R中图节点的二阶邻居

Question

我正在寻找一种有效的方法来查找大图中所有节点的精确度的邻域。即使它将图形存储为稀疏矩阵，igraph::ego也会爆炸：

require(Matrix)
require(igraph)
require(ggplot2)

N <- 10^(1:5)
runtimes <- function(N) {
  g <- erdos.renyi.game(N, 1/N)  
  system.time(ego(g, 2, mindist = 2))[3]
}
runtime <- sapply(N, runtimes)
qplot(log10(N), runtime, geom = "line")

有更有效的方法吗？

Answer 1

直接使用邻接矩阵可以显着改善。

# sparse adjacency-matrix calculation of indirect neighbors -------------------

diff_sparse_mat <- function(A, B) {
  # Difference between sparse matrices.
  #   Input: sparse matrices A and B
  #   Output: C = (A & !B), using element-wise diffing, treating B as logical
  stopifnot(identical(dim(A), dim(B)))
  A <- as(A, "generalMatrix")
  AT <- as.data.table(summary(as(A, "TsparseMatrix")))
  setkeyv(AT, c("i", "j"))
  B <- drop0(B)
  B <- as(B, "generalMatrix")
  BT <- as.data.table(summary(as(B, "TsparseMatrix")))
  setkeyv(BT, c("i", "j"))
  C <- AT[!BT]
  if (length(C) == 2) {
    return(sparseMatrix(i = C$i, j = C$j, dims = dim(A)))
  } else {
    return(sparseMatrix(i = C$i, j = C$j, x = C$x, dims = dim(A)))
  }
}

distance2_peers <- function(adj_mat) {
  # Returns a matrix of indirect neighbors, excluding the diagonal
  #   Input: adjacency matrix A (assumed symmetric)
  #   Output: (A %*% A & !A) with zero diagonal 
  indirect <- forceSymmetric(adj_mat %*% adj_mat) 
  indirect <- diff_sparse_mat(indirect, adj_mat)  # excl. direct neighbors
  indirect <- diff_sparse_mat(indirect, Diagonal(n = dim(indirect)[1]))  # excl. diag.
  return(indirect)  
}    

对于鄂尔多斯仁义的例子，在半分钟内现在可以分析10 ^ 7，而不是10 ^ 5的网络：

N <- 10 ^ (1:7)
runtimes <- function(N) {
  g <- erdos.renyi.game(N, 1 / N, directed = FALSE)
  system.time(distance2_peers(as_adjacency_matrix(g)))[3]
}
runtime <- sapply(N, runtimes)
qplot(log10(N), runtime, geom = "line")

结果矩阵包含（i，j）从i到j的长度为2的路径数（不包括i本身的路径）。