在igraph

时间:2016-08-17 08:17:02

标签: r igraph

我有一个有向的二分图g,有215473个顶点和2326714个边。在创建bipartite.projection g时,我的内存不足(在崩溃之前它使用~35 gig的RAM)。

我尝试按照previous thread on nongnu.org来计算我需要多少内存。

根据此线程中提供的信息,将图形存储在内存成本中(以字节为单位):

(4*|E|+2*|V|) * 8 + 4*|V|

要计算投影需要以下内存(以字节为单位):

16*|V| + (2*|V|+2*|E|) * 8

因此,对于我的图表g,它将花费:

((4*2326714+2*215473) * 8 + 4*215473) + (16*215473 + (2*215473+2*2326714) * 8)

= 78764308 + 44122560
= 122886868 (bytes)
= 122.886868 (mb)

显然,这是不正确的,我一定是做错了。

任何人都可以帮忙弄清楚如何创建我的图表的二分投影吗?

1 个答案:

答案 0 :(得分:1)

使用稀疏矩阵可能会解决您的问题。

# Load tiny toy data as edgelist
df <- data.frame( person =
    c('Sam','Sam','Sam','Greg','Tom','Tom','Tom','Mary','Mary'), group =
    c('a','b','c','a','b','c','d','b','d'), stringsAsFactors = F)

# Transform data to a sparse matrix
library(Matrix)
A <- spMatrix(nrow=length(unique(df$person)),
        ncol=length(unique(df$group)),
        i = as.numeric(factor(df$person)),
        j = as.numeric(factor(df$group)),
        x = rep(1, length(as.numeric(df$person))) )
row.names(A) <- levels(factor(df$person))
colnames(A) <- levels(factor(df$group))

为了进行投影你有多种可能性,这里有两个:

# Use base r
Arow <- tcrossprod(A)
# Alternatively, if you want to project on the other mode:
Acol <- tcrossprod(t(A))

# Use the igraph package, which works with sparse matrices
library(igraph)
g <- graph.incidence(A)

# The command bipartite.projection does both possible projections at once
proj <- bipartite.projection(g)

#proj[[1]]
#proj[[2]]

您还可以使用spMatrix读入数据并在data.table命令中进行转换,这也将加快这些操作。

<强>更新

以下是一个更大的图表和一些内存基准测试的示例:

# Load packages
library(data.table)
library(igraph)

# Scientific collaboration dataset
# Descriptives as reported on https://toreopsahl.com/datasets/#newman2001 
# mode 1 elements: 16726    
# mode 2 elements: 22016    
# two mode ties: 58595
# one mode ties: 47594  
d <- fread("http://opsahl.co.uk/tnet/datasets/Newman-Cond_mat_95-99-two_mode.txt", 
           stringsAsFactors=TRUE, colClasses = "factor", header=FALSE)

# Transform data to a sparse matrix
A <- spMatrix(nrow=length(unique(d[, V1])),
              ncol=length(unique(d[, V2])),
              i = as.numeric(d[, V1]),
              j = as.numeric(d[, V2]),
              x = rep(1, length(as.numeric(d[, V1]))) )
row.names(A) <- levels(d[, V1])
colnames(A) <- levels(d[, V2])

#To do the projection you have acutally multiple possiblities, here are two:

# Use base r
Arow <- tcrossprod(A)
# Alternatively, if you want to project on the other mode:
Acol <- tcrossprod(t(A))

下面概述了使用了多少内存,即稀疏矩阵方法在我的笔记本电脑上用于此网络,但是使用常规矩阵的方法确实会产生内存分配错误(即使从中移除Bcol对象也是如此内存rm(Brow)然后调用垃圾收集器gc()

object.size(A) # Spare matrix: 3108520 bytes
object.size(Arow) # 2713768 bytes
object.size(Acol) # 5542104 bytes

# For comparison
object.size(B <- as.matrix(A)) # Regular matrix: 2945783320 bytes
object.size(Brow <- tcrossprod(B)) # 2239946368 bytes
object.size(Bcol <- tcrossprod(t(B))) # Memory allocation error on my laptop