我正在尝试使 R 函数更加高效。下面找到一个工作示例。
smsurv <- function(Time,Status,X,beta,w,model){
death_point <- sort(unique(subset(Time, Status==1)))
if(model=='ph') coxexp <- exp((beta)%*%t(X[,-1]))
n <- length(death_point)
lambda <- numeric(n)
for(i in 1: n){
if(model=='ph') temp <- sum(as.numeric(Time>=death_point[i])*w*drop(coxexp))
if(model=='aft') temp <- sum(as.numeric(Time>=death_point[i])*w)
lambda[i] <- sum(Status*as.numeric(Time==death_point[i]))/temp
}
HHazard <- numeric()
for(i in 1:length(Time)){
HHazard[i] <- sum(as.numeric(Time[i]>=death_point)*lambda)
if(Time[i]>max(death_point))HHazard[i] <- Inf
if(Time[i]<min(death_point))HHazard[i] <- 0
}
survival <- exp(-HHazard)
list(survival=survival)
}
nr_obs = 50000
Time_input <- rnorm(nr_obs, mean = 100, sd = 36)
Status_input <- sample(c(0,1), replace=TRUE, size=nr_obs)
w_input <- Status_input
# Let's suppose there are 9 variables (first column denotes the intercept)
n_variables <- 9
X_input <- matrix(rnorm(nr_obs*n_variables),nr_obs)
X_input <- cbind(Intercept = rep(1, nrow(X_input)), X_input)
beta_input <- runif(n_variables, min = -1, max = 1)
model_input <- "ph"
output <- smsurv(Time_input,Status_input,X_input,beta_input,w_input,model_input)
Run Code Online (Sandbox Code Playgroud)
我已经尝试用 lapply 和 sapply 替换 for 循环,但这实际上使函数变得更慢:
smsurv2 <- function(Time,Status,X,beta,w,model){
death_point <- sort(unique(subset(Time, Status==1)))
if(model=='ph') coxexp <- exp((beta)%*%t(X[,-1]))
if(model_input=='ph') lambda =unlist(lapply(death_point, function(z) sum(Status_input*as.numeric(Time_input==z))/ sum(as.numeric(Time_input>=z)*w_input*drop(coxexp))))
if(model=='aft') lambda =unlist( lapply(death_point, function(z) sum(Status_input*as.numeric(Time_input==z))/ sum(as.numeric(Time_input>=z)*w_input)))
HHazard <- unlist(lapply(Time, function(t) {sum(as.numeric(t>=death_point)*lambda)}))
HHazard[Time > max(death_point)] <- Inf
HHazard[Time < min(death_point)] <- 0
survival <- exp(-HHazard)
list(survival=survival)
}
smsurv3 <- function(Time, Status, X, beta, w, model){
death_point <- sort(unique(subset(Time, Status==1)))
if(model=='ph') coxexp <- exp((beta)%*%t(X[,-1]))
lambda <- sapply(death_point, function(dp) {return(sum(Status*as.numeric(Time==dp))/sum(as.numeric(Time>=dp)*w*drop(coxexp)))})
HHazard <- sapply(Time, function(t){return(sum(as.numeric(t>=death_point)*lambda))})
HHazard[Time > max(death_point)] <- Inf
HHazard[Time < min(death_point)] <- 0
survival <- exp(-HHazard)
list(survival=survival)
}
Run Code Online (Sandbox Code Playgroud)
考虑到这一点,有人还有其他我可以尝试的建议吗?我最近阅读了有关 rcpp 包的信息,但我不确定如何用 C 代码替换 for 循环。任何建议都非常受欢迎。
Rui*_*das 13
下面的函数smsurv2速度更快,结果与问题的函数相同。
以下是我所做的一些更改。
subset慢,索引子集更快;coxexp我从第一个循环中删除了 的计算,for使循环代码更简单,它总是乘以coxexp,它可以是 1 的向量,从而删除测试if;seq_along1:n比;更安全Inf,这可以在循环外进行向量化;as.numeric永远不需要,因此消除了几个函数调用。
smsurv2 <- function(Time, Status, X, beta, w, model){
death_point <- Time[Status == 1] |> unique() |> sort()
n <- length(death_point)
lambda <- numeric(n)
if(model == 'ph') {
coxexp <- (exp(beta %*% t(X[, -1]))) |> drop()
