Proper way to loop over the length of a dataframe in R

Question

After quite a bit of debugging today, to my dismay i found that:

for (i in 1:0) {
     print(i)
}

Actually prints 1 and 0 respectively in R. The problem came up when writing

for (i in 1:nrow(myframe) {
     fn(i)
}

Which i had intended to not execute at all if nrow(myframe)==0. Is the proper correction just:

if (nrow(myvect) != 0) {
    for (i in 1:nrow(myframe) {
        fn(i)
    }
}

Or is there a more proper way to do what I wanted in R?

Answer 1

You can use seq_along instead:

vec <- numeric() 
length(vec)
#[1] 0

for(i in seq_along(vec)) print(i)   # doesn't print anything

vec <- 1:5

for(i in seq_along(vec)) print(i)
#[1] 1
#[1] 2
#[1] 3
#[1] 4
#[1] 5

Edit after OP update

df <- data.frame(a = numeric(), b = numeric())
> df
#[1] a b
#<0 rows> (or row.names with length 0)

for(i in seq_len(nrow(df))) print(i)    # doesn't print anything

df <- data.frame(a = 1:3, b = 5:7)

for(i in seq_len(nrow(df))) print(i)
#[1] 1
#[1] 2
#[1] 3

Answer 2

Clearly all previous answers do the job.

I like to have something like this:

rows_along <- function(df) seq(nrow(df))

and then

for(i in rows_along(df)) # do stuff

Totally idiosyncratic answer, it is just a wrapper. But I think it is more readable/intuitive.

Answer 3

Regarding the edit, see the counterpart function seq_len(NROW(myframe)). This usage is exactly why you don't use 1:N in a for() loop, incase whatever value ends up replacing N is 0 or negative.

An alternative (which just hides the loop) is to do apply(myframe, 1, FUN = foo) where foo is a function containing the things you want to do to each row of myframe and will probably just be cut and paste from the body of the loop.

Answer 4

For vectors there is seq_along, for DataFrames you may use seq_len

for(i in seq_len(nrow(the.table)){
    do.stuff()
}