rnorm() to generate random numbers from the normal distribution

I use rnorm() a lot, sometimes with good reason and other times when I need some numbers and I really don’t care too much about what they are. 😜

There are three arguments to rnorm(). From the Usage section of the documentation:

rnorm(n, mean = 0, sd = 1)

The n argument is the number of observations we want to generate. The mean and sd arguments show what the default values of the parameters are (note that sd is the standard deviation, not the variance). Not all r functions have defaults to the parameter arguments like this.

To get 5 random numbers from a \(Normal(0, 1)\) (aka the standard normal) distribution we can write code like:

rnorm(5)

[1]  1.7930319 -1.4595365  0.1785649 -0.3500435 -0.1665776

There are a couple things about this code and the output to discuss.

First, the code did get me 5 numbers, which is what I wanted. However, the code itself isn’t particularly clear. What I might refer to as lazy coding on my part can look pretty mysterious to someone reading my code (or to my future self reading my code). Since I used the default values for mean and sd, it’s not clear exactly what distribution I drew the numbers from.

rnorm(n = 5, mean = 0, sd = 1)

[1] -0.36778344  0.42699706 -0.73419911 -1.25398250 -0.06248306

set.seed(16)
rnorm(n = 5, mean = 0, sd = 1)

[1]  0.4764134 -0.1253800  1.0962162 -1.4442290  1.1478293

set.seed(16)
rnorm(n = 5, mean = 0, sd = 1)

[1]  0.4764134 -0.1253800  1.0962162 -1.4442290  1.1478293

rnorm(n = 5, mean = 0, sd = 2)

[1] -0.9368241 -2.0119012  0.1271254  2.0499452  1.1462840

rnorm(n = 5, mean = 50, sd = 20)

[1] 86.94364 52.23867 35.07925 83.16427 64.43441

rnorm(n = 10, mean = c(0, 5, 20), sd = 1)

 [1] -1.6630805  5.5759095 20.4727601 -0.5427317  6.1276871 18.3522024
 [7] -0.3141739  4.8173184 21.4704785 -0.8658988

rnorm(n = 10, mean = c(0, 5, 20), sd = c(1, 5, 20) )

 [1]  1.5274670 10.2708903 40.6014202  0.8401609  6.0848235  6.5494885
 [7]  0.1325985  4.6453633  1.1460906 -1.0220310

rnorm(n = c(2, 10, 10), mean = c(0, 5, 20), sd = c(1, 5, 20) )

[1]  0.2805551  7.7239168 22.6173950

Example of using the simulated numbers from rnorm()

Up to this point we’ve printed the results of each simulation. In reality we’d want to save our vectors as objects in R to use them for some further task.

For example, maybe we want to simulate two unrelated variables and then look to see how correlated they appear to be. This can be a fun exercise to demonstrate how variables can appear to be related by chance even when we know they are not, especially at small sample sizes.

Let’s generate two quantitative vectors of length 10, which I’ll name x and y, and plot the results. I’m using the defaults for mean and sd.

This particular example doesn’t show much of a pattern, but if you take this code and run it may times you’ll see some pretty surprising relationships.

x = rnorm(n = 10, mean = 0, sd = 1)
y = rnorm(n = 10, mean = 0, sd = 1)
plot(y ~ x)

runif() pulls from the uniform distribution

Pulling random numbers from other distributions is extremely similar to using rnorm(), so we’ll go through them more quickly.

I’ve started using runif() pretty regularly, especially when I want to easily generate numbers that are strictly positive but continuous. The uniform distribution is a continuous distribution, with numbers uniformly distributed between some minimum and maximum.

From Usage we can see that by default we pull random numbers between 0 and 1. The first argument, as with all of these r functions, is the number of deviates we want to randomly generate:

runif(n, min = 0, max = 1)

Let’s generate 5 numbers between 0 and 1.

runif(n = 5, min = 0, max = 1)

[1] 0.9994220 0.9432766 0.2496042 0.6482484 0.1125788

What if we want to generate 5 numbers between 50 and 100? We change the values for the parameter arguments.

runif(n = 5, min = 50, max = 100)

[1] 81.56680 66.91604 82.70027 64.29244 54.17023

Example of using the simulated numbers from runif()

I like using runif() for making explanatory variables that have realistic ranges. In multiple regression, having explanatory variables with different magnitudes affects interpretation of regression coefficients.

Let’s generate some data with the response variable (y) pulled from a standard normal distribution and then an explanatory variable with values between 1 and 2. The two variables are unrelated.

