How to assign a list of values to a hashmap in R?

Question

Basically the idea is for the values n=10,20,30,...100 to take the mean of 10,000 random samples, saving the 10,000 means for later usage.

In a language I'm more accustomed to, I would create a hashmap using each n as a key, and a list of means as the value.

In javascript for example:

var mydata
var map = {}

for (int i = 10; i <= 100; i += 10 ) {
  map[i] = [] // create list
  for (int j = 0; j < 10000; j++) {
    map[i][j] = mean(sample(mydata, i))
  }
}

Now I attempted to do this in R (this is my first time using it), and I ended up with:

hashmap  <- new.env()
sunspots <- read.table("sunspots.txt")

for (i in seq(10, 100, by=10)) {
  hashmap[[i]] <- c()
  for (j in 1:10000) {
    hashmap[[i]][j] <- mean(sample(sunspots$x, i))
  }
}

However this throws an error:

wrong args for environment subassignment

Even if it didn't throw this error, I'm not entirely sure if I'm approaching it the right way.

Could someone help me understand the proper way to go about this?

Answer 1

The issue here is that i is a numeric, and environments must be keyed by character strings. Thus your immediate problem can be solved with a simple as.character() coercion on the i variable when it's used to index hashmap.

I would also recommend you refactor the inner loop into a vectorized function call, such as replicate(). Here's how I would do this:

set.seed(1L);
test.data <- 1:200;
N <- 3L;
e <- new.env();
for (i in seq(10L,100L,10L)) e[[as.character(i)]] <- replicate(N,mean(sample(test.data,i)));

Result:

ls(e);
##  [1] "10"  "100" "20"  "30"  "40"  "50"  "60"  "70"  "80"  "90"
for (i in seq(10L,100L,10L)) print(e[[as.character(i)]]);
## [1] 108.3 109.4  82.4
## [1] 108.50  93.65 106.20
## [1] 103.3333  96.0000 101.2333
## [1] 98.075 95.250 83.275
## [1] 106.68  97.48 107.34
## [1]  97.48333 105.95000  98.76667
## [1] 101.8857 102.4857 114.6000
## [1]  99.5875 107.0875  96.0750
## [1]  92.9000 103.0889 100.7889
## [1]  91.19  99.80 101.57

You can change N to 10000 and test.data to sunspots for your real data.

---

Also, here's an alternative that produces a matrix output, built around the convenient feature of sapply() that it returns a matrix for multi-element return values from FUN():

set.seed(1L);
sapply(seq(10L,100L,10L),function(i) replicate(N,mean(sample(test.data,i))));
##       [,1]   [,2]     [,3]   [,4]   [,5]      [,6]     [,7]     [,8]     [,9]  [,10]
## [1,] 108.3 108.50 103.3333 98.075 106.68  97.48333 101.8857  99.5875  92.9000  91.19
## [2,] 109.4  93.65  96.0000 95.250  97.48 105.95000 102.4857 107.0875 103.0889  99.80
## [3,]  82.4 106.20 101.2333 83.275 107.34  98.76667 114.6000  96.0750 100.7889 101.57

Answer 2

Wouldn't this be the same, but simpler and more readable?

set.seed(123)
N = 10000
sunspots <- rnorm(N, 10, 2)

sim <- lapply(seq(10, 100, by=10), function(i){
  sapply(1:N, function(j){
    mean(sample(sunspots, i))
   })
})

lapply(sim, head)

It would make sense, as replicate is just an sapply call.

> replicate
function (n, expr, simplify = "array") 
sapply(integer(n), eval.parent(substitute(function(...) expr)), 
    simplify = simplify)
<bytecode: 0x19b0b7108>
<environment: namespace:base>

EDIT

As mentioned in the comments.

simulation <- function(data, i){
  sapply(1:N, function(j) mean(sample(data, i)))
}

sim <- lapply(seq(10, 100, by=10), function(i) simulation(sunspots, i))

# This would give the same output. 
do.call(cbind, lapply(sim, head))

# You could potentially use sapply on the first level also. 
sim <- sapply(seq(10, 100, by=10), function(i) simulation(sunspots, i))

str(sim)