Say I have two vectors v1 and v2 and that I want to call rbind(v1, v2). However, supposed length(v1) > length(v2). From the documentation I have read that the shorter vector will be recycled. Here is an example of this "recycling":
v1 <- c(1, 2, 3, 4, 8, 5, 3, 11)
v2 <- c(9, 5, 2)
rbind(v1, v2)
#    [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# v1    1    2    3    4    8    5    3   11
# v2    9    5    2    9    5    2    9    5
v2 from being recycled and instead make the remaining entries 0?All help is greatly appreciated!
use the following:
rbind(v1, v2=v2[seq(v1)])
   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
v1    1    2    3    4    8    5    3   11
v2    9    5    2   NA   NA   NA   NA   NA
Why it works: Indexing a vector by a value larger than its length returns a value of NA at that index point.
 #eg: 
{1:3}[c(3,5,1)]
#[1]  3 NA  1
Thus, if you index the shorter one by the indecies of the longer one, you willl get all of the values of the shorter one plus a series of NA's
A generalization:
v <- list(v1, v2)
n <- max(lengths(v))
# same:
# n <- max(sapply(v, length))
do.call(rbind, lapply(v, `[`, seq_len(n)))
Although I think Ricardo has offered a nice solution, something like this would also work applying a function to a list of the vectors you wish to bind. You could specify the character to fill with as well.
test <- list(v1,v2)
maxlen <- max(sapply(test,length))
fillchar <- 0
do.call(rbind,lapply(test, function(x) c(x, rep(fillchar, maxlen - length(x) ) )))
Or avoiding all the do.call(rbind madness:
t(sapply(test, function(x) c(x, rep(fillchar, maxlen - length(x)))))
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
#[1,]    1    2    3    4    8    5    3   11
#[2,]    9    5    2    0    0    0    0    0
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