How to validate a condition in a for loop - r

Question

I am studying R end Data Science. In a question, I need to validate if a number in an array is even.
My code:
vetor <- list(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))
for (i in vetor) {
if (i %% 2 == 0) {
print(i)
}
}
But the result is a warning message:
Warning message:
In if (i%%2 == 0) { :
a condição tem comprimento > 1 e somente o primeiro elemento será usado
Translating:
The condition has a length > 1 and only the first element will be used.
What I need, that each element in a list be verified if is even, and if true, then, print it.
In R, how can I do it?

Answer 1

The wrapper for list is not needed
vetor <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
running the OP's code
for (i in vetor) {
if (i %% 2 == 0) {
print(i)
}
}
#[1] 2
#[1] 4
#[1] 6
#[1] 8
#[1] 10
These are vectorized operations. We don't need a loop
vetor[vetor %% 2 == 0]
#[1] 2 4 6 8 10
When we wrap the vector with list, it returns a list of length 1 and the unit will be the whole vector. The for loop in R is a for each loop and not the traditional counter controlled 3 part expression loop. So, the i will be the whole vetor vector.
Because if/else expects a single element and not a vector of length greater than 1, it results in the warning message
Or if we want to store it in a list with each element of length 1, use as.list
vetor <- as.list(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10))

Answer 2

Let's break down your code and dig into each step to see what happened ...
You should notice that vetor is a list, i.e.,
> vetor
[[1]]
[1] 1 2 3 4 5 6 7 8 9 10
In this case, the iterator i in vetor denotes the array in vetor, which can be seen from
> for (i in vetor) {
+ str(i)
+ }
num [1:10] 1 2 3 4 5 6 7 8 9 10
Therefore, when you have condition i%%2==0, you are indeed running
> for (i in vetor) {
+ print(i %% 2 == 0)
+ }
[1] FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE
which is not a single logic value as a condition for if ... else ... state. That is the reason you got the warnings.
Regarding the workaround, you can refer to #akrun's answer, which could help you a lot