I am trying to build a square adjacency matrix from a data.table. Here is a reproducible example of what I already have :
我正在尝试从data.table中构建一个正方形邻接矩阵。这里有一个可复制的例子说明我已经拥有的:
require(data.table)
require(plyr)
require(reshape2)
# Build a mock data.table
dt <- data.table(Source=as.character(rep(letters[1:3],2)),Target=as.character(rep(letters[4:2],2)))
dt
# Source Target
#1: a d
#2: b c
#3: c b
#4: a d
#5: b c
#6: c b
sry <- ddply(dt, .(Source,Target), summarize, Frequency=length(Source))
sry
# Source Target Frequency
#1 a d 2
#2 b c 2
#3 c b 2
mtx <- as.matrix(dcast(sry, Source ~ Target, value.var="Frequency", fill=0))
rownames(mtx) <- mtx[,1]
mtx <- mtx[,2:ncol(mtx)]
mtx
# b c d
#a "0" "0" "2"
#b "0" "2" "0"
#c "2" "0" "0"
Now, this is very close to what I want to get, except that I would like to have all the nodes represented in both dimensions, like :
现在，这很接近我想要得到的，除了我希望所有的节点都表示在两个维度中，比如:
a b c d
a 0 0 0 2
b 0 0 2 0
c 0 2 0 0
d 0 0 0 0
Note that I am working on quite large data, so I'd like to find an efficient solution for this.
请注意，我正在处理相当大的数据，因此我希望找到一个有效的解决方案。
Thank you for your help.
谢谢你的帮助。
SOLUTIONS (EDIT) :
解决方案(编辑):
Given the quality of the solutions offered and the size of my dataset, I benchmarked all the solutions.
考虑到所提供的解决方案的质量和数据集的大小，我对所有解决方案进行了基准测试。
#The bench was made with a 1-million-row sample from my original dataset
library(data.table)
aa <- fread("small2.csv",sep="^")
dt <- aa[,c(8,9),with=F]
colnames(dt) <- c("Source","Target")
dim(dt)
#[1] 1000001 2
levs <- unique(unlist(dt, use.names=F))
length(levs)
#[1] 2222
Given this data, the desired output is a 2222*2222 matrix (2222*2223 solutions where the first column contains the row names are also obviously acceptable).
考虑到这些数据，所需的输出是2222*2222矩阵(2222*2223解，其中第一列包含行名，显然也可以接受)。
# Ananda Mahto's first solution
am1 <- function() {
table(dt[, lapply(.SD, factor, levs)])
}
dim(am1())
#[1] 2222 2222
# Ananda Mahto's second solution
am2 <- function() {
as.matrix(dcast(dt[, lapply(.SD, factor, levs)], Source~Target, drop=F, value.var="Target", fun.aggregate=length))
}
dim(am2())
#[1] 2222 2223
library(dplyr)
library(tidyr)
# Akrun's solution
akr <- function() {
dt %>%
mutate_each(funs(factor(., levs))) %>%
group_by(Source, Target) %>%
tally() %>%
spread(Target, n, drop=FALSE, fill=0)
}
dim(akr())
#[1] 2222 2223
library(igraph)
# Carlos Cinelli's solution
cc <- function() {
g <- graph_from_data_frame(dt)
as_adjacency_matrix(g)
}
dim(cc())
#[1] 2222 2222
And the result of the benchmark is…
这个基准的结果是……
library(rbenchmark)
benchmark(am1(), am2(), akr(), cc(), replications=75)
# test replications elapsed relative user.self sys.self user.child sys.child
# 1 am1() 75 15.939 1.000 15.636 0.280 0 0
# 2 am2() 75 111.558 6.999 109.345 1.616 0 0
# 3 akr() 75 43.786 2.747 42.463 1.134 0 0
# 4 cc() 75 46.193 2.898 45.532 0.563 0 0
3 个解决方案
#1
7
It sounds like you're just looking for table, but you should make sure that both columns have the same factor levels:
听起来你只是在找表格，但你应该确保两列都有相同的因子水平:
levs <- unique(unlist(dt, use.names = FALSE))
table(lapply(dt, factor, levs))
# Target
# Source a b c d
# a 0 0 0 2
# b 0 0 2 0
# c 0 2 0 0
# d 0 0 0 0
I don't know if it would offer any speed improvements, but you could also use dcast from "data.table":
我不知道它是否能提高速度，但你也可以使用“data.table”中的dcast:
dcast(lapply(dt, factor, levs), Source ~ Target, drop = FALSE,
value.var = "Target", fun.aggregate = length)
#2
3
You can also use igraph. Since you said that you are dealing with large data, igraph has the advantage that it uses sparse matrices:
你也可以使用igraph。既然你说你要处理的是大数据，igraph的优势在于它使用的是稀疏矩阵:
library(igraph)
g <- graph_from_data_frame(dt)
as_adjacency_matrix(g)
4 x 4 sparse Matrix of class "dgCMatrix"
a b c d
a . . . 2
b . . 2 .
c . 2 . .
d . . . .
#3
1
We can use dplyr/tidyr
我们可以使用dplyr / tidyr
library(dplyr)
library(tidyr)
dt %>%
mutate_each(funs(factor(., letters[1:4]))) %>%
group_by(Source, Target) %>%
tally() %>%
spread(Target, n, drop=FALSE, fill=0)
# Source a b c d
# (fctr) (dbl) (dbl) (dbl) (dbl)
#1 a 0 0 0 2
#2 b 0 0 2 0
#3 c 0 2 0 0
#4 d 0 0 0 0
#1
7
It sounds like you're just looking for table, but you should make sure that both columns have the same factor levels:
听起来你只是在找表格，但你应该确保两列都有相同的因子水平:
levs <- unique(unlist(dt, use.names = FALSE))
table(lapply(dt, factor, levs))
# Target
# Source a b c d
# a 0 0 0 2
# b 0 0 2 0
# c 0 2 0 0
# d 0 0 0 0
I don't know if it would offer any speed improvements, but you could also use dcast from "data.table":
我不知道它是否能提高速度，但你也可以使用“data.table”中的dcast:
dcast(lapply(dt, factor, levs), Source ~ Target, drop = FALSE,
value.var = "Target", fun.aggregate = length)
#2
3
You can also use igraph. Since you said that you are dealing with large data, igraph has the advantage that it uses sparse matrices:
你也可以使用igraph。既然你说你要处理的是大数据，igraph的优势在于它使用的是稀疏矩阵:
library(igraph)
g <- graph_from_data_frame(dt)
as_adjacency_matrix(g)
4 x 4 sparse Matrix of class "dgCMatrix"
a b c d
a . . . 2
b . . 2 .
c . 2 . .
d . . . .
#3
1
We can use dplyr/tidyr
我们可以使用dplyr / tidyr
library(dplyr)
library(tidyr)
dt %>%
mutate_each(funs(factor(., letters[1:4]))) %>%
group_by(Source, Target) %>%
tally() %>%
spread(Target, n, drop=FALSE, fill=0)
# Source a b c d
# (fctr) (dbl) (dbl) (dbl) (dbl)
#1 a 0 0 0 2
#2 b 0 0 2 0
#3 c 0 2 0 0
#4 d 0 0 0 0