# How to randomly generate random numbers in one of two intervals

I am trying to generate random numbers with R, uniformly from one of two different intervals. I want the numbers to be generated, for example, in the intervals [-0.8,-0.4] or [0.3,0.9]. I am trying to play by using more runif together, but I have problems when the intervals are not of the same size. Can you please help?

• I don't understand what you want to do with the two intervals. Are these intervals known and you only want to generate points between the intervals? Using your example that would mean picking random numbers in the interval (-0,4, 0.3). – Michael Chernick Mar 30 '17 at 17:23
• @MichaelChernick No, sorry, what I need to generate is random numbers which are either in the first OR in the second interval. Hope this helps! – Enrico Mar 30 '17 at 17:26
• The probabilities are uniformly distributed, if this is what you asking about. – Enrico Mar 30 '17 at 17:33
• I have edited your question. I think it makes what you want to do clear. But I don't understand what "play by using more runif together" means. – Michael Chernick Mar 30 '17 at 17:36
• Since this has been described as a mixture distribution, the general solutions given at stats.stackexchange.com/questions/70855 are applicable. Additional R solutions appear in stats.stackexchange.com/questions/64037 and stats.stackexchange.com/a/122816/919. – whuber Mar 30 '17 at 22:41

From what I understand, you want to to simulate data uniformly from the set $[a,b]\cup [c,d]$ with $a< b < c < d$.

There are several ways to do this, but this will do the trick.

 y <- runif(1, 0, b-a+d-c)
if( y < (b-a) ){
x <- a + y
}else{
x <- c + y - (b-a)
}


Here is a quick example using the numbers from your original post.

• +1 Note that one way to describe this solution is that you are inverting the CDF. – whuber Mar 30 '17 at 20:16

Consider using rejection sampling. Generate realizations of a uniform distribution over the whole interval, then reject any that are in the middle part you're not interested in.

If this is too inefficient (because the rejection region is not small compared to the whole interval), you could do it in two stages, a weighted Bernoulli coin flip (based on the relative size of the regions) to choose which region you sample from, then uniform inside that.

This might look like (for your example, I hard-coded in the boundaries which is not great practice)

n_samp <- 1e5

# Rejection sampling
sample_a <- runif(n_samp, -0.8, 0.9)
accepted_sample_a <- sample_a[!(sample_a > -0.4 & sample_a < 0.3)]

hist(accepted_sample_a)

cat("Sampling efficiency:", length(accepted_sample_a)/length(sample_a))

# Bernoulli-uniform
weight <- (-0.4 - -0.8)/((-0.4 - -0.8) + (0.9 - 0.3))
which_region <- rbinom(n_samp, 1, prob = weight)
sample_b <- runif(n_samp, ifelse(which_region, -0.8, 0.3), ifelse(which_region, -0.4, 0.9))

hist(sample_b)

• I'd go with the second. An alternative that requires only one uniform per sample may be to stick all the intervals end-to-end, pick uniformly from 0 to the total length, then map back to the corresponding interval. – Chris Haug Mar 30 '17 at 17:47

You can also use the following for each range, then join the ranges together. This method avoids rejection of any samples.

n <- 1e5 # set the desired number of samples
a <- (-0.8); b <- (-0.4); c <- 0.3; d <- 0.9 # set ranges

p <- (b-a)/(b-a+c-d)
n1 <- binom(1,n,p)
n2 <- n-n1

low <- runif(n1)*(b-a)+a # create lower random number sample
high <- runif(n2)*(d-c)+c # create upper random sample
rand.unif <- paste(low,high) # join samples together

rand.unif <- sample(rand.unif) # randomize sample after generation.


Edit: n1 and n2 calculations added to correct probability of each selection, with n set to cover the entire dataset. The sample size is based upon the binomial distribution, thanks to an excellent suggestion in the comments.

If time-series is important, you can reorder the vector with rand.unif <- sample(rand.unif) to give you a random time series with a fixed number of samples of each distribution.

• This is incorrect in two ways: (1) it does not select the two ranges with the correct probabilities and (2) it always returns exactly the same number of results in each range. Some might argue that it is incorrect in a third way: the sequence of results is not random, either: all values in one range precede all values in the other. – whuber Mar 30 '17 at 20:17
• I'm afraid none of your objections work. First, the ranges of these two intervals are different. Second, it's incorrect to force your samples to have specified counts within the specified intervals (even though for large $n$ the difference may seem subtle). Third, although you are correct about some applications being immune to reordering of the values, that's not usually what's meant by "generate random numbers": people want a sequence of iid values. – whuber Mar 30 '17 at 20:33
• Thank you. Maybe I should elaborate on why this algorithm is so wrong. When we generate random numbers, sooner or later we compute statistics from them. We wish to guarantee that those statistics will have the expected distribution. Suppose, for instance, someone were to count the number of positive values you generate. Because of how you do it, there would be no variation in the result! That is strong proof of non-randomness in your algorithm. This problem can be fixed by first drawing $n_1$ from a Binomial distribution with parameters $n$ and $p=(b-a)/(b-a+c-d)$ and setting $n_2=n-n_1$. – whuber Mar 30 '17 at 21:08
• @whuber, thank you for the excellent constructive feedback. – Tavrock Mar 30 '17 at 22:17
• You're welcome. You might find it much more efficient to use a calculation like binom(1,n,p) (which computes a single random number) rather than sum(binom(n,1,p)) (which sums $n$ random numbers), though! – whuber Mar 30 '17 at 22:19