You see I’m still writing out my argument names for clarity, but you may be getting a sense how easy it would be to start cutting corners to avoid the extra typing.

set.seed(16)
y = rnorm(n = 100, mean = 0, sd = 1)
x1 = runif(n = 100, min = 1, max = 2)
head(x1)

[1] 1.957004 1.082791 1.710816 1.326998 1.995723 1.449522

Now let’s simulate a second explanatory variable with values between 200 and 300. This variable is also unrelated to the other two.

x2 = runif(n = 100, min = 200, max = 300)
head(x2)

[1] 220.0617 263.4875 209.6036 245.3125 265.1869 257.4817

We can fit a multiple regression linear model via lm(). In this case, the coefficient for the second variable, which has a larger magnitude, is smaller than the first (although neither are actually related to y). The change in y for a “1-unit increase” in x depends on the units of x.

lm(y ~ x1 + x2)

Call:
lm(formula = y ~ x1 + x2)

Coefficients:
(Intercept)           x1           x2  
   0.380887     0.104941    -0.001908  

Discrete counts with rpois()

Let’s look at one last function for generating random numbers, this time for generating discrete integers (including 0) from a Poisson distribution with rpois().

I use rpois() for generating counts for creating data to be fit with generalized linear models. This function has also helped me gain a better understanding of the shape of Poisson distributions with different means.

The Poisson distribution is a single parameter distribution. The function looks like:

rpois(n, lambda)

The single parameter argument, lambda, is the mean. It has no default setting so must always be defined by the user.

Let’s generate five values from a Poisson distribution with a mean of 2.5. Note that mean of the Poisson distribution can be any non-negative value (i.e., it doesn’t have to be an integer) even though the observed values will be discrete integers.

rpois(n = 5, lambda = 2.5)

[1] 2 1 4 1 2

Example of using the simulated numbers from rpois()

Let’s explore the Poisson distribution a little more, seeing how the distribution changes when the mean of the distribution changes. Being able to look at how the Poisson distribution changes with the mean via simulation helped me understand the distribution better, including why it so often does a poor job modeling ecological count data.

We’ll draw 100 values from a Poisson distribution with a mean of 5. We’ll name this vector y and take a look at a summary of those values.

y = rpois(100, lambda = 5)

The vector of values we simulated here fall between 1 and 11.

summary(y)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   1.00    3.00    5.00    4.83    6.00   11.00 

There is mild right-skew when we draw a histogram of the values.

hist(y)

Let’s do the same thing for a Poisson distribution with a mean of 100. The range of values is pretty narrow; there are no values even remotely close to 0.

y = rpois(100, lambda = 100)
summary(y)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  76.00   94.75  102.00  101.31  108.00  124.00 

And the distribution is now pretty symmetric.

hist(y)

An alternative to the Poisson distribution for discrete integers is the negative binomial distribution. Packages MASS has a function called rnegbin() for random number generation from the negative binomial distribution.

Using letters and LETTERS

The first argument of rep() is the vector to be repeated. One option is to write out the character vector you want to repeat. You can also get a simple character vector through the use of letters or LETTERS. These are built in constants in R. letters is the 26 lowercase letters of the Roman alphabet and LETTERS is the 26 uppercase letters.

Letters can be pulled out via the extract brackets ([). I use these built-in constants for pure convenience when I need to make a basic categorical vector and it doesn’t matter what form those categories take. I find it more straightforward to type out the word and brackets than a vector of characters (complete with all those pesky quotes and such 😆).

Here’s the first two letters.

letters[1:2]

[1] "a" "b"

And the last 17 LETTERS.

LETTERS[10:26]

 [1] "J" "K" "L" "M" "N" "O" "P" "Q" "R" "S" "T" "U" "V" "W" "X" "Y" "Z"

Repeat each element of a vector with each

There are three arguments that help us repeat the values in the vector in rep() with different patterns: each, times, and length.out. These can be used individually or in combination.

With each we repeat each unique character in the vector the defined number of times. The replication is done “elementwise”, so the repeats of each unique character are all in a row.

Let’s repeat two characters three times each. The resulting vector is 6 observations long.

This is an example of how I might make a grouping variable for simulating data to be used in a two-sample analysis.

rep(letters[1:2], each = 3)

[1] "a" "a" "a" "b" "b" "b"

Repeat a whole vector with the times argument

The times argument can be used when we want to repeat the whole vector rather than repeating it elementwise.

We’ll make a two-group variable again, but this time we’ll change the repeating pattern of the values in the variable.

rep(letters[1:2], times = 3)

[1] "a" "b" "a" "b" "a" "b"

Set the output vector length with the length.out argument

The length.out argument has rep() repeat the whole vector. However, it repeats the vector only until the defined length is reached. Using length.out is another way to get unbalanced groups.

