从data.frame或data.table建立一个平方邻接矩阵

时间:2015-12-18 12:37:28

标签: r data.table adjacency-matrix

我正在尝试从matrix建立一个正方形邻接data.table。 这是我已有的可重现的例子:

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"

现在,这非常接近我想要的,除了我想在两个维度中表示所有节点,例如:

  a b c d
a 0 0 0 2
b 0 0 2 0
c 0 2 0 0
d 0 0 0 0

请注意,我正在研究相当大的数据,所以我想找到一个有效的解决方案。

感谢您的帮助。

解决方案(编辑):

鉴于所提供解决方案的质量和我的数据集的大小,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

根据这些数据,所需的输出是2222 * 2222矩阵(2222 * 2223解决方案,其中第一列包含行名称也明显可以接受)。

# 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

基准测试的结果是......

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 个答案:

答案 0 :(得分:9)

听起来您只是在寻找$ pwd /Applications/XAMPP/htdocs/project/src $ sift DJANGO_SETTINGS_MODULE Binary file matches: project/__pycache__/wsgi.cpython-35.pyc main_app/wsgi.py:os.environ.setdefault("DJANGO_SETTINGS_MODULE", "main_app.settings.local") Binary file matches: fe_import/__pycache__/CsvReader.cpython-35.pyc fe_import/CsvReader.py: os.environ.setdefault("DJANGO_SETTINGS_MODULE", "main_app.settings.local") ==> This is the script I am running manage.py: os.environ.setdefault("DJANGO_SETTINGS_MODULE", "main_app.settings.local") ,但您应该确保两列都具有相同的因子级别:

table

我不知道它是否会提供任何速度提升,但您也可以使用来自&#34; data.table&#34;的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

dcast

答案 1 :(得分:5)

您也可以使用igraph。既然你说你正在处理大数据,igraph的优势在于它使用稀疏矩阵:

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 . . . .

答案 2 :(得分:2)

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