What Happens When Prob Argument in Sample Sums to Less/Greater Than 1

What happens when prob argument in sample sums to less/greater than 1?

Good question. The docs are unclear on this, but the question can be answered by reviewing the source code.

If you look at the R code, sample always calls another R function, sample.int If you pass in a single number x to sample, it will use sample.int to create a vector of integers less than or equal to that number, whereas if x is a vector, it uses sample.int to generate a sample of integers less than or equal to length(x), then uses that to subset x.

Now, if you examine the function sample.int, it looks like this:

function (n, size = n, replace = FALSE, prob = NULL, useHash = (!replace && 
    is.null(prob) && size <= n/2 && n > 1e+07)) 
{
    if (useHash) 
        .Internal(sample2(n, size))
    else .Internal(sample(n, size, replace, prob))
}

The .Internal means any sampling is done by calling compiled code written in C: in this case, it's the function do_sample, defined here in src/main/random.c.

If you look at this C code, do_sample checks whether it has been passed a prob vector. If not, it samples on the assumption of equal weights. If prob exists, the function ensures that it is numeric and not NA. If prob passes these checks, a pointer to the underlying array of doubles is generated and passed to another function in random.c called FixUpProbs, defined here.

This function examines each member of prob and throws an error if any elements of prob are not positive finite doubles. It then normalises the numbers by dividing each by the sum of all. There is therefore no preference at all for prob summing to 1 inherent in the code. That is, even if prob sums to 1 in your input, the function will still calculate the sum and divide each number by it.

Therefore, the parameter is poorly named. It should be "weights", as others here have pointed out. To be fair, the docs only say that prob should be a vector of weights, not absolute probabilities.

So the behaviour of the prob parameter from my reading of the code should be:

1. prob can be absent altogether, in which case sampling defaults to equal weights.
2. If any of prob's numbers are less than zero, or are infinite, or NA, the function will throw.
3. An error should be thrown if any of the prob values are non-numeric, as they will be interpreted as NA in the SEXP passed to the C code.
4. prob must have the same length as x or the C code throws
5. You can pass a zero probability as one or more elements of prob if you have specified replace=T, as long as you have at least one non-zero probability.
6. If you specify replace=F, the number of samples you request must be less than or equal to the number of non-zero elements in prob. Essentially, FixUpProbs will throw if you ask it to sample with a zero probability.
7. A valid prob vector will be normalised to sum to 1 and used as sampling weights.

As an interesting side effect of this behaviour, this allows you to use odds instead of probabilities if you are choosing between 2 alternatives by setting probs = c(1, odds)

How should I specify argument prob when using sample() for resampling?

Overthinking is devil.

You want to resample these samples, following the original distribution or an empirical distribution. Think about how an empirical CDF is obtained:

plot(sort(x), 1:length(x)/length(x))

In other words, the empirical PDF is just

plot(sort(x), rep(1/length(x), length(x)))

So, we want prob = rep(1/length(x), length(x)) or simply, prob = rep(1, length(x)) as sample normalizes prob internally. Or, just leave it unspecified as equal probability is default.

How does prob argument in rbinom work when prob is a vector?

The vector is recycled over the 17 generated values:

> rbinom(17, 1, c(0,.999))
 [1] 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

Often R will generate a warning if you try recycling two vectors that aren't don't fit into each other:

> (1:10) + (1:3)
 [1]  2  4  6  5  7  9  8 10 12 11
Warning message:
In (1:10) + (1:3) :
  longer object length is not a multiple of shorter object length

but not in this case.

R_Sample with probabilities

sample(...) takes a random sample with probabilities given in prob=..., so you will not get exactly that proportion every time. On the other hand, the proportions get closer to those specified in prob as n increases:

f <- function(n)sample(1:4,n,replace=T,prob=(1:4)/10)
samples <- lapply(10^(2:6),f)
t(sapply(samples,function(x)c(n=length(x),table(x)/length(x))))
#          n        1        2        3        4
# [1,] 1e+02 0.090000 0.220000 0.260000 0.430000
# [2,] 1e+03 0.076000 0.191000 0.309000 0.424000
# [3,] 1e+04 0.095300 0.200200 0.310100 0.394400
# [4,] 1e+05 0.099720 0.199800 0.302250 0.398230
# [5,] 1e+06 0.099661 0.199995 0.300223 0.400121

If you need a random sample with exactly those proportions, use rep(...) and randomize the order.

g <- function(n) rep(1:4,n*(1:4)/10)[sample(1:n,n)]
samples <- lapply(10^(2:6),g)
t(sapply(samples,function(x)c(n=length(x),table(x)/length(x))))
#          n   1   2   3   4
# [1,] 1e+02 0.1 0.2 0.3 0.4
# [2,] 1e+03 0.1 0.2 0.3 0.4
# [3,] 1e+04 0.1 0.2 0.3 0.4
# [4,] 1e+05 0.1 0.2 0.3 0.4
# [5,] 1e+06 0.1 0.2 0.3 0.4

Sample 'randomly' but ensure final sample representative of population?

How about creating copies of all 14 image indices rep(1:14, 4) and then shuffle that array: sample(rep(1:14, 4)).

sample and rbinom functions in R

From the documentation for rbinom:

The numerical arguments other than n are recycled to the length of the result.

This means that in your example the prob vector you pass in will be recycled until it reaches the required length (presumably 5). So the vector which will be used is:

c(0.9, 0.2, 0.3, 0.9, 0.2)

As for the sample function, as @thelatemail pointed out the probabilities do not have to sum to 1. It appears that the prob vector gets normalized to 1 internally.

Understanding code about inverse transform sampling R

It looks like you need to do some basic research on R and programming in general. Here are short answers to your simple questions, but please read on afterward for some broader advice.

• Where is the 1 value when returned? Wherever it is assigned. Here, namely in samples[i] for whichever i that branch is reached.
• Where is this state allocated? In the line for(state in 2:length(p.vec))
• Why is this line names(p.vec)<-1:4 for? Good question. names()<- just assigns names to an object, and I'm not sure why in your context it's useful to have names that are equal to the vector indices, though I could imagine it to be so in some contexts.
• What seq_len means? seq_len(x) creates an integer vector with all the numbers from 1 to x inclusive. See help("seq_len")
• Why samples[i] is not used anymore in the code? Because it's only useful in the for loop.

All this points to a bigger problem though: You don't understand the basics of R. We all started out there, but it means you need to read some basic info and work through some basic tutorials. RStudio provides some resources for learning here.

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