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

Question

我正在使用 pyspark（使用 spark ml 库）降低 Spark DataFrame 和 PCA 模型的维度，如下所示：

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

其中data 是Spark DataFrame，其中一列标记为features，这是一个DenseVector 的3 个维度：

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

拟合后，我对数据进行变换：

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]))

如何提取此 PCA 的特征向量？我如何计算他们解释的方差有多大？

Answer 1

[更新：从 Spark 2.2 开始，PCA 和 SVD 在 PySpark 中都可用 - 请参阅 JIRA 票证 SPARK-6227 和 PCA 和 PCAModel 了解 Spark ML 2.2；下面的原始答案仍然适用于旧版 Spark。]

嗯，这似乎令人难以置信，但确实没有办法从 PCA 分解中提取此类信息（至少从 Spark 1.5 开始）。但同样，也有许多类似的“抱怨”——例如，请参阅 here，因为无法从 CrossValidatorModel 中提取最佳参数。

幸运的是，几个月前，我参加了由 AMPLab（伯克利）和 Databricks （即 Spark 的创建者）举办的 'Scalable Machine Learning' MOOC，作为家庭作业的一部分，我们“手动”实施了完整的 PCA 管道。我从那时起修改了我的函数（请放心，我得到了充分的信任:-)，以便使用与您的格式相同的数据帧作为输入（而不是 RDD）（即 DenseVectors 的行包含数字功能）。

我们首先需要定义一个中间函数estimatedCovariance，如下：

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()

那么，我们可以编写一个mainpca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

让我们先看看现有方法的结果，使用来自 Spark ML PCA documentation 的示例数据（将它们修改为全部为 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])]

让我强调一下，我们不要在我们定义的函数中使用任何 collect() 方法 - score 应该是 RDD。

请注意，我们第二列的符号都与现有方法导出的符号相反；但这不是问题：根据（可免费下载的）An Introduction to Statistical Learning，由 Hastie & Tibshirani 合着，p。 382

每个主成分加载向量都是唯一的，最多一个符号翻转。这 意味着两个不同的软件包将产生相同的主体 组件加载向量，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 向量在 p 维空间中指定一个方向：翻转符号没有 方向不变的效果。 [...] 同样，分数向量是唯一的 最多一个符号翻转，因为 Z 的方差与 -Z 的方差相同。

最后，既然我们有了可用的特征值，编写一个解释方差百分比的函数就很简单了：

 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

[使用 Spark 1.5.0 和 1.5.1 开发和测试]

Answer 2

编辑：

根据已解决的 JIRA 票证SPARK-6227，

PCA 和 SVD 最终都可以在 pyspark 中使用，从 spark 2.2.0 开始。

原答案：

@desertnaut 给出的答案实际上从理论上讲非常好，但我想介绍另一种方法来计算 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，让我们将其应用到一个示例中：

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

您的问题最简单的答案是在您的模型中输入一个单位矩阵。

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件。

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。但是，以利亚萨的答案是缺少对数据进行规范化。所以，我会在 eliasah 的回答中添加以下几行：

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)

实际上，svd.V 和 identity_features.select("pca_features").collect() 应该具有相同的值。

我在blog post 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：

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)