replace a nested for loop with mapply

断了今生、忘了曾经 提交于 2021-01-28 08:24:45

问题


I am a beginner in R. I am working on linear programming using R studio's Cplex to solve a model. One of the constraint in my model is Xl(i,j,t) <= D(i,j,t). I am able to do this with nested for loop with a small dimension (16X16X6). But I want to run my model a lot bigger model, more like 2500X2500X60. I need to save memory and run it faster than the nested for loop. I thought about using apply but I don't know how to make it work. Any help would be greatly appreciated!

location <-16
horizon <-6
Amat <- NULL
Xe_null <- array(0, dim = c(locations, locations, horizon))
Xl_null <- array(0, dim = c(locations, locations, horizon))
Xl      <- array(0, dim = c(locations, locations, horizon))
Xl <- Xl_null
for (t in 1:horizon) {
  for (j in 1:locations) {
    for (i in 1:locations) {
      Xl[i,j,t] <- 1
      Amat <- rbind(Amat, c(as.vector(Xe_null), as.vector(Xl)))
      Xl <- Xl_null 
 } } }
dim(Amat) # 1536 3072

Here is another constraint.

R       <- array(, dim=c(locations, horizon, horizon))
R_null  <- array(, dim=c(locations, horizon, horizon))
R <- R_null
Xe <- Xe_null
Xl <- Xl_null
#
for (t in 1:(horizon-1)) {
  for (tp in (t+1):horizon) {
    for (j in 1:locations)     {
      for (i in 1:locations)      {
        if ((tp-t) ==Travel_time[i,j]) 
        {
          Xe[i,j,t]=1
          Xl[i,j,t]=1
        } 
      }
      R[j,tp,t] = 1
      R[j,tp,t+1] = -1
      Amat <- rbind(Amat, c(as.vector(Xe), as.vector(Xl),as.vector(R)))
      }
  }
}

I try to do this:

Xl = function(ii,jj,tt){1}
t =c(1:horizon)
i =c(1:locations)
j =c(1:locations)
output_Xl = apply(expand.grid(i,j,t),1,function(x,y,h) Xl(x[1],x[2],x[3]))
Xl_new <- array(output_Xl, dim = c(locations, locations, horizon))
Amat <- rbind(Amat, c(as.vector(Xe_null), as.vector(Xl_new)))
dim(Amat) # 1 3072

回答1:


You can get the same output with

T <- horizon*locations*locations
Bmat <- cbind(matrix(0, nrow=T, ncol=T), diag(1, nrow=T, ncol=T))
identical(Amat, Bmat)
# TRUE



回答2:


I think what you need is to make a vectorized function giving the same output (see ?Vectorize). Below code is five hundred times as fast as yours.
At your real problems, maybe you need to use <<- instead of <- (see ?"<<-")

my_func <- function(a, b, c){
  Xl[a, b, c] <- 1
  c(as.vector(Xe_null), as.vector(Xl))
}

vectorized_my_func <- Vectorize(my_func, c("a", "b", "c"))

arg_df <- expand.grid(1:locations, 1:locations, 1:horizon)
res <- vectorized_my_func(arg_df[,1], arg_df[,2], arg_df[,3])

identical(Amat, t(res))    # TRUE

# your code
##   user  system elapsed 
## 77.538  18.293  97.056 

# my code
##   user  system elapsed 
###  0.137   0.051   0.189 


来源:https://stackoverflow.com/questions/47697281/replace-a-nested-for-loop-with-mapply

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!