我正在 R 中执行一些计算要求较高的操作,因此我正在寻找最有效的方法来完成这些操作。我的问题是:
library(dplyr)
library(igraph)
library(bench)
set.seed(123)
edgelist <- data.frame(
node1 = sample(1:2000, 11000, replace = T),
node2 = sample(1:2000, 11000, replace = T),
weight = runif(11000, min = 0, max = 5)
)
g <- graph_from_data_frame(edgelist, directed = F)
#Data.frame
dat <- function() {
dm <- distances(g, weight = E(g)$weight)
UTIndex <- which(upper.tri(dm), arr.ind = T)
df1 <- data.frame(
verticeA = as.numeric(rownames(dm)[UTIndex[, 1]]),
verticeB = as.numeric(colnames(dm)[UTIndex[, 2]]),
path_length = as.numeric(dm[UTIndex])
)
}
#Matrix
mat <- function() {
dm <- distances(g, weight = E(g)$weight)
UTIndex <- which(upper.tri(dm), arr.ind = T)
df1 <- cbind(
verticeA = as.numeric(rownames(dm)[UTIndex[, 1]]),
verticeB = as.numeric(colnames(dm)[UTIndex[, 2]]),
path_length = as.numeric(dm[UTIndex])
)
}
####
results <- bench::mark(
dat = dat(),
mat = mat(),
check = F
)
t1 <- system.time({
df1 <- dat()
})
rm(df1)
t2 <- system.time({
df1 <- mat()
})
rm(df1)
这是
t1
、t2
和 results
的输出:
> results
# A tibble: 2 × 13
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time
<bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int> <dbl> <bch:tm>
1 dat 2.89s 2.89s 0.346 269MB 1.39 1 4 2.89s
2 mat 2.83s 2.83s 0.353 315MB 1.41 1 4 2.83s
# ℹ 4 more variables: result <list>, memory <list>, time <list>, gc <list>
> t1
user system elapsed
2.78 0.04 2.81
> t2
user system elapsed
3.12 0.08 3.21
当您进行基准测试时,您没有充分隔离您正在调查的元素(data.frame() 和 cbind() 之间的差异)函数调用;您运行的任一测试中的大部分计算都是除 data.frame 和 cbind() 之外的所有内容,并且您进行的测试比较相对较少,这意味着您可能会将随机变化误认为是显着差异。
下面我隔离出了与基准测试无关的公共部分,仅保留相关部分;但我更进一步展开关于 R 内存管理的讨论。
让我们从代码重写开始:
library(dplyr)
library(igraph)
library(bench)
set.seed(123)
edgelist <- data.frame(
node1 = sample(1:2000, 11000, replace = T),
node2 = sample(1:2000, 11000, replace = T),
weight = runif(11000, min = 0, max = 5)
)
g <- graph_from_data_frame(edgelist, directed = F)
dm <- distances(g, weight = E(g)$weight)
UTIndex <- which(upper.tri(dm), arr.ind = T)
verticeA <- as.numeric(rownames(dm)[UTIndex[, 1]])
verticeB <- as.numeric(colnames(dm)[UTIndex[, 2]])
path_length <- as.numeric(dm[UTIndex])
#Data.frame
datpure <- function(verticeA,verticeB,path_length) {
data.frame(
verticeA=verticeA,
verticeB=verticeB,
path_length =path_length
)
}
datpure(verticeA,verticeB,path_length)
#Matrix
matpure <- function(verticeA,verticeB,path_length) {
cbind(
verticeA=verticeA,
verticeB=verticeB,
path_length =path_length
)
}
matpure(verticeA,verticeB,path_length)
####
adjust_to_1s <- function(x){
x[,] <- 1
x
}
(results <- bench::mark(
dat_pure = dat(verticeA,verticeB,path_length),
mat_pure = mat(verticeA,verticeB,path_length),
dat_1 = adjust_to_1s(dat(verticeA,verticeB,path_length)),
mat_1 = adjust_to_1s(mat(verticeA,verticeB,path_length)),
check = F,
iterations = 100L,
time_unit = 'ms'
))
+ ))
# A tibble: 4 × 13
expression min median `itr/sec` mem_alloc `gc/sec` n_itr n_gc total_time result memory time gc
<bch:expr> <dbl> <dbl> <dbl> <bch:byt> <dbl> <int> <dbl> <dbl> <list> <list> <list> <list>
1 dat_pure 0.169 0.178 5350. 0B 54.0 99 1 18.5 <NULL> <Rprofmem [11 × 3]> <bench_tm> <tibble>
2 mat_pure 10.7 12.4 72.9 45.8MB 0.736 99 1 1358. <NULL> <Rprofmem [2 × 3]> <bench_tm> <tibble>
3 dat_1 182. 254. 4.00 292.8MB 0.347 92 8 23023. <NULL> <Rprofmem> <bench_tm> <tibble>
4 mat_1 30.3 37.9 23.8 53.4MB 0.241 99 1 4153. <NULL> <Rprofmem [2 × 3]> <bench_tm> <tibble>
从中我们可以看出,data.frame() 调用本身比 cbind 快了一个数量级。我认为为什么是内存布局? 对于 data.frame() R 只需要一个 list/data.frame() 并将现有向量与名称相关联; R 是只写时复制,否则通过引用工作,因此 data.frame 构造本质上是对元数据的微不足道的更改。 而 cbind 创建一个矩阵,它本质上是一个向量,因此必须复制数据并布局。
我添加了一个变体,在最初的纯 data.frame 和 cbind 调用之后,我们从根本上改变了对象(将每个条目设置为 1) 现在,在这两种情况下,R 都必须写入内存,并且速度都会变慢。 data.frame 表现更差。