Lecture 16: Functions as Objects

20 October 2014


  • Writing our own functions
  • Dividng labor with multiple functions
  • Refactoring to create higher-level operations
  • Using apply, sapply, etc., to avoid iteration


  • Functions are objects, and can be arguments to other functions
  • Functions are objects, and can be returned by other functions
  • Example: surface

Reading: Sections 7.5, 7.11 and 7.13 of Matloff

Optional Recommended Reading: Chapter 3 of Chambers

Functions as Objects

  • In R, functions are objects, just like everything else

  • This means that they can be passed to functions as arguments
    and returned by functions as outputs as well

Functions of Functions: Computationally

  • We often want to do very similar things to many different functions

  • The procedure is the same, only the function we're working with changes

  • \( \therefore \) Write one function to do the job, and pass the function as an argument

  • Because R treats a function like any other object, we can do this simply: invoke the function by its argument name in the body

  • We have already seen examples

R Functions That Take Functions as Arguments

  • apply(), sapply(), etc.: Take this function and use it on all of these objects
  • nlm(): Take this function and try to make it small, starting from here
  • ks.test(): Compare these data to this cumulative distribution function
  • curve(): Evaluate this function over that range, and plot the results

Some R Syntax Facts About Functions

  • Typing a function's name, without parentheses, in the terminal gives you its source code:
function (x, size, replace = FALSE, prob = NULL) 
    if (length(x) == 1L && is.numeric(x) && x >= 1) {
        if (missing(size)) 
            size <- x
        sample.int(x, size, replace, prob)
    else {
        if (missing(size)) 
            size <- length(x)
        x[sample.int(length(x), size, replace, prob)]
<bytecode: 0x102d004a0>
<environment: namespace:base>

Some R Syntax Facts About Functions

  • Functions are their own class in R:
[1] "function"
[1] "function"
resample <- function(x) { sample(x, size=length(x), replace=TRUE) }
[1] "function"

Some R Syntax Facts About Functions

  • Functions can be put into lists or even arrays
  • A call to function returns a function object
    • body executed; access with body(foo)
    • arguments required: access with formals(foo)
      gives argument list of foo: names are argument names, values are expressions for defaults (if any)
    • parent environment: access with environment(foo)

Some R Syntax Facts About Functions

  • R has separate types for built-in functions and for those written in R:
[1] "closure"
[1] "closure"
[1] "builtin"

Why closure for written-in-R functions? Because expressions are “closed” by referring to the parent environment

There's also a 2nd class of built-in functions called primitive

Anonymous Functions

  • function() returns an object of class function
  • So far we've assigned that object to a name
  • If we don't have an assignment, we get an anonymous function
  • Usually part of some larger expression:
[1] 0.1192 0.2689 0.5000 0.7311 0.8808

Anonymous Functions

  • Often handy when connecting other pieces of code
    • especially in things like apply and sapply
  • Won't cluttering the workspace
  • Can't be examined or re-used later

Example: grad()

  • Problems in stats. come down to optimization
    So do lots of problems in econ., physics, CS, bio, …

  • Lots of optimization problems require the gradient of the objective function

  • Gradient of \( f \) at \( x \): \[ \nabla f(x) = \left[\left.\frac{\partial f}{\partial x_1}\right|_{x} \ldots \left.\frac{\partial f}{\partial x_p}\right|_{x}\right] \]

Example: grad()

  • We do the same thing to get the gradient of \( f \) at \( x \) no matter what \( f \) is:
  find the partial derivative of f with respect to each component of x
  return the vector of partial derivatives
  • It makes no sense to re-write this every time we change \( f \)!

  • \( \therefore \) write code to calculate the gradient of an arbitrary function

  • We could write our own, but there are lots of tricky issues

    • Best way to calculate partial derivative
    • What if \( x \) is at the edge of the domain of \( f \)?
  • Fortunately, someone has already done this

Example: grad()

From the package numDeriv

grad(func, x, ...) # Plus other arguments
  • Assumes func is a function which returns a single floating-point value
  • Assumes x is a vector of arguments to func
    • If x is a vector and func(x) is also a vector, then it's assumed func is vectorized and we get a vector of derivatives
  • Extra arguments in ... get passed along to func
  • Other functions in the package for the Jacobian of a vector-valued function, and the matrix of 2nd partials (Hessian)

Example: grad()

  • Does it work as advertized?
just_a_phase <- runif(n=1,min=-pi,max=pi)
[1] TRUE
phases <- runif(n=10,min=-pi,max=pi)
[1] TRUE
grad(func=function(x){x[1]^2+x[2]^3}, x=c(1,-1))
[1] 2 3

Note: grad is perfectly happy with func being an anonymous function!


