本文结合实例介绍如何用代数方法求PageRank。
Table of Contents
1. PageRank
博文《网页排序算法PageRank》介绍了PageRank,计算PageRank可以用迭代的方法也可以用代数的方法,其背后的数学基本运算是一样的,即:
$$
PR(p_i) = \frac{1-d}{N} + d \sum_{p_j \in B(p_i)} \frac{PR(p_j)}{L(p_j)}
$$
下文结合图1介绍如何用代数方法求PageRank。
Fig. 1: PageRanks for a simple network (image from Wikipedia).
为了便于讨论,将图1下方的节点分别标上G, H, I, J, K,如下图所示:
Fig. 2: Draw Fig. 1 in NetworkX.
2. 代数方法
根据1中的等式,把所有节点都放在一块,可以得到:
$$
\begin{bmatrix}
PR(p_1) \\
PR(p_2) \\
\vdots \\
PR(p_N)
\end{bmatrix} =
\begin{bmatrix}
{(1-d)/ N} \\
{(1-d) / N} \\
\vdots \\
{(1-d) / N}
\end{bmatrix}
+ d
\begin{bmatrix}
\ell(p_1,p_1) & \ell(p_1,p_2) & \cdots & \ell(p_1,p_N) \\
\ell(p_2,p_1) & \ddots & & \vdots \\
\vdots & & \ell(p_i,p_j) & \\
\ell(p_N,p_1) & \cdots & & \ell(p_N,p_N)
\end{bmatrix}
\cdot
\begin{bmatrix}
PR(p_1) \\
PR(p_2) \\
\vdots \\
PR(p_N)
\end{bmatrix}
$$
上述等式可以缩写为:
$$
\mathbf{R} = \frac{1-d}{N} \mathbf{1} + d \mathcal{M}\mathbf{R}
$$
其中,\(\mathbf{1}\)为\(N\)维的列向量,所有元素皆为1。以图1为例,该列向量为,
N = len(G.nodes()) # N = 11
column_vector = np.ones((N, 1), dtype=np.int)
[[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[1]]
2.1 Adjacency function
邻接函数(adjacency function)\(\ell(p_1,p_2)\)组成了矩阵\(\mathcal{M}\),
$$
\mathcal{M}_{ij} =
\ell(p_i,p_j) =
\begin{cases}
1 /L(p_j) , & \mbox{if }j\mbox{ links to }i. L(p_j)\mbox{是指从}p_j\mbox{链出去的网页数目} \\
0, & \mbox{otherwise}
\end{cases}
$$
这样矩阵每一行乘以\(\mathbf{R}\),就得到了新的PR值,比如第二行(图1的节点B
),
$$
\begin{split}
M_{2j} & = \ell(p_2,p_1) \cdot PR(p_2) + \ell(p_2,p_2) \cdot PR(p_2) + \cdots + \ell(p_2,p_N) \cdot PR(p_2) \\
&= 0 \mbox{ (‘A’)} + 0 \mbox{ (‘B’)} + 1 \mbox{ (‘C’)} + \frac{1}{2} \mbox{ (‘D’)} + \frac{1}{3} \mbox{ (‘E’)} + \frac{1}{2} \mbox{ (‘F’)} \\
& + \frac{1}{2} \mbox{ (‘G’)} + \frac{1}{2} \mbox{ (`H’)} + \frac{1}{2} \mbox{ (‘I’)} + 0 \mbox{ (‘J’)} + 0 \mbox{ (‘K’)}
\end{split}
$$
以节点G
为例,G
给B
和E
投票,所以B
得到1/2
。
矩阵\(\mathcal{M}\)每一列加起来都是1(值得注意的是,对于没有出链的节点,列加起来等于0,比如图1的节点A
),即\(\sum_{i = 1}^N \ell(p_i,p_j) = 1\)。事实上,\(\mathcal{M}\)是一个转移矩阵transition matrix(也叫概率矩阵probability matrix,马尔可夫矩阵Markov matrix)。因此,PageRank是eigenvector centrality的一个变体。
2.2 矩阵\(\mathcal{M}\)
事实上,\(\mathcal{M}\)可以被看成normalized的图邻接矩阵,即:
$$
\mathcal{M} = (K^{-1} A)^T
$$
其中,\(\mathcal{A}\)为图的邻接矩阵,以图1为例,
# Get adjacency matrix
nodelist = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K'] # sorted(G.nodes())
A = nx.to_numpy_matrix(G, nodelist)
'A' 'B' 'C' 'D' 'E' 'F' 'G' 'H' 'I' 'J' 'K'
[[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 1. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 1. 0. 1. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0.]]
