Store values from a loop - r

I am simulating dice throws, and would like to save the output in a single object, but cannot find a way to do so. I tried looking here, here, and here, but they do not seem to answer my question.
Here is my attempt to assign the result of a 20 x 3 trial to an object:
set.seed(1)
Twenty = for(i in 1:20){
trials = sample.int(6, 3, replace = TRUE)
print(trials)
i = i+1
}
print(Twenty)
What I do not understand is why I cannot recall the function after it is run?
I also tried using return instead of print in the function:
Twenty = for(i in 1:20){
trials = sample.int(6, 3, replace = TRUE)
return(trials)
i = i+1
}
print(Twenty)
or creating an empty matrix first:
mat = matrix(0, nrow = 20, ncol = 3)
mat
for(i in 1:20){
mat[i] = sample.int(6, 3, replace = TRUE)
print(mat)
i = i+1
}
but they seem to be worse (as I do not even get to see the trials).
Thanks for any hints.

There are several things wrong with your attempts:
1) A loop is not a function nor an object in R, so it doesn't make sense to assign a loop to a variable
2) When you have a loop for(i in 1:20), the loop will increment i so it doesn't make sense to add i = i + 1.
Your last attempt implemented correctly would look like this:
mat <- matrix(0, nrow = 20, ncol = 3)
for(i in 1:20){
mat[i, ] = sample.int(6, 3, replace = TRUE)
}
print(mat)
I personally would simply do
matrix(sample.int(6, 20 * 3, replace = TRUE), nrow = 20)
(since all draws are independent and with replacement, it doesn't matter if you make 3 draws 20 times or simply 60 draws)

Usually, in most programming languages one does not assign objects to for loops as they are not formally function objects. One uses loops to interact iteratively on existing objects. However, R maintains the apply family that saves iterative outputs to objects in same length as inputs.
Consider lapply (list apply) for list output or sapply (simplified apply) for matrix output:
# LIST OUTPUT
Twenty <- lapply(1:20, function(x) sample.int(6, 3, replace = TRUE))
# MATRIX OUTPUT
Twenty <- sapply(1:20, function(x) sample.int(6, 3, replace = TRUE))
And to see your trials, simply print out the object
print(Twenty)
But since you never use the iterator variable, x, consider replicate (wrapper to sapply which by one argument can output a matrix or a list) that receives size and expression (no sequence inputs or functions) arguments:
# MATRIX OUTPUT (DEFAULT)
Twenty <- replicate(20, sample.int(6, 3, replace = TRUE))
# LIST OUTPUT
Twenty <- replicate(20, sample.int(6, 3, replace = TRUE), simplify = FALSE)

You can use list:
Twenty=list()
for(i in 1:20){
Twenty[[i]] = sample.int(6, 3, replace = TRUE)
}

Related

Sample draw in sapply without replacement

How does one draw a sample within a sapply function without replacement? Consider the following MWE below. What I am trying to achieve is for a number in idDRAW to receive a letter from chrSMPL (given the sample size of chrSMPL). Whether a number from idDRAW receives a letter is determined by the respective probabilities, risk factors and categories. This is calculated in the sapply function and stored in tmp.
The issue is sample replacement, leading to a number being named with a letter more than once. How can one avoid replacement whilst still using the sapply function? I have tried to adjust the code from this question (Alternative for sample) to suit my needs, but no luck. Thanks in advance.
set.seed(3)
chr<- LETTERS[1:8]
chrSMPL<- sample(chr, size = 30, replace = TRUE)
idDRAW<- sort(sample(1:100, size = 70, replace = FALSE))
p_mat<- matrix(runif(16, min = 0, max = 0.15), ncol = 2); rownames(p_mat) <- chr ## probability matrix
r_mat <- matrix(rep(c(0.8, 1.2), each = length(chr)), ncol = 2); rownames(r_mat) <- chr ## risk factor matrix
r_cat<- sample(1:2, 70, replace = TRUE) ## risk categories
# find number from `idDRAW` to be named a letter:
Out<- sapply(chrSMPL, function(x){
tmp<- p_mat[x, 1] * r_mat[x, r_cat]
sample(idDRAW, 1, prob = tmp)
})
> sort(Out)[1:3]
G B B
5 5 5
I managed with an alternative solution using a for loop as seen below. If anyone can offer suggestions on how the desired result can be achieved without using a for loop it would be greatly appreciated.
set.seed(3)
Out <- c()
for(i in 1:length(chrSMPL)){
tmp <- p_mat[chrSMPL[i], 1] * r_mat[chrSMPL[i], r_cat]
Out <- c(Out, sample(idDRAW, 1, prob = tmp))
rm <- which(idDRAW == Out[i])
idDRAW <- idDRAW[-rm]
r_cat <- r_cat[-rm]
}
names(Out) <- chrSMPL
sort(Out)[1:3]

