how to append an element to a list without keeping track of the index?
There is a function called append
:
ans <- list()
for (i in 1992:1994){
n <- 1 #whatever the function is
ans <- append(ans, n)
}
ans
## [[1]]
## [1] 1
##
## [[2]]
## [1] 1
##
## [[3]]
## [1] 1
##
Note: Using apply
functions instead of a for loop is better (not necessarily faster) but it depends on the actual purpose of your loop.
Answering OP's comment: About using ggplot2
and saving plots to a list, something like this would be more efficient:
plotlist <- lapply(seq(2,4), function(i) {
require(ggplot2)
dat <- mtcars[mtcars$cyl == 2 * i,]
ggplot() + geom_point(data = dat ,aes(x=cyl,y=mpg))
})
Thanks to @Wen for sharing Comparison of c()
and append()
functions:
Concatenation (c) is pretty fast, but append is even faster and therefor preferable when concatenating just two vectors.
mylist <- list()
for (i in 1:100){
n <- 1
mylist[[(length(mylist) +1)]] <- n
}
This seems to me the faster solution.
x <- 1:1000
aa <- microbenchmark({xx <- list(); for(i in x) {xx <- append(xx, values = i)} })
bb <- microbenchmark({xx <- list(); for(i in x) {xx <- c(xx, i)} } )
cc <- microbenchmark({xx <- list(); for(i in x) {xx[(length(xx) + 1)] <- i} } )
sapply(list(aa, bb, cc), (function(i){ median(i[["time"]]) / 10e5 }))
#{append}=4.466634 #{c}=3.185096 #{this.one}=2.925718
mylist <- list()
for (i in 1:100) {
df <- 1
mylist <- c(mylist, df)
}
There is: mylist <- c(mylist, df)
but that's usually not the recommended way in R. Depending on what you're trying to achieve, lapply()
is often a better option.