在多依赖函数中加速计算的函数的可能形式

Question

我有以下列表，并希望将每个子列表的每个元素添加到第二个列表的每个元素中，并将累积和向前进行。在我每次移动一个列表时，请注意，列表中删除的元素数量增加了一个。计算非常耗时，并想知道是否有可能制定最后的多个函数，以加快该函数的速度。

DF3 <- list( c ( 12 ,35 ,90 ,33 ,51 ) , c (  44 , 3 ,88 ,35 ,51 ) , c(12 ,16  ,6 ,10 ,3 ,12 ,2 ,6 ,9 ,4 ,4 ,51 ,13 ,22 ,51 ) , c( 44 ,3 ,37 ,51 ,35 ,51 ) , c( 12, 16 , 6 ,10 , 3 ,12 , 2 , 6 , 9 , 4 , 4 ,51 ,13 , 8 , 3 , 5 , 6 ,51 ) , c( 12 ,16,  6, 10,  3, 37, 51, 35, 51 ) , c( 12, 16, 16, 3, 37, 51, 35, 51 ))

DF3
[[1]]
12 35 90 33 51 
[[2]]
44 3 88 35 51 
[[3]]
12 16 6 10 3 12 2 6 9 4 4 51 13 22 51 
[[4]]
44  3 37 51 35 51 
[[5]]
12 16 6 10 3 12 2 6 9 4 4 51 13 8 3 5 6 51 
[[6]]
12 16 6 10 3 37 51 35 51 
[[7]]
12 16 16 3 37 51 35 51 

# Obtain the list of elements to add to the prior list of elements sequentially. 

fun <- function (x) tail ( DF3[[x]] , length ( DF3[[x]] ))
S <- lapply ( seq ( length ( DF3 ))[ 1 : ( max (length ( DF3 )))] , fun )[-1]
fun <- function (x) tail (S[[x]] , length( S[[x]])-x)
SS <- c( DF3[[1]][[1]], lapply ( seq ( length ( DF3 )-1), fun ))

# Every result element from previous sum will be added to next list of elements rolling one possition forward for each new list than in the prior list.  

D1  <-  ( SS[[1]] + SS[[2]] ) 
a   <- seq(nrow(expand.grid(SS[[1]]+SS[[2]],SS[[3]])))
fun <- function (x) sum(expand.grid ( SS[[1]] + SS[[2]] , SS[[3]] )[,1][x], expand.grid ( SS[[1]] + SS[[2]] , SS[[3]] )[,2][x] )
D2  <- unlist(lapply (a , fun)) 
b   <- seq(nrow(expand.grid(D2,SS[[4]])))
fun <- function (x) sum(expand.grid (expand.grid(D2,SS[[4]])[,1][x], expand.grid(expand.grid(D2,SS[[4]]))[,2][x] ))
D3  <- unlist(lapply (b , fun)) 
c   <- seq(nrow(expand.grid(D3,SS[[5]])))
fun <- function (x) sum(expand.grid (expand.grid(D3,SS[[5]])[,1][x], expand.grid(expand.grid(D3,SS[[5]]))[,2][x] ))
D4  <- unlist(lapply (c , fun)) 
d   <- seq(nrow(expand.grid(D4,SS[[6]])))
fun <- function (x) sum(expand.grid (expand.grid(D4,SS[[6]])[,1][x], expand.grid(expand.grid(D4,SS[[6]]))[,2][x] ))
D5  <- unlist(lapply (d , fun))
e   <- seq(nrow(expand.grid(D5,SS[[7]])))
fun <- function (x) sum(expand.grid (expand.grid(D5,SS[[7]])[,1][x], expand.grid(expand.grid(D5,SS[[7]]))[,2][x] ))
D6  <- unlist(lapply (d , fun))

是否有一种方法可以更简洁地写出最后的表达，也有一种方法可以通过另一种方式加速，因为这是非常缓慢的。

编辑

根据最终结果D6 (cs来自下面的答案)。我想找出每个DF3列表中的位置索引，当顺序累加到221时。

# Taking the answer for the sake of time

f <- function(x, y) rep(x, length(y)) + rep(y, each = length(x))
cs <- SS[[1]]
for(i in 2:length(SS)) {
cs <- f(cs, SS[[i]])
}

# Those cs that add up to 221
which ( cs == 221 ) 

Answer 1

Beyond the response from danas above you could use a fast expanding grid 

library(data.table)
# data.table::CJ is a fast expand.grid
# get all value combinations 
DT <- do.call(CJ, args = input_list)  # get all combination



# See link - http://stackoverflow.com/questions/35418479/tracing-the-location-index-of-the-components-of-a-cumulative-sum-of-sequential-l

Answer 2

expand.grid是你的杀手。我的解决方案：

f <- function(x, y) rep(x, length(y)) + rep(y, each = length(x))
cs <- SS[[1]]
for(i in 2:length(SS)) {
  cs <- f(cs, SS[[i]])
}