# R – Vectorize the thinking: Vector Operations in R

rvector

So earlier I answered my own question on thinking in vectors in R. But now I have another problem which I can't 'vectorize.' I know vectors are faster and loops slower, but I can't figure out how to do this in a vector method:

I have a data frame (which for sentimental reasons I like to call my.data) which I want to do a full marginal analysis on. I need to remove certain elements one at a time and 'value' the data frame then I need to do the iterating again by removing only the next element. Then do again… and again… The idea is to do a full marginal analysis on a subset of my data. Anyhow, I can't conceive of how to do this in a vector efficient way.

I've shortened the looping part of the code down and it looks something like this:

``````for (j in my.data\$item[my.data\$fixed==0]) { # <-- selects the items I want to loop
#     through
my.data.it <- my.data[my.data\$item!= j,] # <-- this kicks item j out of the list
sum.data <-aggregate(my.data.it, by=list(year), FUN=sum, na.rm=TRUE) #<-- do an
# aggregation

do(a.little.dance) && make(a.little.love) -> get.down(tonight) # <-- a little
#  song and dance

delta <- (get.love)                                         # <-- get some love
delta.list<-append(delta.list, delta, after=length(delta.list)) #<-- put my love
#    in a vector
}
``````

So obviously I hacked out a bunch of stuff in the middle, just to make it less clumsy. The goal would be to remove the j loop using something more vector efficient. Any ideas?

#### Best Solution

Here's what seems like another very R-type way to generate the sums. Generate a vector that is as long as your input vector, containing nothing but the repeated sum of n elements. Then, subtract your original vector from the sums vector. The result: a vector (isums) where each entry is your original vector less the ith element.

``````> (my.data\$item[my.data\$fixed==0])
 1 1 3 5 7
> sums <- rep(sum(my.data\$item[my.data\$fixed==0]),length(my.data\$item[my.data\$fixed==0]))
> sums
 17 17 17 17 17
> isums <- sums - (my.data\$item[my.data\$fixed==0])
> isums
 16 16 14 12 10
``````