Rather than defining the number of repeats like we did with each and times we define the length of the output vector.

Here we’ll make a two-group variable of length 5. This means the second group will have one less value than the first.

rep(letters[1:2], length.out = 5)

[1] "a" "b" "a" "b" "a"

Repeat each element a different number of times

Unlike each and length.out, we can use times with a vector of values. This allows us to repeat each element of the character vector a different number of times. This is one way to simulate unbalanced groups.

Using times with a vector repeats each element like each does. I found this a little confusing as it makes it harder to remember which argument repeats “elementwise” and which “vectorwise”. But length.out always repeats “vectorwise”, so that’s something.

Let’s repeat the first element twice and the second four times.

rep(letters[1:2], times = c(2, 4) )

[1] "a" "a" "b" "b" "b" "b"

Combining each with times

As your simulation situation get more complicated, like if you are simulating data from a blocked design or with multiple sizes of experimental units, you may need more complicated patterns for your categorical variable. The each argument can be combined with times to first repeat each value elementwise (via each) and then repeat that whole pattern (via times).

When using times this way it will only take a single value and not a vector.

Let’s repeat each value twice, 3 times.

rep(letters[1:2], each = 2, times = 3)

 [1] "a" "a" "b" "b" "a" "a" "b" "b" "a" "a" "b" "b"

Combining each with length.out

Similarly we can use each with length.out. This can lead to some imbalance.

Here we’ll repeat the two values twice each and then repeat that pattern until we hit a total final vector length of 7.

rep(letters[1:2], each = 2, length.out = 7)

[1] "a" "a" "b" "b" "a" "a" "b"

Note you can’t use length.out and times together (if you try, length.out will be given priority and times ignored).

Simulate data with no differences among two groups

Let’s start by simulating data that we would use in a simple two-sample analysis with no difference between groups. We’ll make a total of 6 observations, three in each group.

We’ll be using the tools we reviewed above but will now name the output and combine them into a data.frame. This last step isn’t always necessary, but can help you keep things organized.

First we’ll make separate vectors for the continuous and categorical data and then bind them together via data.frame().

Notice there is no need to use cbind() here, which is commonly done by R beginners (I know I did!). Instead we can use data.frame() directly.

group = rep(letters[1:2], each = 3)
response = rnorm(n = 6, mean = 0, sd = 1)
data.frame(group,
           response)

  group   response
1     a  0.4933614
2     a  0.5234101
3     a  1.2365975
4     b  0.3563153
5     b  0.5748968
6     b -0.4222890

When I make a data.frame like this I prefer to make my vectors and the data.frame simultaneously to avoid having a lot of variables cluttering up my R Environment.

I often teach/blog with all the steps clearly delineated as I think it’s easier when you are starting out, so (as always) use the method that works for you.

data.frame(group = rep(letters[1:2], each = 3),
           response = rnorm(n = 6, mean = 0, sd = 1) )

  group   response
1     a  0.4024228
2     a  0.9585800
3     a -1.8763844
4     b -0.2115171
5     b  1.4374372
6     b  0.3855285

Now let’s add another categorical variable to this dataset.

Say we are in a situation involving two factors, not one. We have a single observations for every combination of the two factors (i.e., the two factors are crossed).

The second factor, which we’ll call factor, will take on the values “C”, “D”, and “E”.

LETTERS[3:5]

[1] "C" "D" "E"

We need to repeat the values in a way that every combination of group and factor is present in the dataset exactly one time.

Remember the group factor is repeated elementwise.

rep(letters[1:2], each = 3)

[1] "a" "a" "a" "b" "b" "b"

We need to repeat the three values twice. But what argument do we use in rep() to do so?

rep(LETTERS[3:5], ?)

Does each work?

rep(LETTERS[3:5], each = 2)

[1] "C" "C" "D" "D" "E" "E"

No, if we use each then each element is repeated twice and some of the combinations of group and the new variable will be missing.

This is a job for the times or length.out arguments, so the whole vector is repeated. We can repeat the whole vector twice using times, or via length.out = 6. I decided to do the former.

In the result below we can see every combination of the two factors is present once.

data.frame(group = rep(letters[1:2], each = 3),
           factor = rep(LETTERS[3:5], times = 2),
           response = rnorm(n = 6, mean = 0, sd = 1) )

  group factor   response
1     a      C  0.4255576
2     a      D  0.2903586
3     a      E -0.3638877
4     b      C  1.9778117
5     b      D  1.0869069
6     b      E -0.5869200

Simulate data with a difference among groups

The dataset above is one with “no difference” among groups. What if we want data where the means are different between groups? Let’s make two groups of three observations where the mean of one group is 5 and the other is 10. The two groups have a shared variance (and so standard deviation) of 1.

