R has many *apply functions which are ably described in the help files (e.g. ?apply). There are enough of them, though, that beginning useRs may have difficulty deciding which one is appropriate for their situation or even remembering them all. They may have a general sense that "I should be using an *apply function here", but it can be tough to keep them all straight at first.
Despite the fact (noted in other answers) that much of the functionality of the *apply family is covered by the extremely popular plyr package, the base functions remain useful and worth knowing.
apply - When you want to apply a function to the rows or columns
of a matrix (and higher-dimensional analogues); not generally advisable for data frames as it will coerce to a matrix first.
# Two dimensional matrix
M <- matrix(seq(1,16), 4, 4)
# apply min to rows
apply(M, 1, min)
[1] 1 2 3 4
# apply max to columns
apply(M, 2, max)
[1] 4 8 12 16
# 3 dimensional array
M <- array( seq(32), dim = c(4,4,2))
# Apply sum across each M[*, , ] - i.e Sum across 2nd and 3rd dimension
apply(M, 1, sum)
# Result is one-dimensional
[1] 120 128 136 144
# Apply sum across each M[*, *, ] - i.e Sum across 3rd dimension
apply(M, c(1,2), sum)
# Result is two-dimensional
[,1] [,2] [,3] [,4]
[1,] 18 26 34 42
[2,] 20 28 36 44
[3,] 22 30 38 46
[4,] 24 32 40 48
If you want row/column means or sums for a 2D matrix, be sure to
investigate the highly optimized, lightning-quick colMeans,
rowMeans, colSums, rowSums.
lapply - When you want to apply a function to each element of a
list in turn and get a list back.
This is the workhorse of many of the other *apply functions. Peel
back their code and you will often find lapply underneath.
x <- list(a = 1, b = 1:3, c = 10:100)
lapply(x, FUN = length)
$a
[1] 1
$b
[1] 3
$c
[1] 91
lapply(x, FUN = sum)
$a
[1] 1
$b
[1] 6
$c
[1] 5005
sapply - When you want to apply a function to each element of a
list in turn, but you want a vector back, rather than a list.
If you find yourself typing unlist(lapply(...)), stop and consider
sapply.
x <- list(a = 1, b = 1:3, c = 10:100)
# Compare with above; a named vector, not a list
sapply(x, FUN = length)
a b c
1 3 91
sapply(x, FUN = sum)
a b c
1 6 5005
In more advanced uses of sapply it will attempt to coerce the
result to a multi-dimensional array, if appropriate. For example, if our function returns vectors of the same length, sapply will use them as columns of a matrix:
sapply(1:5,function(x) rnorm(3,x))
If our function returns a 2 dimensional matrix, sapply will do essentially the same thing, treating each returned matrix as a single long vector:
sapply(1:5,function(x) matrix(x,2,2))
Unless we specify simplify = "array", in which case it will use the individual matrices to build a multi-dimensional array:
sapply(1:5,function(x) matrix(x,2,2), simplify = "array")
Each of these behaviors is of course contingent on our function returning vectors or matrices of the same length or dimension.
vapply - When you want to use sapply but perhaps need to
squeeze some more speed out of your code or want more type safety.
For vapply, you basically give R an example of what sort of thing
your function will return, which can save some time coercing returned
values to fit in a single atomic vector.
x <- list(a = 1, b = 1:3, c = 10:100)
#Note that since the advantage here is mainly speed, this
# example is only for illustration. We're telling R that
# everything returned by length() should be an integer of
# length 1.
vapply(x, FUN = length, FUN.VALUE = 0L)
a b c
1 3 91
mapply - For when you have several data structures (e.g.
vectors, lists) and you want to apply a function to the 1st elements
of each, and then the 2nd elements of each, etc., coercing the result
to a vector/array as in sapply.
This is multivariate in the sense that your function must accept
multiple arguments.
#Sums the 1st elements, the 2nd elements, etc.
mapply(sum, 1:5, 1:5, 1:5)
[1] 3 6 9 12 15
#To do rep(1,4), rep(2,3), etc.
mapply(rep, 1:4, 4:1)
[[1]]
[1] 1 1 1 1
[[2]]
[1] 2 2 2
[[3]]
[1] 3 3
[[4]]
[1] 4
Map - A wrapper to mapply with SIMPLIFY = FALSE, so it is guaranteed to return a list.
Map(sum, 1:5, 1:5, 1:5)
[[1]]
[1] 3
[[2]]
[1] 6
[[3]]
[1] 9
[[4]]
[1] 12
[[5]]
[1] 15
rapply - For when you want to apply a function to each element of a nested list structure, recursively.
To give you some idea of how uncommon rapply is, I forgot about it when first posting this answer! Obviously, I'm sure many people use it, but YMMV. rapply is best illustrated with a user-defined function to apply:
# Append ! to string, otherwise increment
myFun <- function(x){
if(is.character(x)){
return(paste(x,"!",sep=""))
}
else{
return(x + 1)
}
}
#A nested list structure
l <- list(a = list(a1 = "Boo", b1 = 2, c1 = "Eeek"),
b = 3, c = "Yikes",
d = list(a2 = 1, b2 = list(a3 = "Hey", b3 = 5)))
# Result is named vector, coerced to character
rapply(l, myFun)
# Result is a nested list like l, with values altered
rapply(l, myFun, how="replace")
tapply - For when you want to apply a function to subsets of a
vector and the subsets are defined by some other vector, usually a
factor.
The black sheep of the *apply family, of sorts. The help file's use of
the phrase "ragged array" can be a bit confusing, but it is actually
quite simple.
A vector:
x <- 1:20
A factor (of the same length!) defining groups:
y <- factor(rep(letters[1:5], each = 4))
Add up the values in x within each subgroup defined by y:
tapply(x, y, sum)
a b c d e
10 26 42 58 74
More complex examples can be handled where the subgroups are defined
by the unique combinations of a list of several factors. tapply is
similar in spirit to the split-apply-combine functions that are
common in R (aggregate, by, ave, ddply, etc.) Hence its
black sheep status.
Best Answer
You must put
colour = rhg_colsinsideaes(). As far as I can tell, you want to apply gradient to bars (in barplot) with factor variable on the abscissa? Then usefill- try this instead:or try to achieve approximate replica with:
Notice that in second case a continuous variable is passed to
fillaesthetics, thereforescale_fill_gradientis passed afterwards. If you pass afactorto thefillaes, you must stick withscale_fill_manual(values = rhg_cols).