Returning an existing data frame with four new columns

Question

I'm trying to implement a function that given a data frame returns the same data frame with four columns added. These new four columns are: for each row, I get the maximum element and its index and put them as two new columns. I do the same with the second maximum element. I don't care if they are repeated.

add_2max <- function(x)
{
  max1 = max(x, na.rm=TRUE)
  indmax1 = which.max(x)
  y=x[-c(indmax1)]
  max2 = max(y, na.rm=TRUE)
  indmax2 = which(x==max2)
  indmax2 = ifelse(max1==max2, indmax2[2], indmax2[1])
  x=c(x, max1, max2, indmax1, indmax2)
  return (x)
}


add_2max_df <- function(DF)
{
  NewDF=t(apply(DF, 1, add_2max))
  return(NewDF)
} 

I'm sure this code can be improved. What do you recommend in order to do that? Is it fast enough?

Answer 1

Here is a more compact version:

add_2maxBrian <- function(x) {
    r <- order(x, decreasing=TRUE)
    c(x, x[r[1:2]], r[1:2])
}

With some sample data (different than Tommy's because I want the chance of ties):

set.seed(123)
DF <- as.data.frame(matrix(sample(1:20, 10000, replace=TRUE), 2500, 4))

It's not as fast a Tommy's solution, but still faster than the original.

library("rbenchmark")

benchmark(a1 <- t(apply(DF, 1, add_2max)),
          a2 <- t(apply(DF, 1, add_2maxFaster)),
          a3 <- t(apply(DF, 1, add_2maxBrian)),
          order = "relative")
#                                   test replications elapsed relative user.self
#2 a2 <- t(apply(DF, 1, add_2maxFaster))          100   6.537 1.000000     6.441
#3  a3 <- t(apply(DF, 1, add_2maxBrian))          100   8.259 1.263424     8.073
#1       a1 <- t(apply(DF, 1, add_2max))          100  12.168 1.861404    12.038
#  sys.self user.child sys.child
#2    0.067          0         0
#3    0.082          0         0
#1    0.089          0         0
identical(a1, a2) #TRUE
identical(a1, a3) #TRUE

Answer 2

Here's a faster way:

add_2maxFaster <- function(x)
{
  imax1 <- which.max(x)
  imax2 <- which.max(x[-imax1])
  if (imax2 >= imax1) imax2 <- imax2 + 1L
  c(x, x[imax1], x[imax2], imax1, imax2) 
}

set.seed(42)
m <- matrix(runif(1e6), 1e4)

# Compare speed:
system.time( a1<-apply(m, 1, add_2max) )        # 0.38 secs
system.time( a2<-apply(m, 1, add_2maxFaster) )  # 0.15 secs

# ...And compare results
all.equal(a1,a2) # TRUE