} else if(model == 'aft') {
coxexp <- rep(1, length(Time))
}
for(i in seq_along(death_point)){
temp <- sum((Time >= death_point[i]) * w * coxexp)
lambda[i] <- sum(Status * (Time == death_point[i])) / temp
}
HHazard <- numeric(length(Time))
for(i in seq_along(Time)){
HHazard[i] <- sum((Time[i] >= death_point) * lambda)
}
HHazard[ Time > max(death_point) ] <- Inf
HHazard[ Time < min(death_point) ] <- 0
survival <- exp(-HHazard)
list(survival = survival)
}
nr_obs = 50000
Time_input <- rnorm(nr_obs, mean = 100, sd = 36)
Status_input <- sample(c(0,1), replace=TRUE, size=nr_obs)
w_input <- Status_input
# Let's suppose there are 9 variables (first column denotes the intercept)
n_variables <- 9
X_input <- matrix(rnorm(nr_obs*n_variables),nr_obs)
X_input <- cbind(Intercept = rep(1, nrow(X_input)), X_input)
beta_input <- runif(n_variables, min = -1, max = 1)
model_input <- "ph"
system.time(
output <- smsurv(Time_input,Status_input,X_input,beta_input,w_input,model_input)
)
#> user system elapsed
#> 25.08 4.95 33.39
system.time(
output2 <- smsurv2(Time_input,Status_input,X_input,beta_input,w_input,model_input)
)
#> user system elapsed
#> 16.92 1.45 19.86
identical(output, output2)
#> [1] TRUE
Run Code Online (Sandbox Code Playgroud)
创建于 2023-09-30,使用reprex v2.0.2
cumsum这里的要点是在排序后使用Time以避免循环。然后一切都被矢量化,并且几乎立即运行。
由于这个问题得到了如此多的关注,我添加了一个更长的解释来解释为什么for这里的循环如此缓慢。滚动到底部即可。
使用data.table:
library(data.table)
smsurv2 <- function(Time, Status, X, beta, w, model){
list(
survival = setorder(
setorder(
data.table(Time, Status)[,r := .I],
-Time, Status
)[
,`:=`(
death_point = Time != shift(Time, -1L, Time[1] - 1) & Status,
temp = cumsum((if (model=='ph') w*drop(exp(tcrossprod(beta, X[,-1]))) else w)[r]),
lambda = cumsum(Status)
)
][
death_point == TRUE, lambda := c(lambda[1], diff(lambda))/temp
][
death_point == FALSE, lambda := 0
][
,survival := exp(-rev(cumsum(rev(lambda))))
][
seq_len(which.max(death_point) - 1L), survival := 0
][, c(1:2, 4:6) := NULL],
r
)[[2]]
)
}
Run Code Online (Sandbox Code Playgroud)
使用 OP 的示例数据比较时间:
system.time({
output <- smsurv(Time_input,Status_input,X_input,beta_input,w_input,model_input)
})
#> user system elapsed
#> 36.19 4.97 42.89
system.time({
output2 <- smsurv2(Time_input,Status_input,X_input,beta_input,w_input,model_input)
})
#> user system elapsed
#> 0.07 0.00 0.07
all.equal(output, output2)
#> [1] TRUE
Run Code Online (Sandbox Code Playgroud)
for循环考虑第一个循环中的第一行for:
if(model=='ph') temp <- sum(as.numeric(Time>=death_point[i])*w*drop(coxexp))
Run Code Online (Sandbox Code Playgroud)
这里,death_point是取自 的值向量Time,w*coxexp是与 长度相同 (50000) 的值向量Time。该行只是简单地求和w*coxexp,其中对应的值Time大于或等于death_point[i]。这涉及 50000 次比较,然后进行两次(向量化)乘法,并对结果求和。如果death_point长度为 1,那么这样做是有意义的。然而,death_point在例子中长度是24920,所以这一行涉及50000*24920~12.5亿次比较!相反,我们实际上可以通过一次排序来完成所有这些比较:将Time和 的结果w*coxexp放入表中,按 降序排序Time,计算 的累积和w*coxexp,然后获取标记为death_points 的时间的值。我们现在拥有了所有需要的值temp,并且只需要两个向量化函数调用(setorder和cumsum)。所有其他计算都smsurv2可以轻松执行,因为表已经排序。
| 归档时间: |
|
| 查看次数: |
805 次 |
| 最近记录: |