Replace element in a vector with a list in R

Question

So, I have a vector, points, that I want to populate with a pair of points, whose x and y coordinates are uniformly distributed between 0 and 1.

Here is my code so far:

n <- 10000
points <- rep(0, n)
for (i in 1:n) {
  a <- list(x=runif(1, 0, 1), y=runif(1, 0, 1))
  b <- list(x=runif(1, 0, 1), y=runif(1, 0, 1))
  replace(points, i, list(a, b))
}

I have tried the following:

n <- 10000
points <- rep(0, n)
for (i in 1:n) {
  a <- list(x=runif(1, 0, 1), y=runif(1, 0, 1))
  b <- list(x=runif(1, 0, 1), y=runif(1, 0, 1))
  points[i] <- list(a, b)
}

As you can see, a is the first point with uniformly distributed x and y, and b is the second point with uniformly distributed x and y. For each i in 1:n, I'm trying to replace the ith element in points with list(a, b); however, I keep on getting the warning "number of items to replace is not a multiple of replacement length", indicating that it's trying to replace the ith element in points with a and b as separate elements, and not replacing it with a single list element containing a and b. Is there any way I can coax it to do what I want?

It seems like `replicate(n, runif(2), simplify = FALSE)` gives the output you want (a list of n vectors of length 2 with each element a random uniform variable between 0 and 1). — IceCreamToucan, Sep 11 '18 at 17:09

Kansrekenaar · Answer 1 · 2018-09-11T17:30:25.160

1

Try to initiate points as a list, using points <- vector(mode="list",length=n).

Edit: And use double square brackets for the replacement (see comment)

edited Sep 11 '18 at 17:30

answered Sep 11 '18 at 17:04

Kansrekenaar

21
2

Sorry, that kept giving me the same warnings and problems... thank you, though! – user3742846 Sep 11 '18 at 17:23
You should also use double brackets for replacement in a list. That is, use `points[[i]] <- list(a, b)` in your last line. For more details see https://stackoverflow.com/questions/1169456/the-difference-between-bracket-and-double-bracket-for-accessing-the-elem – Kansrekenaar Sep 11 '18 at 17:27
I think you need `points = replicate(n, vector('list', length = 2), simplify = FALSE)` because each element of points is itself a list of length 2. And double square brackets as Kansrekenaar said. – konvas Sep 11 '18 at 17:28

score 0 · Answer 2 · answered Sep 11 '18 at 17:25

The function runif is already vectorized, so your code will be much faster if you call runif once. Essentially what you want is to get 4 * n numbers which are uniformly distributed between 0 and 1, so all you need is numbers <- runif(4 * n). You can then structure them as required. Do you actually need to have a list or are you just using it to make the for loop more efficient? For example matrix(numbers, ncol = 4), will give you a matrix (you can interpret the columns as the x coordinate of a, the y coordinate of a, the x coordinate of b, and the y coordinate of b, it doesn't really matter as they are independently sampled). I can be more specific if you post an example with your desired output.

score 0 · Answer 3 · answered Sep 11 '18 at 19:18

This should work. I defined each point a and b as a data frame, as it is more convenient. point is a list of points

points <- list(a = data.frame(x = runif(10000,0,1),y = runif(10000,0,1)), b =data.frame(x = runif(10000,0,1),y = runif(10000,0,1)))

Here the result with 10 points:

> points
$a
            x         y
1  0.69817192 0.4931020
2  0.08508172 0.5187833
3  0.71519413 0.7871234
4  0.87653502 0.0803383
5  0.57951306 0.4365260
6  0.14673983 0.8060857
7  0.80315435 0.2124616
8  0.79790622 0.6328104
9  0.22130415 0.6239957
10 0.99642466 0.2076050

$b
           x           y
1  0.3798732 0.419808119
2  0.5833667 0.002143209
3  0.8335923 0.708020370
4  0.1923296 0.322810176
5  0.9421873 0.190519505
6  0.6633320 0.143258369
7  0.7465269 0.071942890
8  0.6919004 0.627667826
9  0.5362267 0.345617737
10 0.1725097 0.476865279

Replace element in a vector with a list in R

3 Answers3