But assign will do this:

for(i in 1:n) assign(paste('obj', i, sep='.'), mylist[[i]])

WARNING: An operation like this can land you in Circle 3 ([page 17] failing to vectorize|this example is anti-vectorizing) and/or the heresy of Circle 6 (page 35).

As we have just seen with get, assigning to a name like 'myobj$comp' is not going to get you where you want to go|it will create an object with a nonstandard name rather than modifying the comp component of myobj. Create a copy of the object, change the component in the copy, then assign the name to the copy.

8.1.41string not the name (still)

A formula can easily be created out of a character string:

>myvars <- paste('V', 1:9, sep='')

>myvars

[1] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9"

> as.formula(paste('y ~ ', paste(myvars[c(3,5,8)], + collapse=' + ')))

y ~ V3 + V5 + V8

8.1.42name not the argument

You may want to produce a plot or other output that states the dataset that was used. You might try something like:

myfun <- function(x, ...) f

plot(x, main=x, ...)

g

But that is going to produce a less than pleasing main title. The substitute function along with deparse is probably what you are looking for.

myfun2 <- function(x, ...) f

plot(x, main=deparse(substitute(x)), ...)

g

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8.1.43unexpected else

Error: unexpected 'else' in "else"

If you aren't expecting 'else' in 'else', then where would you expect it?

While you may think that R is ludicrous for giving you such an error message, R thinks you are even more ludicrous for expecting what you did to work.

R takes input until it gets a complete statement, evaluates that statement, then takes more input. Here is how to get that error:

if(any(abs(x) > 1)) atan(x) else asin(x)

When R gets to the end of the rst line, it has a perfectly complete statement so it evaluates it. Now it nds a statement starting with 'else'|this makes no sense. Proper formatting is the key. If it is convenient, you can put the whole statement on one line:

if(any(abs(x) > 1)) atan(x) else asin(x)

Alternatively, use curly braces (and a well-placed 'else'):

if(any(abs(x) > 1)) f atan(x)

g else f asin(x)

g

8.1.44dropping dimensions

>xmat <- array(1:4, c(2,2))

>xmat[1,] # simple vector, not a matrix [1] 1 3

>xmat[1, , drop=FALSE] # still a matrix [,1] [,2]

[1,] 1 3

By default dimensions of arrays are dropped when subscripting makes the dimension length 1. Subscripting with drop=FALSE overrides the default.

NOTE: Failing to use drop=FALSE inside functions is a major source of bugs. You only test the function when the subscript has length greater than 1. The function fails once it hits a situation where the subscript is length 1|somewhere downstream a matrix is expected and there is a simple vector there instead.

NOTE: Failing to use drop=FALSE inside functions is a major source of bugs.

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8.1.45drop data frames

The drop function has no e ect on a data frame. If you want dropping to be done in a data frame, you need to use the drop argument in subscripting.

Dropping in data frames can be surprising (but it is logical).

>xdf <- data.frame(a=1:2, b=c('v', 'w'))

>xdf[1,] # data frame

ab 1 1 v

>drop(xdf[1,]) # data frame

ab

1 1 v

>xdf[1, , drop=TRUE] # list $a

[1] 1 $b [1] v

Levels: v w

>xdf[,1] # numeric vector [1] 1 2

>xdf[, 1, drop=FALSE] # data frame a

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2 2

> drop(xdf[, 1, drop=FALSE]) # data frame a

11

2 2

8.1.46losing row names

The row names of a data frame are lost through dropping:

> xdf2 <- data.frame(a=1:4, b=42:45, + row.names=LETTERS[1:4])

>xdf2[, 1] [1] 1 2 3 4

>as.matrix(xdf2)[, 1] A B C D

1 2 3 4

Coercing to a matrix rst will retain the row names, but possibly at the expense of not getting the right values if the column types of the data frame are mixed.

> xdf2b <- data.frame(a=1:4, b=letters[21:24],

+row.names=LETTERS[1:4])

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> as.matrix(xdf2b)[,1] A B C D

"1" "2" "3" "4"

> drop(as.matrix(xdf2b[, 1, drop=FALSE])) A B C D

1 2 3 4

The nal incantation, though a bit complicated, will give you the right thing.

8.1.47apply function returning a vector

If you use apply with a function that returns a vector, that becomes the rst dimension of the result. This is likely not what you naively expect if you are operating on rows:

> matrix(15:1, 3)

[,1] [,2] [,3] [,4] [,5]

> apply(matrix(15:1, 3), 1, sort) [,1] [,2] [,3]

The naive expectation is really arrived at with:

t(apply(matrix(15:1, 3), 1, sort))

But note that no transpose is required if you operate on columns|the naive expectation holds in that case.

8.1.48empty cells in tapply

If there are combinations of levels that do not appear, then tapply gives NA as the answer (or NULL if simplify=FALSE):

>tapply(9, factor(1, levels=1:2), sum) 1 2

9 NA

>tapply(9, factor(1, levels=1:2), sum, simplify=FALSE) $ 8 1 8

[1] 9 $ 8 2 8

NULL

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