Using user-defined "for loop" function to construct a data frame

I am trying to generate a data frame based on a user-defined function. My problem is that in the output only the first row is being filled.
Here is an example of the function I am using:
df <- data.frame(cs=rep(c("T1","T2","T3","T4"),each=16),yr=rep(c(1:4), times = 4, each = 4))
sp.df <- data.frame(matrix(sample.int(100, size = 20*64,replace=T), ncol = 20, nrow = 64))
myfunc<-function(X, system, Title)
{
for(i in 1:4){
Col_T <- data.frame(matrix(NA, ncol = length(X), nrow = 4))
Col_T[i,] <- colSums(X[which(df$yr==i & df$cs==system),])
return (Col_T)}}
myfunc(X=sp.df, system="T1", Title="T1")
I would welcome any suggestion to resolve this issue.
Thank you very much.
There are two problems with the function:
You're overwriting Col_T with all NAs as the first statement inside the for loop.
You're returning from the function inside the for loop.
Rewrite it as follows:
myfunc <- function(X, system, Title ) {
Col_T <- data.frame(matrix(NA, ncol=length(X), nrow=4 ));
for (i in 1:4)
Col_T[i,] <- colSums(X[which(df$yr==i & df$cs==system),]);
return(Col_T);
};

Create a function that takes in a vector and returns a matrix in R

I am trying to create a function that will take in a vector k and return to me a matrix with dimensions length(distMat[1,]) by length(k). distMat is a huge matrix and indSpam is a long vector. In particular to my situation, length(distMat[1,]) is 2412. When I enter in k as a vector of length one, I get a vector of length 2412. I want to be able to enter in k as a vector of length two and get a matrix of 2412x2. I am trying to use a while loop to let it go through the length of k, but it only returns to me a vector of length 2412. What am I doing wrong?
predNeighbor = function(k, distMat, indSpam){
counter = 1
while (counter<(length(k)+1))
{
preMatrix = apply(distMat, 1, order)
orderedMatrix = t(preMatrix)
truncate = orderedMatrix[,1:k[counter]]
checking = indSpam[truncate]
checking2 = matrix(checking, ncol = k[counter])
number = apply(checking2, 1, sum)
return(number[1:length(distMat[1,])] > (k[counter]/2))
counter = counter + 1
}
}
I am trying to create a function that will take in a vector k and return to me a matrix with dimensions length(distMat[1,]) by length(k)
Here's a function that does this.
foo <- function(k, distMat) {
return(matrix(0, nrow = length(distMat[1, ]), ncol = length(k)))
}
If you have other requirements, please describe them in words.
Based on your comment, I think I understand better your goal. You have a function that returns a vector of length k and you want to save it's output as rows in a matrix. This is a pretty common task. Let's do a simple example where k starts out as 1:10, and say we want to add some noise to it with a function foo() and see how the rank changes.
In the case where the input to the function is always the same, replicate() works very well. It will automatically put everything in a matrix
k <- 1:10
noise_and_rank <- function(k) {
rank(k + runif(length(k), min = -2, max = 2))
}
results <- replicate(n = 8, expr = {noise_and_rank(k)})
In the case where you want to iterate, i.e., the output from the one go is the input for the next, a for loop is good, and we just pre-allocate a matrix with 0's, to fill in one column/row at a time
k <- 1:10
n.sim <- 8
results <- matrix(0, nrow = length(k), ncol = n.sim)
results[, 1] <- k
for(i in 2:n.sim) {
results[, i] <- noise_and_rank(results[, i - 1])
}
What your original question seems to be about is how to do the pre-allocation. If the input is always the same, using replicate() means you don't worry about it. If the input is is different each time, then pre-allocate using matrix(), you don't need to write any special function.

