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compare adjacent elements of the same vector (avoiding loops)

I managed to write a for loop to 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 the function Comp() I can check if a letter is matching to 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)
}

Resulting in;

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

This is working, however its verry slow with large numbers. Therefor i tried sapply()

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

Unfortunately its not working... (Error in i + 1 : non-numeric argument to binary operator) I have trouble figuring out how to access the preceding letter in the vector to compare it. Also the length(data)-1, to "not compare" the last letter might become a problem.

Thank you all for the help!

Cheers Lucky

like image 791
LuckyLion Avatar asked Aug 05 '11 11:08

LuckyLion


3 Answers

Just "lag" test and use ==, which is vectorized.

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:

Also, your Comp function is slow because you don't specify the length of output when you initialize it. I suspect you were trying to pre-allocate, but vector() creates a zero-length vector that must be expanded during every iteration of your loop. Your Comp function is significantly faster if you change the call to vector() to 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
like image 141
Joshua Ulrich Avatar answered Sep 28 '22 08:09

Joshua Ulrich


As @Joshua wrote, you should of course use vectorization - it is way more efficient. ...But just for reference, your Comp function can still be optimized a bit.

The result of a comparison is TRUE/FALSE which is glorified versions of 1/0. Also, ensuring the result is integer instead of numeric consumes half the memory.

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

...and the speed difference:

> 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
like image 23
Tommy Avatar answered Sep 28 '22 10:09

Tommy


Have a 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

Might work faster.

like image 33
AntoniosK Avatar answered Sep 28 '22 10:09

AntoniosK