\(\mathcal{A}\)是对角矩阵,对角线上的元素是对应节点的出度。
nodelist = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K'] # sorted(G.nodes())
list_outdegree = map(operator.itemgetter(1), sorted(G.out_degree().items()))
K = np.diag(list_outdegree)
[[0 0 0 0 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 2 0 0 0 0 0 0 0]
[0 0 0 0 3 0 0 0 0 0 0]
[0 0 0 0 0 2 0 0 0 0 0]
[0 0 0 0 0 0 2 0 0 0 0]
[0 0 0 0 0 0 0 2 0 0 0]
[0 0 0 0 0 0 0 0 2 0 0]
[0 0 0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 0 0 1]]
\(\mathbf{K}\)的逆矩阵\(\mathbf{K^{-1}}\)为,
K_inv = np.linalg.pinv(K)
[[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0.5 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0.33 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0.5 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0.5 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0.5 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0.5 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 1. ]]
那么,根据公式\(\mathcal{M} = (K^{-1} A)^T\)就可以求得\(\mathcal{M}\),如下,
M = (K_inv * A).transpose()
[[ 0. 0. 0. 0.5 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 1. 0.5 0.33 0.5 0.5 0.5 0.5 0. 0. ]
[ 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0.33 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0.5 0.5 0.5 0.5 1. 1. ]
[ 0. 0. 0. 0. 0.33 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. ]]
2.3 求解
\(\mathbf{R}\)是2.1等式的特征向量(eigenvector),求解等式得:
$$
\mathbf{R} = (\mathbf{I}-d \mathcal{M})^{-1} \frac{1-d}{N} \mathbf{1}
$$
其中\(\mathbf{I}\)是单位矩阵。
d = 0.85
I = np.identity(N)
R = np.linalg.pinv(I - d*M) * (1-d)/N * column_vector
[[ 0.028]
[ 0.324]
[ 0.289]
[ 0.033]
[ 0.068]
[ 0.033]
[ 0.014]
[ 0.014]
[ 0.014]
[ 0.014]
[ 0.014]]
咦,结果怎么跟图1不一样。得到\(\mathbf{R}\)需要normalized,如此,所有节点的PR加起来才能等于1。
R = R/sum(R) # normalized R, so that page ranks sum to 1.
[[ 0.033]
[ 0.384]
[ 0.343]
[ 0.039]
[ 0.081]
[ 0.039]
[ 0.016]
[ 0.016]
[ 0.016]
[ 0.016]
[ 0.016]]
用NetworkX作出来的图,是这样的:
Fig. 3: PageRanks for a simple network
本文涉及到的源代码已经分享到我的GitHub,pagerank目录下的pagerank_algebraic.py
.
3. Python源代码
NetworkX实现了PageRank的代数计算方法nx.pagerank_numpy
,源代码在这里。
def pagerank_numpy(G, alpha=0.85, personalization=None, weight='weight', dangling=None):
"""Return the PageRank of the nodes in the graph.
"""
if len(G) == 0:
return {}
M = google_matrix(G, alpha, personalization=personalization,
weight=weight, dangling=dangling)
# use numpy LAPACK solver
eigenvalues, eigenvectors = np.linalg.eig(M.T)
ind = eigenvalues.argsort()
# eigenvector of largest eigenvalue at ind[-1], normalized
largest = np.array(eigenvectors[:, ind[-1]]).flatten().real
norm = float(largest.sum())
return dict(zip(G, map(float, largest / norm)))
References:
[1] StackOverflow: Incorrect PageRank calculation result
Series: 网页排序算法PageRank (3/3)
- 网页排序算法PageRank
- 迭代方法求PageRank
- 代数方法求PageRank
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个人有一些问题,向请教请教;对于本章第2部分代数方法表示中,等式可以缩写为:R=((1−d)/N)*I+dmR,个人觉得有错,因为第一个R与第二个R的取值不一样,它是将等式后边的数值赋给R,这两个数值都不一样,不能左提取R进行进一步运算