Replacing for by Lapply function in R

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Replacing for by Lapply function in R

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#1 by (4 votes)
#2 by (1 votes)

4

This is my list

lista1<-list(c(1, 2, 3, 4, 5), c(6, 7, 8, 9, 10), c(11, 12, 13, 14, 15
), c(16, 17, 18, 19, 20), c(21, 22, 23, 24, 25))

I build a new list of the way below.

novalista <- vector(mode = "list", length=5)

for (i in 1:5) {
  for(k in 1:5) {     
    novalista[[i]][k] <- lista1[[i]][k]*8
  }
}

novalista

I've been wanting to make code simpler / cleaner using the lapply function. That is, I want to replace these two "for's" with the lapply function.

The difficulty is to use the function lapply with two different indices (i and k)

Any help?

r

asked by anonymous 27.08.2018 / 03:49

2 answers

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score 4 · Answer 1

First, R is a vectorized programming language. This means that the double for cycle is not necessary. Since each member of lista1 is a vector, we can multiply the integer vector by 8.

for (i in 1:5) {
    novalista[[i]] <- lista1[[i]]*8
}

And the code is already simpler and faster.

In the example of the question, you can do even better. What the question asks is to apply a function to each member of the lista1 list, or else that's exactly what lapply does. No need to index list members . Just apply the function directly to each of them. It is also necessary to take into account that lapply does not eliminate the cycle, lapply is also a cycle form.

novalista2 <- lapply(lista1, '*', 8)

identical(novalista, novalista2)
#[1] TRUE

Here the function is the multiplication, * , with the extra argument 8 , the multiplier.

This almost always gives simpler code, but unlike many R users, it's not always faster. To test this I will use the microbenchmark package. The two ways to create novalista are written in function form and then tested.

f <- function(lst){
    novalista <- vector(mode = "list", length=length(lst))
    for (i in seq_along(lst)) {
        novalista[[i]] <- lst[[i]]*8
    }
    novalista
}

g <- function(lst) lapply(lst, '*', 8)

First with the short list of the question.

microbenchmark::microbenchmark(f(lista1), g(lista1), times = 1e4)

Now with a large list.

lista2 <- lapply(1:1000, function(i) sample(1000, 100))
microbenchmark::microbenchmark(f(lista2), g(lista2))

As you can see, in both cases for was faster.

score 1 · Answer 2

Run:

lista1<-list(c(1, 2, 3, 4, 5), c(6, 7, 8, 9, 10), c(11, 12, 13, 14, 15
), c(16, 17, 18, 19, 20), c(21, 22, 23, 24, 25))

novalista<-lapply(lista1,'*',8)
novalista

#[[1]]
#[1]  8 16 24 32 40

#[[2]]
#[1] 48 56 64 72 80

#[[3]]
#[1]  88  96 104 112 120

#[[4]]
#[1] 128 136 144 152 160

#[[5]]
#[1] 168 176 184 192 200