adding values to the vector inside for loop in R

Question

I have just started learning R and I wrote this code to learn on functions and loops.

squared<-function(x){
  m<-c()
  for(i in 1:x){
    y<-i*i
    c(m,y)
  }
  return (m)
}
squared(5)  

NULL

Why does this return NULL. I want i*i values to append to the end of mand return a vector. Can someone please point out whats wrong with this code.

I guess you want `m <- c(m, y)` instead of just `c(m, y)`. That being said, you really shouldn't build vectors by appending one element at a time -- see the second circle of [the R inferno](http://www.burns-stat.com/pages/Tutor/R_inferno.pdf) for details. Part of the beauty of R is its vectorized operations -- you can just do `squared <- function(x) x^2`. This is both much less typing and much more efficient. — josliber, Sep 03 '15 at 04:01

Rich Scriven · Accepted Answer · 2015-09-03T04:19:57.367

You haven't put anything inside m <- c() in your loop since you did not use an assignment. You are getting the following -

m <- c()
m
# NULL

You can change the function to return the desired values by assigning m in the loop.

squared <- function(x) {
    m <- c()
    for(i in 1:x) {
        y <- i * i
        m <- c(m, y)
    }
    return(m)
}

squared(5)
# [1]  1  4  9 16 25

But this is inefficient because we know the length of the resulting vector will be 5 (or x). So we want to allocate the memory first before looping. This will be the better way to use the for() loop.

squared <- function(x) {
    m <- vector("integer", x)
    for(i in seq_len(x)) {
        m[i] <- i * i
    }
    m
}

squared(5)
# [1]  1  4  9 16 25

Also notice that I have removed return() from the second function. It is not necessary there, so it can be removed. It's a matter of personal preference to leave it in this situation. Sometimes it will be necessary, like in if() statements for example.

I know the question is about looping, but I also must mention that this can be done more efficiently with seven characters using the primitive ^, like this

(1:5)^2
# [1]  1  4  9 16 25

^ is a primitive function, which means the code is written entirely in C and will be the most efficient of these three methods

`^`
# function (e1, e2)  .Primitive("^")

@thelatemail - I feel like I might have over-done it there. Haha — Rich Scriven, Sep 03 '15 at 04:20
@RichardScriven Thank You for the explanation. I would like to know why `<-` are used and can't `=` be used in R? That is can't I code as `y=i*i`? Is it necessary to be as `y<-i*i` — sam_rox, Sep 03 '15 at 05:14
@sam_rox either works here. You can read more about the differences here: http://stackoverflow.com/questions/1741820/assignment-operators-in-r-and — josliber, Sep 03 '15 at 06:29

stevec · Answer 2 · 2022-01-05T22:51:18.303

Here's a general approach:

# Create empty vector
vec <- c()

for(i in 1:10){

  # Inside the loop, make one or elements to add to vector
  new_elements <- i * 3
  
  # Use 'c' to combine the existing vector with the new_elements
  vec <- c(vec, new_elements)

}

vec
#  [1]  3  6  9 12 15 18 21 24 27 30

If you happen to run out of memory (e.g. if your loop has a lot of iterations or vectors are large), you can try vector preallocation which will be more efficient. That's not usually necessary unless your vectors are particularly large though.

adding values to the vector inside for loop in R

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