Question:

##
```
Nh<-matrix(c(17,26,30,17,23, 17 ,24, 23), nrow=2, ncol=4); Nh
Sh<-matrix(c(8.290133, 6.241174, 6.096808, 7.4449672, 6.894924, 7.692115,
4.540521, 7.409122), nrow=2, ncol=4); Sh
NhSh<-as.matrix(Nh*Sh); NhSh
rh<-c( 0.70710678, 0.40824829, 0.28867513, 0.22360680, 0.18257419,
0.15430335, 0.13363062, 0.11785113, 0.10540926, 0.09534626); rh
pv <- c()
for (j in 1:2) {
for (i in 1:4) {
pv <- rbind(pv, NhSh[j,i]*rh)
}
}
pv
row.names(pv) <- rep(c(1:2), each = 4)
lst<-lapply(split(seq_len(nrow(pv)), as.numeric(row.names(pv))), function(i)
pv[i,])
data<-40
nlargest <- function(x, data)
{
res <- order(x)[seq_len(data)];
pos <- arrayInd(res, dim(x), useNames = TRUE);
list(values = pv[res], position = pos)
}
out <- lapply(lst, nlargest, data = 40)
```

In continuation of above code Is there any brief way of repeating the following steps for each out$’k’$position for k in 1:2?

```
s1<-c(1,1,1,1); ch<-c(5,7,10,5); C<-150; a<-out$'1'$position
for (j in a[40:1, "row"] )
{
s1[j] <- s1[j]+1;
cost1 <- sum(ch*s1);
if (cost1>=C) break
}
s1; cost1
#Output [1] 5 6 6 5
# [1] 152
```

I have to get 2 values for 's' and 'cost' for out$k$position. I tried

```
mat = replicate (2,{x = matrix(data = rep(NA, 80), ncol = 2)}); mat
for (k in 1:2)
{
mat[,,k]<-out$'k'$position
}
mat
```

<strong>Error</strong> in mat[, , k] <- out$k$position :number of items to replace is not a multiple of replacement length

```
for (k in 1:2)
{
for (j in mat[,,k][40:1] ) {
s[j] <- s[j]+1
cost <- sum(ch*s)
if (cost>=C) break
}
}
s; cost
```

<strong>Error</strong> : Error in s[j] <- s[j] + 1 : NAs are not allowed in subscripted assignments

Please anyone help in resolving these errors.

Answer1:We could apply the function directly by looping over the `list`

. Note that each element of the `list`

is a `matrix`

```
sapply(lst, is.matrix)
# 1 2
#TRUE TRUE
```

so, there is no need to `unlist`

and create a `matrix`

```
out <- lapply(lst, nlargest, data = 40)
```

-checking with the OP's results

```
out1 <- nlargest(sub1, 40)
identical(out[[1]], out1)
#[1] TRUE
```

<h3>Update2</h3>
Based on the second update, we need to initialize 'cost' and 'sl' with the same length as 'k' elements. Here, we initialize 'sl' as a `list`

of `vectors`

```
sl <- rep(list(c(1, 1, 1, 1)), 2)
C <- 150
cost <- numeric(2)
for (k in 1:2){
for (j in mat[,,k][40:1, 1] ) {
sl[[k]][j] <- sl[[k]][j]+1
cost[k] <- sum(ch*sl[[k]])
if (cost[k] >=C) break
}
}
sl
#[[1]]
#[1] 5 7 6 4
#[[2]]
#[1] 6 5 5 7
cost
#[1] 154 150
```