Remembering how rnorm() works with a vector of means is key here. The function draws iteratively from each distribution.

response = rnorm(n = 6, mean = c(5, 10), sd = 1)
response

[1]  4.413753 12.484499  4.740506 10.273258  5.369074 10.024199

How do we get the group pattern correct?

group = rep(letters[1:2], ?)

We need to repeat the whole vector three times instead of elementwise.

To get the groups in the correct order we need to use times or length.out in rep(). With length.out we define the output length of the vector, which is 6. Alternatively we could use times = 3 to repeat the whole vector 3 times.

group = rep(letters[1:2], length.out = 6)
group

[1] "a" "b" "a" "b" "a" "b"

These can then be combined into a data.frame. Working out this process is another reason why sometimes we build each vector separately prior to combining them into a data.frame.

data.frame(group,
            response)

  group  response
1     a  4.413753
2     b 12.484499
3     a  4.740506
4     b 10.273258
5     a  5.369074
6     b 10.024199

Multiple quantitative variables with groups

For our last dataset we’ll have two groups, with 10 observations per group.

rep(LETTERS[3:4], each = 10)

 [1] "C" "C" "C" "C" "C" "C" "C" "C" "C" "C" "D" "D" "D" "D" "D" "D" "D" "D" "D"
[20] "D"

Let’s make a dataset that has two quantitative variables, unrelated to both each other and the groups. One variable ranges from 10 and 15 and one from 100 and 150.

How many observations should we draw from each uniform distribution?

runif(n = ?, min = 10, max = 15)

We had 2 groups with 10 observations each and 2*10 = 20. So we need to use n = 20 in runif().

Here is the dataset made in a single step.

data.frame(group = rep(LETTERS[3:4], each = 10),
           x = runif(n = 20, min = 10, max = 15),
           y = runif(n = 20, min = 100, max = 150))

   group        x        y
1      C 13.20331 126.7004
2      C 13.91440 137.0772
3      C 12.72031 134.8689
4      C 14.27637 122.7582
5      C 11.72933 118.0189
6      C 14.59640 108.4544
7      C 12.81629 142.1792
8      C 13.45864 104.0355
9      C 13.94198 107.0199
10     C 12.15106 145.4622
11     D 12.01762 117.3381
12     D 13.66006 121.2900
13     D 14.78914 145.2424
14     D 11.70157 120.1363
15     D 13.25180 139.8288
16     D 10.76914 106.6282
17     D 13.97263 147.6383
18     D 14.91621 112.6612
19     D 13.53270 104.8965
20     D 14.40677 119.5231

What happens if we get the number wrong? If we’re lucky we get an error that will help us troubleshoot the problem.

data.frame(group = rep(LETTERS[3:4], each = 10),
           x = runif(n = 15, min = 10, max = 15),
           y = runif(n = 15, min = 100, max = 150))

Error in data.frame(group = rep(LETTERS[3:4], each = 10), x = runif(n =
15, : arguments imply differing number of rows: 20, 15

But if we get things wrong and the number we use happens to go into the number we need evenly, R will recycle the vector to the end of the data.frame().

This is a hard mistake to catch. If you look carefully through the output below you can see that the continuous variables start to repeat on line 10 because I used n = 10 instead of n = 20.

data.frame(group = rep(LETTERS[3:4], each = 10),
           x = runif(n = 10, min = 10, max = 15),
           y = runif(n = 10, min = 100, max = 150))

   group        x        y
1      C 10.82006 112.9378
2      C 12.42877 114.4226
3      C 11.57143 123.7625
4      C 14.28623 139.3863
5      C 11.46303 130.4657
6      C 14.11453 138.4861
7      C 13.78674 102.5954
8      C 11.17580 116.3792
9      C 13.75111 116.8000
10     C 11.35357 147.0055
11     D 10.82006 112.9378
12     D 12.42877 114.4226
13     D 11.57143 123.7625
14     D 14.28623 139.3863
15     D 11.46303 130.4657
16     D 14.11453 138.4861
17     D 13.78674 102.5954
18     D 11.17580 116.3792
19     D 13.75111 116.8000
20     D 11.35357 147.0055

Simple example of replicate()

Let’s say we want to simulate some values from a normal distribution, which we can do using the rnorm() function as above. But now instead of drawing some number of values from a distribution one time we want to do it many times. This could be something we’d do when demonstrating the central limit theorem, for example.

Doing the random number generation many times is where replicate() comes in. It allows us to run the function in expr exactly n times.

