Compare neighboring elements of the same vector (avoiding loops)

Question

I managed to write for loopto compare letters in the following vector:

bases <- c("G","C","A","T")
test <- sample(bases, replace=T, 20)

test will return

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

with function Comp()I can check if the letter matches the next letter

Comp <- function(data)
{
    output <- vector()
    for(i in 1:(length(data)-1))
    {
    if(data[i]==data[i+1])
        {
        output[i] <-1
        }
        else
        {
        output[i] <-0
        }
    }
    return(output)
}

Result:

> Comp(test)
 [1] 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0

This works, however its slow slow with large numbers. So I triedsapply()

Comp <- function(x,i) if(x[i]==x[i+1]) 1 else 0
unlist(lapply(test, Comp, test))

Unfortunately, it does not work ... ( Error in i + 1 : non-numeric argument to binary operator) I find it difficult to understand how to access the previous letter in the vector to compare it. Also, the problem of length(data)-1not “comparing” the last letter can be a problem.

Thank you all for your help!

Cheers Lucky

Answer 1

"lag" test ==, .

bases <- c("G","C","A","T")
set.seed(21)
test <- sample(bases, replace=TRUE, 20)
lag.test <- c(tail(test,-1),NA)
#lag.test <- c(NA,head(test,-1))
test == lag.test

Update:

, Comp , output . , , vector() , . Comp , vector() vector(length=NROW(data)-1).

set.seed(21)
test <- sample(bases, replace=T, 1e5)
system.time(orig <- Comp(test))
#    user  system elapsed 
#  34.760   0.010  34.884 
system.time(prealloc <- Comp.prealloc(test))
#    user  system elapsed 
#    1.18    0.00    1.19 
identical(orig, prealloc)
# [1] TRUE

Answer 2

@Joshua, , , - . ... Comp .

TRUE/FALSE, 1/0. , .

Comp.opt <- function(data)
{
    output <- integer(length(data)-1L)
    for(i in seq_along(output))
    {
        output[[i]] <- (data[[i]]==data[[i+1L]])
    }
    return(output)
}

... :

> system.time(orig <- Comp(test))
   user  system elapsed 
  21.10    0.00   21.11 
> system.time(prealloc <- Comp.prealloc(test))
   user  system elapsed 
   0.49    0.00    0.49 
> system.time(opt <- Comp.opt(test))
   user  system elapsed 
   0.41    0.00    0.40 
> all.equal(opt, orig) # opt is integer, orig is double
[1] TRUE

Answer 3

Look at this:

> x = c("T", "G", "T", "G", "G","T","T","T")
> 
> res = sequence(rle(x)$lengths)-1
> 
> dt = data.frame(x,res)
> 
> dt
  x res
1 T   0
2 G   0
3 T   0
4 G   0
5 G   1
6 T   0
7 T   1
8 T   2

May work faster.