我编写了一个代码来将函数应用于数据框input:
set.seed(1234)
n = 5000000
input <- as.matrix(data.frame(c1 = sample(1:10, n, replace = T), c2 = sample(1:10, n, replace = T), c3 = sample(1:10, n, replace = T), c4 = sample(1:10, n, replace = T)))
system.time(
test <- input %>%
split(1:nrow(input)) %>%
map(~ func1(.x, 2, 2, "test_1")) %>%
do.call("rbind", .))
## Here is the function used:
func1 <- function(dataC, PR, DB, MT){
c1 <- as.vector(dataC[1])
c2 <- as.vector(dataC[2])
c3 <- as.vector(dataC[3])
c4 <- as.vector(dataC[4])
newc1 <- -999
newc2 <- -999
if(MT=="test_1"){
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 + 1) : (c2 + PR)))
V1 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - 1) : (c1 + 1)), y = c((c2 + 1) : (c2 + PR)))
V2 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 + 1) : (c2 + PR)))
V3 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 - 1) : (c2 + 1)))
V4 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
V5 <- 0
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 - 1) : (c2 + 1)))
V6 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 - PR) : (c2 - 1)))
V7 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - 1) : (c1 + 1)), y = c((c2 - PR) : (c2 - 1)))
V8 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 - PR) : (c2 - 1)))
V9 <- mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
} else if(MT=="test_2"){
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 + 1) : (c2 + PR)))
V1 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - 1) : (c1 + 1)), y = c((c2 + 1) : (c2 + PR)))
V2 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 + 1) : (c2 + PR)))
V3 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 - 1) : (c2 + 1)))
V4 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
V5 <- 0
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 - 1) : (c2 + 1)))
V6 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 - PR) : (c1 - 1)), y = c((c2 - PR) : (c2 - 1)))
V7 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
listC <- expand.grid(x = c((c1 - 1) : (c1 + 1)), y = c((c2 - PR) : (c2 - 1)))
V8 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * DB
listC <- expand.grid(x = c((c1 + 1) : (c1 + PR)), y = c((c2 - PR) : (c2 - 1)))
V9 <- harmonic.mean(sample(1:10, size = dim(listC)[1], replace = TRUE)) * sqrt(2) * DB
}
tot <- sum(c(1/V1, 1/V2, 1/V3, 1/V4, 1/V6, 1/V7, 1/V8, 1/V9), na.rm = TRUE)
mat_V <- matrix(data = c((1/V1)/tot, (1/V2)/tot, (1/V3)/tot, (1/V4)/tot, V5,
(1/V6)/tot, (1/V7)/tot, (1/V8)/tot, (1/V9)/tot), nrow = 3, ncol = 3, byrow = TRUE)
while((newc1 == -999 && newc2 == -999) || (c3 == newc1 && c4 == newc2)){
if(c3 == newc1 && c4 == newc2){
mat_V[choiceC[1], choiceC[2]] <- NaN
## print(mat_V)
}
choiceC <- which(mat_V == max(mat_V, na.rm = TRUE), arr.ind = TRUE)
## print(choiceC)
## If there are several maximum values
if(nrow(choiceC) > 1){
choiceC <- choiceC[sample(1:nrow(choiceC), 1), ]
}
if(choiceC[1]==1 & choiceC[2]==1){
newC <- matrix(c(x = c1 - 1, y = c2 + 1), ncol = 2)
} else if(choiceC[1]==1 & choiceC[2]==2){
newC <- matrix(c(x = c1, y = c2 + 1), ncol = 2)
} else if(choiceC[1]==1 & choiceC[2]==3){
newC <- matrix(c(x = c1 + 1, y = c2 + 1), ncol = 2)
} else if(choiceC[1]==2 & choiceC[2]==1){
newC <- matrix(c(x = c1 - 1, y = c2), ncol = 2)
} else if(choiceC[1]==2 & choiceC[2]==3){
newC <- matrix(c(x = c1 + 1, y = c2), ncol = 2)
} else if(choiceC[1]==3 & choiceC[2]==1){
newC <- matrix(c(x = c1 - 1, y = c2 - 1), ncol = 2)
} else if(choiceC[1]==3 & choiceC[2]==2){
newC <- matrix(c(x = c1, y = c2 - 1), ncol = 2)
} else if(choiceC[1]==3 & choiceC[2]==3){
newC <- matrix(c(x = c1 + 1, y = c2 - 1), ncol = 2)
}
newc1 <- as.vector(newC[,1])
newc2 <- as.vector(newC[,2])
}
return(newC)
}
该代码适用于小型数据集,但当数据框包含超过100万行时,它非常慢.我认为在函数(例如条件if else)中重复有许多代码行会降低速度.有没有办法立即在函数中进行所有计算？我真的很感激任何建议.
