rmulti
The current implementation of rmulti() is experimental.
It has six arguments:
nthe number of samples requireddistA named vector of the distributions of each variableparamsA list of lists of the arguments to pass to each distribution functionrthe correlations among the variables (can be a single number, vars*vars matrix, vars*vars vector, or a vars*(vars-1)/2 vector)empiricallogical. If true, params specify the sample parameters, not the population parametersas.matrixlogical. If true, returns a matrix
By default, it returns a data frame with 100 rows of two normally distributed values with means of 0, standard deviations of 1, and a correlation of 0.
dat <- rmulti()
get_params(dat)
#> n var A B mean sd
#> 1 100 A 1.00 -0.06 0.05 0.99
#> 2 100 B -0.06 1.00 0.05 0.93The n, r, empirical and
as.matrix arguments work the same as the do for
rnorm_multi().
dat <- rmulti(n = 200, r = 0.5,
empirical = TRUE,
as.matrix = FALSE)
get_params(dat)
#> n var A B mean sd
#> 1 200 A 1.0 0.5 0 1
#> 2 200 B 0.5 1.0 0 1Distributions and parameters
You can set the distribution for each variable with the
dist argument. The options are any distribution function in
the {stats} package, such as “norm”, “pois”, “binom”, and “unif”, plus
the “truncnorm” distribution from the {truncnorm} package and the
“likert” distribution from {faux}.
Set the params argument to a named list with a vector of
named arguments for the random generation function for each
distribution. For example, use ?rpois to find out what
arguments the rpois() function needs to simulate values
from a poisson distribution. Don’t set n in
params; it will be set by the nin the
rmulti() function.
dat <- rmulti(n = 1000,
dist = c(uniform_var = "unif",
poisson_var = "pois"),
params = list(uniform_var = c(min = 0, max = 100),
poisson_var = c(lambda = 3)),
r = 0.5)
get_params(dat)
#> n var uniform_var poisson_var mean sd
#> 1 1000 uniform_var 1.00 0.49 50.62 28.67
#> 2 1000 poisson_var 0.49 1.00 2.98 1.75
You can also simulate more than two variables. Set the correlations using the upper right triangle or a correlation matrix (e.g., from a pilot dataset you’re trying to simulate).
dat <- rmulti(
n = 1000,
dist = c(N = "norm",
T = "truncnorm",
L = "likert"),
params = list(
N = list(mean = 100, sd = 10),
T = list(a = 1, b = 7, mean = 3.5, sd = 2.1),
L = list(prob = c(`much less` = .10,
`less` = .20,
`equal` = .35,
`more` = .25,
`much more` = .10))
),
r = c(-0.5, -0.6, 0.7)
)
# convert likert-scale variable to integer
dat$L <- as.integer(dat$L)
get_params(dat)
#> n var N T L mean sd
#> 1 1000 N 1.00 -0.47 -0.57 100.32 9.70
#> 2 1000 T -0.47 1.00 0.67 3.80 1.43
#> 3 1000 L -0.57 0.67 1.00 3.08 1.08
Impossible correlations
Not all correlations are possible for a given pair of distributions. At the extreme, it’s obvious that a normal distribution and a uniform distribution can’t be correlated 1.0, because they wouldn’t be different distributions then.
If you ask rmulti() for a correlation that is too high
or low, you will get a message telling you the maximum and minimum
correlations that can be generated.
dat <- rmulti(
dist = c(A = "binom", B = "pois", C = "norm"),
params = list(A = list(size = 1, prob = 0.5),
B = list(lambda = 3),
C = list(mean = 100, sd = 10)),
r = c(0.8, 0.9, 0.5)
)
#> Error in rmulti(dist = c(A = "binom", B = "pois", C = "norm"), params = list(A = list(size = 1, : Some of the correlations are not possible:
#> * A&B (-0.776 to 0.776)
#> * A&C (-0.798 to 0.798)