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?
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
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
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.
For vectors there is seq_along, for DataFrames you may use seq_len
for(i in seq_len(nrow(the.table)){
do.stuff()
}
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.
I think the most proper way in R is to use an apply function. More often than not, there's an apply function that does that. And more often than not, you don't need a sequence.
Here's an example that applies diff to each column, or each row.
> d <- data.frame(x = 1:5, y = 6:10)
over the columns,
> lapply(d, diff)
$x
[1] 1 1 1 1
$y
[1] 1 1 1 1
across the rows,
> apply(d, 1, diff)
[1] 5 5 5 5 5
over the columns again, returning a matrix
> sapply(d, diff)
x y
[1,] 1 1
[2,] 1 1
[3,] 1 1
[4,] 1 1
See this link for a most excellent explanation about apply