Moo*_*per 10
首先是一些强烈的爱,但我强烈建议你覆盖你的基础,你的代码是不良做法的集中,通过花费一些时间研究矢量化等你将获得巨大的投资回报率...考虑也发布在https上://codereview.stackexchange.com/questions/tagged/r下次因为这是一个更合适的问题.
你的瓶颈不是嵌套的ifs,而是使用不足expand.grid.
您在代码中创建数据框,通过expand.grid您不正确地调用listC(它们不是列表).然后,这个代价高昂的data.frame只用于它的行数,你dim(listC)[1]会得到更多的惯用类型nrow(listC).
此值(dim(listC)[1])只能PR^2或3*PR在实践中,因此您可以先计算它们,然后重复使用它们.
嵌套的ifs可以用嵌套的switch语句替换,更具可读性,并且只有在我们提高效率时才测试第一个选择.
它允许我们看到您在代码中忘记了一个条件.请参阅下面的改进代码.
当它看起来更整洁时,我们看到我们实际上可以简单地替换它newC <- c(c1 - 2 + choice[2], c2 + 2 - choice[1]).
补充意见
c2 <- as.vector(dataC[2]) 可以替换为 c2 <- dataC[[2]]t(c(1,2))代替matrix(c(x = 1, y = 2), ncol = 2),但如果你打算使用as.vector到底就可以了,做c(1,2)摆在首位修改后的代码
func1 <- function(dataC, PR, DB, MT){
c1 <- dataC[[1]]
c2 <- dataC[[2]]
c3 <- dataC[[3]]
c4 <- dataC[[4]]
fun <- if(MT=="test_1") mean else if(MT=="test_2") harmonic.mean
fun2 <- function(size,mult)
fun(sample(1:10, size = size, replace = TRUE)) * mult
pr_sq <- PR^2
pr_3 <- 3*PR
sqrt_2_DB <- sqrt(2) * DB
V1 <- fun2(pr_sq, sqrt_2_DB)
V2 <- fun2(pr_3, DB)
V3 <- fun2(pr_sq, sqrt_2_DB)
V4 <- fun2(pr_3, DB)
V5 <- 0
V6 <- fun2(pr_3, DB)
V7 <- fun2(pr_sq, sqrt_2_DB)
V8 <- fun2(pr_3, DB)
V9 <- fun2(pr_sq, sqrt_2_DB)
inv <- 1/c(V1, V2, V3, V4, V6, V7, V8, V9)
tot <- sum(inv, na.rm = TRUE)
mat_V <- matrix(data = c(inv[1:4], V5, inv[5:8]) / tot,
nrow = 3, ncol = 3, byrow = TRUE)
newC <- NULL
while(is.null(newC) || identical(c(c3,c4), newC)){
if(identical(c(c3,c4), newC)){
mat_V[choiceC[1], choiceC[2]] <- NaN
## print(mat_V)
}
choiceC <- which(mat_V == max(mat_V, na.rm = TRUE), arr.ind = TRUE)
## print(choiceC)
## If there are several maximum values
if(nrow(choiceC) > 1){
choiceC <- choiceC[sample(1:nrow(choiceC), 1), ]
}
newC <- c(c1 - 2 + choiceC[2], c2 + 2 - choiceC[1])
# using switch it would have been
# newC <- switch(choiceC[1],
# `1` = switch(choiceC[2],
# `1` = c(x = c1 - 1, y = c2 + 1),
# `2` = c(x = c1, y = c2 + 1),
# `3` = c(x = c1 + 1, y = c2 + 1)),
# `2` = switch(choiceC[2],
# `1` = c(x = c1 - 1, y = c2),
# `2` = c(x = c1, y = c2), # you were missing this one
# `3` = c(x = c1 + 1, y = c2)),
# `3` = switch(choiceC[2],
# `1` = c(x = c1 - 1, y = c2 - 1),
# `2` = c(x = c1, y = c2 - 1),
# `3` = c(x = c1 + 1, y = c2 - 1)))
}
t(newC)
}