Assign single output for multiple-output function to new function - r

Question

I have a function that gives me a single output which is however composed of two elements. Example for it would be:
example <- function(x){
sin <- sin(x)
cos <- cos(x)
output <- cbind(sin, cos)
return(output)
}
Now my idea is to plot separately sin and cos, each as functions of x. I would like to avoid writing a separate function in this context since the two objects are better to be calculated all at once.
If I try :
x_grid = seq(0,1,0,0.05)
plot(x_grid, sapply(x_grid, FUN = example[1]))
I get the following error message :
Error in example[1] : object of type 'closure' is not subsettable
How to proceed then? (notice that I use sapply because I need my function to deal with more than a single value of x in my real case).

Answer 1

If you're looking for a non-base graphics solution:
library(ggplot2)
example3 <- function(x){
data.frame(
x = x,
sin = sin(x),
cos = cos(x)
)
}
x_grid=seq(0,1,0.05)
ggplot(data = example3(x_grid),
aes(x=x)) +
geom_line(aes(y = sin), color = "blue") +
geom_line(aes(y = cos), color = "red")
With the output:

Answer 2

Your function is vectorized so you can input a vector and extract each column by example(x_grid)[, "sin"] or example(x_grid)[, "cos"].
example(x_grid)
# sin cos
# [1,] 0.000000000 1.000000000
# [2,] 0.049979169 0.998750260
# [3,] 0.099833417 0.995004165
example(x_grid)[, "sin"]
# [1] 0.000000000 0.049979169 0.099833417 0.149438132 0.198669331
# [6] 0.247403959 0.295520207 0.342897807 0.389418342 0.434965534
Note: In this case, sapply is not recommended because the function itself has been vectorized. sapply will make it inefficient. Here is an illustration by benchmark:
library(microbenchmark)
bm <- microbenchmark(
basic = example(x_grid)[, 1],
sapply = sapply(x_grid, function(x) example(x)[1]),
times = 1000L
)
ggplot2::autoplot(bm)
If you want to plot both the two functions, matplot() can plot each column of one matrix.
x_grid <- seq(0, 10, 0.05)
matplot(x_grid, example(x_grid), type = "l")

Answer 3

Appears to be an extra parameter to seq
x_grid <- seq(0, 1, 0.05)
Slight modification to pass variable to function and then subset
plot(x_grid, sapply(x_grid, function(x) example(x)[1]))
Another approach for function which uses a list and then the function can be subset by name
example2 <- function(x) {
within(list(), {
sin <- sin(x)
cos <- cos(x)
})
}
plot(x_grid, sapply(x_grid, function(x) example2(x)$sin))
Unless the example is simplified, the following works without sapply
plot(x_grid, example2(x_grid)$sin)
Plotting both results
lapply(example2(x_grid), plot, x_grid)