8.3.18invisibility cloak

In rare circumstances the visibility of a result may not be as expected:

>myfun6 <- function(x) x

>myfun6(zz <- 7)

>.Last.value

[1] 7

>a6 <- myfun6(zz <- 9)

>a6

[1] 9

>myfun6(invisible(11))

>myfun7 <- function(x) 1 * x

>myfun7(invisible(11))

[1]11

8.3.19evaluation of default arguments

Consider:

>myfun2 <- function(x, y=x) x + y

>x <- 100

>myfun2(2)

[1] 4

> myfun2(2, x) [1] 102

Some people expect the result of the two calls above to be the same. They are not. The default value of an argument to a function is evaluated inside the function, not in the environment calling the function.

Thus writing a function like the following will not get you what you want.

>myfun3 <- function(x=x, y) x + y

>myfun3(y=3)

Error in myfun3(y = 3) : recursive default argument reference

(The actual error message you get may be di erent in your version of R.)

The most popular error to make in this regard is to try to imitate the default value of an argument. Something like:

>myfun5 <- function(x, n=xlen) f xlen <- length(x); ...g

>myfun5(myx, n=xlen-2)

xlen is de ned inside myfun5 and is not available for you to use when calling myfun5.

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8.3.20sapply simpli cation

The sapply function \simpli es" the output of lapply. It isn't always so simple. That is, the simpli cation that you get may not be the simpli cation you expect. This uncertainty makes sapply not so suitable for use inside functions. The vapply function is sometimes a safer alternative.

8.3.21one-dimensional arrays

Arrays can be of any positive dimension (modulo memory and vector length limits). In particular, one-dimensional arrays are possible. Almost always these look and act like plain vectors. Almost.

Here is an example where they don't:

> df2 <- data.frame(x=rep(1, 3), y=tapply(1:9,

+ factor(rep(c('A', 'B', 'C'), each=3)), sum))

>df2 x y A 1 6

B 1 15 C 1 24

>tapply(df2$y, df2$x, length)

1

3

>by(df2$y, df2$x, length) INDICES: 1

[1] 1

>by(as.vector(df2$y), df2$x, length) INDICES: 1

[1] 3

tapply returns an array, in particular it can return a one-dimensional array| which is the case with df2$y. The by function in this case when given a onedimensional array produces the correct answer to a question that we didn't think we were asking.

One-dimensional arrays are neither matrices nor (exactly) plain vectors.

8.3.22by is for data frames

The by function is essentially just a pretty version of tapply for data frames. The \for data frames" is an important restriction. If the rst argument of your call to by is not a data frame, you may be in for trouble.

>tapply(array(1:24, c(2,3,4)), 1:24 %% 2, length) 0 1 12 12

>by(array(1:24, c(2,3,4)), 1:24 %% 2, length)

110

Error in tapply(1:2L, list(INDICES = c(1, 0, 1, 0, 1, 0, 1, : arguments must have same length

In this example we have the good fortune of an error being triggered, we didn't have that for the problem in Circle 8.3.21.

8.3.23stray backquote

A stray backquote in a function de nition can yield the error message:

symbol print-name too long

The backquote is sometimes (too) close to the tab key and/or the escape key. It is also close to minimal size and hence easy to overlook.

8.3.24array dimension calculation

There are times when the creation of matrices fails to be true to the intention:

Notice that the matrix is created|that is a warning not an error|the matrix is merely created inappropriately. If you ignore the warning, there could be consequences down the line.

Let's investigate the ingredients:

>(2 - .1) / .1 [1] 19

>(2 - .1) / .1 - 19 [1] -3.552714e-15

>as.integer((2 - .1) / .1) [1] 18

When R coerces from a oating point number to an integer it truncates rather than rounds.

The moral of the story is that round can be a handy function to use. In a sense this problem really belongs in Circle 1 (page 9), but the subtlety makes it di cult to nd.

8.3.25replacing pieces of a matrix

We have two matrices:

>m6 <- matrix(1:6, 3)

>m4 <- matrix(101:104, 2)

>m6

111

[,1] [,2]

[,1] [,2] [1,] 101 103 [2,] 102 104

Our task is to create a new matrix similar to m6 where some of the rows are replaced by the rst row of m4. Here is the natural thing to do:

>m6new <- m6

>m6new[c(TRUE, FALSE, TRUE), ] <- m4[1,]

>m6new

[,1] [,2] [1,] 101 101 [2,] 2 5 [3,] 103 103

We are thinking about rows being the natural way of looking at the problem. The problem is that that isn't the R way, despite the context.

One way of getting what we want is:

>s6 <- c(TRUE, FALSE, TRUE)

>m6new[s6, ] <- rep(m4[1,], each=sum(s6))

>m6new

[,1] [,2] [1,] 101 103 [2,] 2 5 [3,] 101 103

8.3.26reserved words

R is a language. Because of this, there are reserved words that you can not use as object names. Perhaps you can imagine the consequences if the following command actually worked:

FALSE <- 4

You can see the complete list of reserved words with:

?Reserved

8.3.27return is a function

Unlike some other languages return is a function that takes the object meant to be returned. The following construct does NOT do what is intended:

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