Now we can use this as a piece of a larger machine:

gradient.descent <- function(f,x,max.iterations,step.scale,
  stopping.deriv,...) {
  for (iteration in 1:max.iterations) {
    gradient <- grad(f,x,...)
    if(all(abs(gradient) < stopping.deriv)) { break() }
    x <- x - step.scale*gradient
  fit <- list(argmin=x,final.gradient=gradient,final.value=f(x,...),
  • Works equally well whether f is mean squared error of a regression, \( \psi \) error of a regression, (negative log) likelihood, cost of a production plan, …


  • Scoping f takes values for all names which aren't its arguments from the environment where it was defined, not the one where it is called (e.g., not from inside grad or gradient.descent)
  • Debugging If f and g are both complicated, avoid debugging g(f) as a block; divide the work by writing very simple f.dummy to debug/test g, and debug/test the real f separately

Returning Functions: A trivial example

Functions can be return values like anything else

make.noneuclidean <- function(ratio.to.diameter=pi) {
  circumference <- function(d) { return(ratio.to.diameter*d) }

Returning Functions: A trivial example (cont'd.)

kings.i <- make.noneuclidean(3)
[1] 30
    return(ratio.to.diameter * d)
<environment: 0x100b588d8>

A Less Trivial Example

Create a linear predictor, based on sample values of two variables

make.linear.predictor <- function(x,y) {
  linear.fit <- lm(y~x)
  predictor <- function(x) {

The predictor function persists and works, even when the data we used to create it is gone

A Less Trivial Example

library(MASS); data(cats)
vet_predictor <- make.linear.predictor(x=cats$Bwt,y=cats$Hwt)
rm(cats)            # Data set goes away
vet_predictor(3.5)  # My cat's body mass in kilograms

A more mathematical example

  • Instead of finding \( \nabla f(x) \), find the function \( \nabla f \):
nabla <- function(f,...) {
  g <- function(x,...) { grad(func=f,x=x,...) }

Exercise: Write a test case!

Example: curve()

  • You learned to use curve in the first week (because you did all of the assigned reading, including section 2.3.3 of the textbook)

  • A call to curve looks like this:

curve(expr, from = a, to = b, ...)

expr is some expression involving a variable called x
which is swept from the value a to the value b
... are other plot-control arguments

  • curve feeds the expression a vector x and expects a numeric vector back, e.g. curve(x^2 * sin(x)) is fine

Using curve() with our own functions

  • If we have defined a function already, we can use it in curve:
psi <- function(x,c=1) {ifelse(abs(x)>c,2*c*abs(x)-c^2,x^2)}

Try this! Also try


and explain it to yourself

Using curve() with our own functions

  • If our function doesn't take vectors to vectors, curve becomes unhappy
mse <- function(y0,a,Y=gmp$pcgmp,N=gmp$pop) {
   mean((Y - y0*(N^a))^2)
> curve(mse(a=x,y0=6611),from=0.10,to=0.15)
Error in curve(mse(a = x, y0 = 6611), from = 0.1, to = 0.15) : 
  'expr' did not evaluate to an object of length 'n'
In addition: Warning message:
In N^a : longer object length is not a multiple of shorter object length

How do we solve this?

Using curve() with our own functions

  • Define a new, vectorized function, say with sapply:
[1] 154701953 102322974  68755654  64529166 104079527 207057513
[1] 154701953
mse.plottable <- function(a,...){ return(sapply(a,mse,...)) }
[1] 154701953 102322974  68755654  64529166 104079527 207057513

Using curve() with our own functions


plot of chunk unnamed-chunk-19

Using curve() with our own functions

  • Alternate strategy: Vectorize() returns a new, vectorized function
mse.vec <- Vectorize(mse, vectorize.args=c("y0","a"))
[1] 154701953 102322974  68755654  64529166 104079527 207057513
[1] 134617132  74693733  63732256

Using curve() with our own functions


plot of chunk unnamed-chunk-21

Example: surface()

  • curve takes an expression and, as a side-effect, plots a 1-D curve by sweeping over x

  • Suppose we want something like that but sweeping over two variables

  • Built-in plotting function contour:

    contour(x,y,z, [[other stuff]])

    x and y are vectors of coordinates, z is a matrix of the corresponding shape
    (see help(contour) for graphical options)

  • Strategy: surface should make x and y sequences, evaluate the expression at each combination to get z, and then call contour

First attempt at surface()

  • Only works with vector-to-number functions:
surface.1 <- function(f,from.x=0,to.x=1,from.y=0,to.y=1,n.x=101,
  n.y=101,...) {
  x.seq <- seq(from=from.x,to=to.x,length.out=n.x)
  y.seq <- seq(from=from.y,to=to.y,length.out=n.y)
  plot.grid <- expand.grid(x=x.seq,y=y.seq)
  z.values <- apply(plot.grid,1,f)
  z.matrix <- matrix(z.values,nrow=n.x)

First attempt at surface()


plot of chunk unnamed-chunk-23

Expressions and Evaluation

  • curve doesn't require us to write a function every time — what's it's trick?

  • Expressions are just another class of R object, so they can be created and manipulated

  • One manipulation is evaluation


evaluates the expression expr in the environment envir, which can be a data frame or even just a list

  • When we type something like x^2+y^2 as an argument to surface.1, R tries to evaluate it prematurely

  • substitute returns the unevaluted expression

  • curve uses first substitute(expr) and then eval(expr,envir), having made the right envir

Second attempt at surface()

surface.2 <- function(expr,from.x=0,to.x=1,from.y=0,to.y=1,n.x=101,
  n.y=101,...) {
  x.seq <- seq(from=from.x,to=to.x,length.out=n.x)
  y.seq <- seq(from=from.y,to=to.y,length.out=n.y)
  plot.grid <- expand.grid(x=x.seq,y=y.seq)
  unevaluated.expression <- substitute(expr)
  z.values <- eval(unevaluated.expression,envir=plot.grid)
  z.matrix <- matrix(z.values,nrow=n.x)

Second attempt at surface()