There is a shorter/elegant/efficient way of writing this? - r

Question

This R code works, but the for loop looks too long and ugly, and I had read that using for loops is not advised in R.
What I want to do is to copy vectors of varying length from the list of vectors HaarData#W to the rows of the matrix MyMatrix.
Since the vectors length is shorter than the number of columns in the matrix, I want to duplicate the values to fill the row.
The vectors have length 2z z ∈ ℤ , and the matrix row length needs to be n such 2z ≤ n
library(wavelets)
Data <- seq(1, 16)
n <- as.integer(log2(length(Data)))
#Data <- seq(1, 2 ^ n, 1)
HaarData <- dwt(Data, filter = "haar")
#Square matrix to write data
MyMatrix <- matrix(, nrow = n, ncol = 2 ^ n)
row <- 0 #row counter
for (vector in HaarData#W) {
row <- row + 1
duplication <- (2 ^ n) / length(vector)
newRow <- c(rep(vector, each = duplication))
MyMatrix[row,] <- newRow
}

Answer 1

I am not sure why you want to do the operation in first place, nevertheless the following would be my approach:
library(wavelets)
library(microbenchmark)
Data <- seq(1, 32)
n <- as.integer(log2(length(Data)))
HaarData <- dwt(as.numeric(Data), filter = "haar")
# Abstract operation in the loop in a function, no side effects
duplicate_coefs <- function(filter_coefs, n){
rep(filter_coefs, each = `^`(2, n - as.integer(log2(length(filter_coefs))) ))
}
microbenchmark(
old = {
#Square matrix to write data
MyMatrix <- matrix(, nrow = n, ncol = 2 ^ n)
row <- 0 #row counter
for (vector in HaarData#W) {
row <- row + 1
duplication <- (2 ^ n) / length(vector)
newRow <- c(rep(vector, each = duplication))
MyMatrix[row,] <- newRow
}
}
,
new = {
n_len <- length(HaarData#W)
new_result <- matrix(unlist( lapply(HaarData#W, duplicate_coefs, n_len) )
, nrow = n_len
, byrow = TRUE)
)
identical(MyMatrix, new_result)
On my machine you get about 50x speed-up
Unit: microseconds
expr min lq mean median uq max neval
old 2891.967 2940.0550 3203.14740 2982.5360 3110.3985 6472.223 100
new 48.519 50.8065 59.04673 56.4805 60.8905 302.845 100
Hope this helps