

Answer 2

编辑:

PCA并SVD终于在这两个可用pyspark开始火花2.2.0根据本解决JIRA票SPARK-6227.

原始答案:

从理论的角度来看,@ desertnaut给出的答案实际上非常出色,但我想提出另一种方法来研究如何计算SVD并提取特征向量.

from pyspark.mllib.common import callMLlibFunc, JavaModelWrapper
from pyspark.mllib.linalg.distributed import RowMatrix

class SVD(JavaModelWrapper):
    """Wrapper around the SVD scala case class"""
    @property
    def U(self):
        """ Returns a RowMatrix whose columns are the left singular vectors of the SVD if computeU was set to be True."""
        u = self.call("U")
        if u is not None:
        return RowMatrix(u)

    @property
    def s(self):
        """Returns a DenseVector with singular values in descending order."""
        return self.call("s")

    @property
    def V(self):
        """ Returns a DenseMatrix whose columns are the right singular vectors of the SVD."""
        return self.call("V")

这定义了我们的SVD对象.我们现在可以使用Java Wrapper定义我们的computeSVD方法.

def computeSVD(row_matrix, k, computeU=False, rCond=1e-9):
    """
    Computes the singular value decomposition of the RowMatrix.
    The given row matrix A of dimension (m X n) is decomposed into U * s * V'T where
    * s: DenseVector consisting of square root of the eigenvalues (singular values) in descending order.
    * U: (m X k) (left singular vectors) is a RowMatrix whose columns are the eigenvectors of (A X A')
    * v: (n X k) (right singular vectors) is a Matrix whose columns are the eigenvectors of (A' X A)
    :param k: number of singular values to keep. We might return less than k if there are numerically zero singular values.
    :param computeU: Whether of not to compute U. If set to be True, then U is computed by A * V * sigma^-1
    :param rCond: the reciprocal condition number. All singular values smaller than rCond * sigma(0) are treated as zero, where sigma(0) is the largest singular value.
    :returns: SVD object
    """
    java_model = row_matrix._java_matrix_wrapper.call("computeSVD", int(k), computeU, float(rCond))
    return SVD(java_model)

现在,让我们将其应用于一个例子:

from pyspark.ml.feature import *
from pyspark.mllib.linalg import Vectors

data = [(Vectors.dense([0.0, 1.0, 0.0, 7.0, 0.0]),), (Vectors.dense([2.0, 0.0, 3.0, 4.0, 5.0]),), (Vectors.dense([4.0, 0.0, 0.0, 6.0, 7.0]),)]
df = sqlContext.createDataFrame(data,["features"])

pca_extracted = PCA(k=2, inputCol="features", outputCol="pca_features")

model = pca_extracted.fit(df)
features = model.transform(df) # this create a DataFrame with the regular features and pca_features

# We can now extract the pca_features to prepare our RowMatrix.
pca_features = features.select("pca_features").rdd.map(lambda row : row[0])
mat = RowMatrix(pca_features)

# Once the RowMatrix is ready we can compute our Singular Value Decomposition
svd = computeSVD(mat,2,True)
svd.s
# DenseVector([9.491, 4.6253])
svd.U.rows.collect()
# [DenseVector([0.1129, -0.909]), DenseVector([0.463, 0.4055]), DenseVector([0.8792, -0.0968])]
svd.V
# DenseMatrix(2, 2, [-0.8025, -0.5967, -0.5967, 0.8025], 0)

Answer 3

您问题的最简单答案是将单位矩阵输入到您的模型中。

identity_input = [(Vectors.dense([1.0, .0, 0.0, .0, 0.0]),),(Vectors.dense([.0, 1.0, .0, .0, .0]),), \
              (Vectors.dense([.0, 0.0, 1.0, .0, .0]),),(Vectors.dense([.0, 0.0, .0, 1.0, .0]),),
              (Vectors.dense([.0, 0.0, .0, .0, 1.0]),)]
df_identity = sqlContext.createDataFrame(identity_input,["features"])
identity_features = model.transform(df_identity)

这应该为您提供主要组成部分。

我认为 Eliasah 的答案在 Spark 框架方面更好，因为 Desertnaut 正在通过使用 numpy 的函数而不是 Spark 的操作来解决问题。然而，埃利亚萨的答案缺少对数据的标准化。因此，我将以下几行添加到 Eliasah 的答案中：

from pyspark.ml.feature import StandardScaler
standardizer = StandardScaler(withMean=True, withStd=False,
                          inputCol='features',
                          outputCol='std_features')
model = standardizer.fit(df)
output = model.transform(df)
pca_features = output.select("std_features").rdd.map(lambda row : row[0])
mat = RowMatrix(pca_features)
svd = computeSVD(mat,5,True)

显然， svd.V 和 Identity_features.select("pca_features").collect() 应该具有相同的值。

我在这篇博文中总结了 PCA 及其在 Spark 和 sklearn 中的使用。

Answer 4

在spark 2.2+中，您现在可以轻松获得解释的方差为：

from pyspark.ml.feature import VectorAssembler
assembler = VectorAssembler(inputCols=<columns of your original dataframe>, outputCol="features")
df = assembler.transform(<your original dataframe>).select("features")
from pyspark.ml.feature import PCA
pca = PCA(k=10, inputCol="features", outputCol="pcaFeatures")
model = pca.fit(df)
sum(model.explainedVariance)