Assign single output for multiple-output function to new function

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).

krfurlong · Accepted Answer · 2020-02-11 15:17:47Z

2

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:

edited Feb 11, 2020 at 15:17

answered Feb 11, 2020 at 15:08

krfurlong

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Comments

Darren Tsai · Accepted Answer · 2020-02-11 20:55:04Z

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")

manotheshark · Accepted Answer · 2020-02-11 15:07:32Z

1

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)

edited Feb 11, 2020 at 15:07

answered Feb 11, 2020 at 15:02

manotheshark

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1 Comment

user9875321__ Over a year ago

your second line did the job. Simple and effective!

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