Fill elements of a list without looping

I am trying not to use a for loop to assign values to the elements of a list.
Here, I create an empty list, gives it a length of 20 and name each of the 20 elements.
mylist <- list()
length(mylist) <- 20
names(mylist) <- paste0("element", 1:20, sep = "")
I want each element of mylist to contain samples drawn from a pool of randomly generated numbers denoted as x:
x <- runif(100, 0, 1)
I tried the following codes, which do not get to the desired result:
mylist[[]] <- sample(x = x, size = 20, replace = TRUE) # Gives an error
mylist[[1:length(mylist)]] <- sample(x = x, size = 20, replace = TRUE) # Does not give the desired result
mylist[1:length(mylist)] <- sample(x = x, size = 20, replace = TRUE) # Gives the same undesired result as the previous line of code
mylist[] <- sample(x = x, size = 20, replace = TRUE) # Gives the same undesired result as the previous line of code
P.S. As explained above, the desired result is a list of 20 elements, which individually contains 20 numeric values. I can do it using a for loop, but I would like to become a better R user and use vectorized operations as much as possible.
Thank you for your help.
Maybe replicate is what you're looking for.
mylist <- replicate(20, sample(x = x, size = 20, replace = TRUE), simplify=FALSE)
names(mylist) <- paste0("element", 1:20, sep = "")
Note that there is no need to first create a list, replicate will do it for you.
Since you're using replace=TRUE you could also generate all 400 at once and then split them up. If you were doing this many times, this probably would be faster than replicate. For only 20 times, the speed difference won't matter hardly at all and tje code using replicate is perhaps easier to read and understand and so might be preferred for that reason.
foo <- sample(x = x, size = 20*20, replace = TRUE)
mylist <- split(foo, rep(1:20, each=20))
Alternatively, you could split them by converting to a data frame first. Not sure which would be faster.
mylist <- as.list(as.data.frame(matrix(foo, ncol=20)))

R: Sample a vector with replacement multiple times

I'd like to sample a vector x of length 7 with replacement and sample that vector 10 separate times. I've tried the something like the following but can't get the resulting 7x10 output I'm looking for. This produces a 1x7 vector but I can't figure out to get the other 9 vectors
x <- runif(7, 0, 1)
for(i in 1:10){
samp <- sample(x, size = length(x), replace = T)
}
This is a very convenient way to do this:
replicate(10,sample(x,length(x),replace = TRUE))
Since you seem to want to sample with replacement, you can just get the 7*10 samples at once (which is more efficient for large sizes):
x <- runif(7)
n <- 10
xn <- length(x)
matrix(x[sample.int(xn, xn*n, replace=TRUE)], nrow=xn)
# Or slightly shorter:
matrix(sample(x, length(x)*n, replace=TRUE), ncol=n)
The second version uses sample directly, but there are some issues with that: if x is a numeric of length 1, bad things happen. sample.int is safer.
x <- c(pi, -pi)
sample(x, 5, replace=T) # OK
x <- pi
sample(x, 5, replace=T) # OOPS, interpreted as 1:3 instead of pi...
Looks like you got a suitable answer, but here's an approach that's similar to your first attempt. The difference is that we define samp with the appropriate dimensions, and then iteratively index into that object and fill it one row at a time:
samp <- matrix(NA, ncol = 7, nrow = 10)
for(i in 1:10){
samp[i,] <- sample(x, size = length(x), replace = T)
}

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