How can i create a function that returns a vector in R?
I want to create a function that returns its result as a vector, more specifically a function that returns the divisors of an input value and places them inside a vector.
divisors<-function(n){
i <- 2
c<-1
x<-c()
while(i <= n) {
if(n%%i==0) {
x[c]<-i
}
i <- i + 1
c<-c+1
x
}
}
r
add a comment |
I want to create a function that returns its result as a vector, more specifically a function that returns the divisors of an input value and places them inside a vector.
divisors<-function(n){
i <- 2
c<-1
x<-c()
while(i <= n) {
if(n%%i==0) {
x[c]<-i
}
i <- i + 1
c<-c+1
x
}
}
r
add a comment |
I want to create a function that returns its result as a vector, more specifically a function that returns the divisors of an input value and places them inside a vector.
divisors<-function(n){
i <- 2
c<-1
x<-c()
while(i <= n) {
if(n%%i==0) {
x[c]<-i
}
i <- i + 1
c<-c+1
x
}
}
r
I want to create a function that returns its result as a vector, more specifically a function that returns the divisors of an input value and places them inside a vector.
divisors<-function(n){
i <- 2
c<-1
x<-c()
while(i <= n) {
if(n%%i==0) {
x[c]<-i
}
i <- i + 1
c<-c+1
x
}
}
r
r
asked Nov 16 '18 at 1:50
graphboigraphboi
61
61
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You just need a return
statement at the end of the function. You should also have c <- c+1
be inside the if
statement. Here's an improved version of your function:
divisors <- function(n) {
i <- 2
c <- 1
x <- c()
while(i <= n) {
if(n %% i==0) {
x[c] <- i
c <- c+1
}
i <- i + 1
}
return (x)
}
A faster version might look like this:
divisors <- function(n) {
x <- n / (n-1):1
x[round(x) == x]
}
which doesn't use the return
statement, but returns the last evaluated expression (namely x[round(x) == x]
).
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You just need a return
statement at the end of the function. You should also have c <- c+1
be inside the if
statement. Here's an improved version of your function:
divisors <- function(n) {
i <- 2
c <- 1
x <- c()
while(i <= n) {
if(n %% i==0) {
x[c] <- i
c <- c+1
}
i <- i + 1
}
return (x)
}
A faster version might look like this:
divisors <- function(n) {
x <- n / (n-1):1
x[round(x) == x]
}
which doesn't use the return
statement, but returns the last evaluated expression (namely x[round(x) == x]
).
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
add a comment |
You just need a return
statement at the end of the function. You should also have c <- c+1
be inside the if
statement. Here's an improved version of your function:
divisors <- function(n) {
i <- 2
c <- 1
x <- c()
while(i <= n) {
if(n %% i==0) {
x[c] <- i
c <- c+1
}
i <- i + 1
}
return (x)
}
A faster version might look like this:
divisors <- function(n) {
x <- n / (n-1):1
x[round(x) == x]
}
which doesn't use the return
statement, but returns the last evaluated expression (namely x[round(x) == x]
).
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
add a comment |
You just need a return
statement at the end of the function. You should also have c <- c+1
be inside the if
statement. Here's an improved version of your function:
divisors <- function(n) {
i <- 2
c <- 1
x <- c()
while(i <= n) {
if(n %% i==0) {
x[c] <- i
c <- c+1
}
i <- i + 1
}
return (x)
}
A faster version might look like this:
divisors <- function(n) {
x <- n / (n-1):1
x[round(x) == x]
}
which doesn't use the return
statement, but returns the last evaluated expression (namely x[round(x) == x]
).
You just need a return
statement at the end of the function. You should also have c <- c+1
be inside the if
statement. Here's an improved version of your function:
divisors <- function(n) {
i <- 2
c <- 1
x <- c()
while(i <= n) {
if(n %% i==0) {
x[c] <- i
c <- c+1
}
i <- i + 1
}
return (x)
}
A faster version might look like this:
divisors <- function(n) {
x <- n / (n-1):1
x[round(x) == x]
}
which doesn't use the return
statement, but returns the last evaluated expression (namely x[round(x) == x]
).
answered Nov 16 '18 at 1:55
mickeymickey
1,4402418
1,4402418
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
add a comment |
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
thank you, i see where i was confused :)
– graphboi
Nov 16 '18 at 2:03
add a comment |
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