How would you program Pascal's triangle in R?

Question

I am reading about programming, and one exercise involved programming Pascal's triangle in R. My first idea was to make a list and then append things to it, but that didn't work too well. Then I thought of starting with a vector, and making a list out of that, at the end. Then I thought of making a matrix, and making a list out of that at the end.

Not sure which way to even approach this.

Wayless

The most convenient storing structure would probably depend on the algorithm that you want to use. After all, R does have a choose function. – Primipara

Answer 1

There is one solution on Rosetta Code:

pascalTriangle <- function(h) {
  for(i in 0:(h-1)) {
    s <- ""
    for(k in 0:(h-i)) s <- paste(s, "  ", sep="")
    for(j in 0:i) {
      s <- paste(s, sprintf("%3d ", choose(i, j)), sep="")
    }
    print(s)
  }
}

I would store this in a list if I was developing it myself, since that is the most natural data structure to handle variable length rows. But you really would need to clarify a use case before making that decision. Are you intending on doing analysis on the data after it has been generated?

Edit:

Here is the Rosetta solution rewritten with less looping, and storing the results as a list:

pascalTriangle <- function(h) {
  lapply(0:h, function(i) choose(i, 0:i))
}

Smoke

I always check Rosetta Code first for something as generic as that. Although the solutions usually aren't optimal, they can help as a first step. – Smoke

pascalTriangle <- function(h) { sapply(0:h, function(i) choose(0:h, i)) } returns the triangle as a square matrix – Unbodied

Answer 2

using a property of the Pascal triangle:

x <- 1
print(x)
for (i in 1:10) { x <- c(0, x) + c(x, 0); print(x) }

I suppose this code is very fast.

Manatee

Answer 3

Here's a solution avoiding loops (R is not a big fan of loops):

sapply(1:10, function(n) sapply(0:n, function(k) choose(n, k)))

You can replace 1:10 with any vector, even noncontiguous ones:

R> sapply(c(5, 10), function(n) sapply(0:n, function(k) choose(n, k)))
[[1]]
[1]  1  5 10 10  5  1

[[2]]
[1]   1  10  45 120 210 252 210 120  45  10   1

Samuella