List of string to list of vectors of characters

Question

After defining

> Seq.genes <- as.list(c("ATGCCCAAATTTGATTT","AGAGTTCCCACCAACG"))

I have a list of strings :

> Seq.genes[1:2]
[[1]]
[1] "ATGCCCAAATTTGATTT"

[[2]]
[1] "AGAGTTCCCACCAACG"

I would like to convert it in a list of vectors :

>Seq.genes[1:2]
[[1]]
[1]"A" "T" "G" "C" "C" "C" "A" "A" "A" "T" "T" "T" "G" "A" "T" "T" "T"

[[2]]
[1] "A" "G" "A" "G" "T" "T" "C" "C" "C" "A" "C" "C" "A" "A" "C" "G"

I tried something like :

for (i in length(Seq.genes)){
  x <- Seq.genes[i]
  Seq.genes[i] <- substring(x, seq(1,nchar(x),2), seq(1,nchar(x),2))

}

score 7 · Accepted Answer · answered Nov 02 '17 at 10:00

7

It may be better to have the strings in a vector rather than in a list. So, we could unlist, then do an strsplit

strsplit(unlist(Seq.genes), "")

answered Nov 02 '17 at 10:00

akrun

874,273
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662

Do note that the `strsplit` itself returns a list object, with an entry for each element in the input vector (`unlist(Seq.genes)`). The code you provided will give the exact output as OP requested. – MrGumble Nov 02 '17 at 10:02
@MrGumble Yes, it returns a list object. In this case it may be better to have a list – akrun Nov 02 '17 at 10:05

pzhao · Answer 2 · 2017-11-02T11:37:24.903

-1

sapply(Seq.genes, strsplit, split = '')

or

lapply(Seq.genes, strsplit, split = '')

edited Nov 02 '17 at 11:37

answered Nov 02 '17 at 10:12

pzhao

335
1
10

Why sapply loop? – zx8754 Nov 02 '17 at 10:18
it is not a loop. sapply() and lapply() just apply a function to a list. – pzhao Nov 02 '17 at 10:32
5

@dapeng Not quite. See [this](https://stackoverflow.com/questions/28983292/is-the-apply-family-really-not-vectorized) – David Arenburg Nov 02 '17 at 11:08

List of string to list of vectors of characters

2 Answers2