> fibo <- function(n) {
+   if ( n < 2 ) n
+   else fibo(n-1) + fibo(n-2)
+ }
> system.time(for(i in 0:26) fibo(i))
   user  system elapsed
   7.48    0.00    7.52
> system.time(sapply(0:26, fibo))
   user  system elapsed
   7.50    0.00    7.54
> system.time(lapply(0:26, fibo))
   user  system elapsed
   7.48    0.04    7.54
> library(plyr)
> system.time(ldply(0:26, fibo))
   user  system elapsed
   7.52    0.00    7.58

library(snow)
cl <- makeSOCKcluster(c("localhost","localhost"))
parSapply(cl, 1:20, get("+"), 3)

parLapply(cl, x, fun, ...)
parSapply(cl, X, FUN, ..., simplify = TRUE, USE.NAMES = TRUE)
parApply(cl, X, MARGIN, FUN, ...)
parRapply(cl, x, fun, ...)
parCapply(cl, x, fun, ...)

> df <- 1:10
> # *apply example
> lapply(2:3, function(i) df <- df * i)
> df
 [1]  1  2  3  4  5  6  7  8  9 10
> # for loop example
> for(i in 2:3) df <- df * i
> df
 [1]  6 12 18 24 30 36 42 48 54 60

set.seed(1)  #for reproducability of the results

# The data
X <- rnorm(100000)
Y <- as.factor(sample(letters[1:5],100000,replace=T))
Z <- as.factor(sample(letters[1:10],100000,replace=T))

# the function forloop that averages X over every combination of Y and Z
forloop <- function(x,y,z){
# These ones are for optimization, so the functions
#levels() and length() don't have to be called more than once.
  ylev <- levels(y)
  zlev <- levels(z)
  n <- length(ylev)
  p <- length(zlev)

  out <- matrix(NA,ncol=p,nrow=n)
  for(i in 1:n){
      for(j in 1:p){
          out[i,j] <- (mean(x[y==ylev[i] & z==zlev[j]]))
      }
  }
  rownames(out) <- ylev
  colnames(out) <- zlev
  return(out)
}

# Used on the generated data
forloop(X,Y,Z)

# The same using tapply
tapply(X,list(Y,Z),mean)

> system.time(forloop(X,Y,Z))
   user  system elapsed
   0.94    0.02    0.95

> system.time(tapply(X,list(Y,Z),mean))
   user  system elapsed
   0.06    0.00    0.06

> system.time({z <- numeric(1e6); for(i in y) z[i] <- foo(i)})
   user  system elapsed
   3.54    0.00    3.53
> system.time(z <- lapply(y, foo))
   user  system elapsed
   2.89    0.00    2.91
> system.time(z <- vapply(y, foo, numeric(1)))
   user  system elapsed
   1.35    0.00    1.36

foo <- function(x) {
   x <- x+1
 }
y <- numeric(1e6)
system.time({z <- numeric(1e6); for(i in y) z[i] <- foo(i)})
#   user  system elapsed
#  4.967   0.049   7.293
system.time(z <- sapply(y, foo))
#   user  system elapsed
#  5.256   0.134   7.965
system.time(z <- lapply(y, foo))
#   user  system elapsed
#  2.179   0.126   3.301

set.seed(1)  #for reproducability of the results

# The data - copied from Joris Meys answer
X <- rnorm(100000)
Y <- as.factor(sample(letters[1:5],100000,replace=T))
Z <- as.factor(sample(letters[1:10],100000,replace=T))

# an R way to generate tapply functionality that is fast and
# shows more general principles about fast R coding
YZ <- interaction(Y, Z)
XS <- split(X, YZ)
m <- vapply(XS, mean, numeric(1))
m <- matrix(m, nrow = length(levels(Y)))
rownames(m) <- levels(Y)
colnames(m) <- levels(Z)
m

df <- data.frame(id = rep(letters[1:10], 100000),
                 value = rnorm(1000000))

f1 <- function(x)
  tapply(x$value, x$id, sum)

f2 <- function(x){
  res <- 0
  for(i in seq_along(l <- unique(x$id)))
    res[i] <- sum(x$value[x$id == l[i]])
  names(res) <- l
  res
}

library(microbenchmark)

> microbenchmark(f1(df), f2(df), times=100)
Unit: milliseconds
   expr      min       lq   median       uq      max neval
 f1(df) 28.02612 28.28589 28.46822 29.20458 32.54656   100
 f2(df) 38.02241 41.42277 41.80008 42.05954 45.94273   100

mat <- matrix(rnorm(1000000), nrow=1000)

f3 <- function(x)
  apply(x, 2, sum)

f4 <- function(x){
  res <- 0
  for(i in 1:ncol(x))
    res[i] <- sum(x[,i])
  res
}

> microbenchmark(f3(mat), f4(mat), times=100)
Unit: milliseconds
    expr      min       lq   median       uq      max neval
 f3(mat) 14.87594 15.44183 15.87897 17.93040 19.14975   100
 f4(mat) 12.01614 12.19718 12.40003 15.00919 40.59100   100

f5 <- function(x)
  colSums(x)

> microbenchmark(f5(mat), times=100)
Unit: milliseconds
    expr      min       lq   median       uq      max neval
 f5(mat) 1.362388 1.405203 1.413702 1.434388 1.992909   100