Here I’ll generate 5 values from a standard normal distribution three times. Notice the addition of simplify = FALSE to get a list as output.

The output below is a list of three vectors. Each vector is from a unique run of the function, so contains five random numbers drawn from the normal distribution with a mean of 0 and standard deviation of 1.

set.seed(16)
replicate(n = 3, 
          expr = rnorm(n = 5, mean = 0, sd = 1), 
          simplify = FALSE )

[[1]]
[1]  0.4764134 -0.1253800  1.0962162 -1.4442290  1.1478293

[[2]]
[1] -0.46841204 -1.00595059  0.06356268  1.02497260  0.57314202

[[3]]
[1]  1.8471821  0.1119334 -0.7460373  1.6582137  0.7217206

Note if I don’t use simplify = FALSE I will get a matrix of values instead of a list. Each column in the matrix is the output from one run of the function.

In this case there will be three columns in the output, one for each run, and 5 rows. This can be a useful output type for some simulations. I focus on list output throughout the rest of this post only because that’s what I have been using recently for simulations.

set.seed(16)
replicate(n = 3, 
          expr = rnorm(n = 5, mean = 0, sd = 1) )

           [,1]        [,2]       [,3]
[1,]  0.4764134 -0.46841204  1.8471821
[2,] -0.1253800 -1.00595059  0.1119334
[3,]  1.0962162  0.06356268 -0.7460373
[4,] -1.4442290  1.02497260  1.6582137
[5,]  1.1478293  0.57314202  0.7217206

An equivalent for() loop example

A for() loop can be used in place of replicate() for simulations. With time and practice I’ve found replicate() to be much more convenient in terms of writing the code. However, in my experience some folks find for() loops intuitive when they are starting out in R. I think it’s because for() loops are more explicit on the looping process: the user can see the values that i takes and the output for each i iteration is saved into the output object because the code is written out explicitly.

In my example I’ll save the output of each iteration of the loop into a list called list1. I initialize this as an empty list prior to starting the loop. To match what I did with replicate() I do three iterations of the loop (i in 1:3), drawing 5 values via rnorm() each time.

The result is identical to my replicate() code above. It took a little more code to do it but the process is very clear since it is explicitly written out.

set.seed(16)
list1 = list() # Make an empty list to save output in
for (i in 1:3) { # Indicate number of iterations with "i"
    list1[[i]] = rnorm(n = 5, mean = 0, sd = 1) # Save output in list for each iteration
}
list1

[[1]]
[1]  0.4764134 -0.1253800  1.0962162 -1.4442290  1.1478293

[[2]]
[1] -0.46841204 -1.00595059  0.06356268  1.02497260  0.57314202

[[3]]
[1]  1.8471821  0.1119334 -0.7460373  1.6582137  0.7217206

Using replicate() to repeatedly make a dataset

Earlier we were making datasets with random numbers and some grouping variables. Our code looked like:

data.frame(group = rep(letters[1:2], each = 3),
           response = rnorm(n = 6, mean = 0, sd = 1) )

  group   response
1     a -1.6630805
2     a  0.5759095
3     a  0.4727601
4     b -0.5427317
5     b  1.1276871
6     b -1.6477976

We could put this process as the expr argument in replicate() to get many simulated datasets. I would do something like this if I wanted to compare the long-run performance of two different statistical tools using the exact same random datasets.

I’ll replicate things 3 times again to easily see the output. I still use simplify = FALSE to get things into a list.

simlist = replicate(n = 3, 
          expr = data.frame(group = rep(letters[1:2], each = 3),
                            response = rnorm(n = 6, mean = 0, sd = 1) ),
          simplify = FALSE)

We can see this result is a list of three data.frames.

str(simlist)

List of 3
 $ :'data.frame':   6 obs. of  2 variables:
  ..$ group   : chr [1:6] "a" "a" "a" "b" ...
  ..$ response: num [1:6] -0.314 -0.183 1.47 -0.866 1.527 ...
 $ :'data.frame':   6 obs. of  2 variables:
  ..$ group   : chr [1:6] "a" "a" "a" "b" ...
  ..$ response: num [1:6] 1.03 0.84 0.217 -0.673 0.133 ...
 $ :'data.frame':   6 obs. of  2 variables:
  ..$ group   : chr [1:6] "a" "a" "a" "b" ...
  ..$ response: num [1:6] -0.943 -1.022 0.281 0.545 0.131 ...

Here is the first one.

simlist[[1]]

  group   response
1     a -0.3141739
2     a -0.1826816
3     a  1.4704785
4     b -0.8658988
5     b  1.5274670
6     b  1.0541781