In NONMEM, $MIXTURE will allow you to estimate mixture models, where individuals are classified in to two or more populations with a certain probability. It is straightforward to simulate from models like these in your mrgsolve model code.

1 Two Populations

Let’s imagine there were two populations in the mixture model, with the second having smaller clearance than the first. In this example, we will develop some code for a simple model and then extend it to implement the mixture model component.

A simple model might be:

library(mrgsolve)
library(dplyr)

code <- '
$PARAM TVCL = 1, V = 30, KA=1.2, POP = 1, THETA1 = 0.5

$PKMODEL cmt="GUT CENT", depot=TRUE

$MAIN
double CL = TVCL;

if(POP==2) CL = TVCL * THETA1;
'

In this model, we created a parameter for the population indicator (POP) and if POP is 2 then clearance is lower than it would be otherwise.

Compile this model and run

mod <- mcode_cache("A", code) %>% update(end=72, delta=0.1)

idata <- data_frame(POP=c(1,2))

e <- ev(amt=100)

mod %>% mrgsim(idata=idata,events=e) %>% plot

The profile in pink was for POP==2 or the lower clearance profile and blue was for POP==1.

1.1 Modify the model to simulate a population mixture

In the get-started example model, we hard-coded POP as a parameter and we had to supply the value of POP in the input data set (in this case, it was via idata).

For the mixture model, we want POP to be simulated and we want the simulated value to be 1 with a probability of, say, 0.8 and 2 with a probability of 0.2.

To make this happen, we need to simulate a binary variate for each individual. Random numbers are easy to simulate with mrgsolve when you use $PLUGIN.

code <- '
$PLUGIN Rcpp

$MAIN
if(NEWIND <=1) {
  int POP = 1 + R::rbinom(1,0.2);
}

$CAPTURE POP
'

mod <- mcode_cache("B", code)

Here, we invoked the Rcpp plugin that allows us to call R::binom(1,0.8). R::binom is just like the regular R version, but it only draws one variate (n=1).

Let’s test it out

set.seed(222)
out <- mrgsim(mod, nid=10000, end=-1)

head(out)
.   ID time POP
. 1  1    0   2
. 2  2    0   1
. 3  3    0   1
. 4  4    0   1
. 5  5    0   2
. 6  6    0   2

Here, we’ve got 20% of the people in the population with POP of 2:

mean(out$POP==2)
. [1] 0.1973

Now, let’s modify the model again to incorporate our random POP calculation with the PK model. I have also included a home-brewed ETA using R::rnorm as another example and to make the summary a little more interesting.

code <- '
$PLUGIN Rcpp

$PARAM TVCL = 1, V = 30, KA=1.2, THETA1 = 0.5

$PKMODEL cmt="GUT CENT", depot=TRUE

$MAIN
if(NEWIND <=1) {
  int POP = 1 + R::rbinom(1,0.2);
  double myETA = R::rnorm(0,sqrt(0.09));
}

double CL = TVCL;

if(POP==2) CL = TVCL * THETA1;

double CLi = CL*exp(myETA);

$CAPTURE POP CL CLi
'

mod <- mcode_cache("C", code)

And simulate again

set.seed(444)
out <- mrgsim(mod,nid=10000, end=72, events=e,obsonly=TRUE)
head(out)
.   ID time        GUT     CENT POP CL       CLi
. 1  1    0  0.0000000  0.00000   1  1 0.7642753
. 2  1    1 30.1194212 68.50511   1  1 0.7642753
. 3  1    2  9.0717953 86.89259   1  1 0.7642753
. 4  1    3  2.7323722 90.25855   1  1 0.7642753
. 5  1    4  0.8229747 89.17134   1  1 0.7642753
. 6  1    5  0.2478752 86.81173   1  1 0.7642753
mean(out$CL==0.5)
. [1] 0.1977
out %>% 
  filter(time==0) %>%
  group_by(POP) %>% 
  summarise(N=n(), Median = median(CLi))
. # A tibble: 2 × 3
.     POP     N Median
.   <dbl> <int>  <dbl>
. 1     1  8023  1.00 
. 2     2  1977  0.499

2 Three (or more) Populations

There are probably several ways to simulate three populations. Here is one way. We’ll drop the PK model for now and focus on generating POP.

code <- '
$PARAM p1 = 0.33, p2 = 0.6

$PLUGIN Rcpp

$MAIN
if(NEWIND <=1) {
  double mixv = R::runif(0,1);
  int POP = 1;
  if(mixv > p1) POP = 2;
  if(mixv > (p1+p2)) POP = 3;
}

$CAPTURE POP mixv
'

Here’s what we did

Now, let’s compile and test it out

mod <- mcode_cache("D", code)

set.seed(333)
out <- mrgsim(mod, nid=10000, end=-1)
head(out)
.   ID time POP       mixv
. 1  1    0   2 0.46700066
. 2  2    0   1 0.08459815
. 3  3    0   3 0.97348527
. 4  4    0   2 0.57130558
. 5  5    0   1 0.02011937
. 6  6    0   2 0.72355739

And check that the population is properly configured

out %>% as_tibble() %>% count(POP) %>% mutate(p = n/nrow(out))
. # A tibble: 3 × 3
.     POP     n      p
.   <dbl> <int>  <dbl>
. 1     1  3225 0.322 
. 2     2  6053 0.605 
. 3     3   722 0.0722

And we get back the 33% in population 1, 60% in population 2, and the remaining 7% in population 3.

As a final note: remember to call set.seed() prior to simulating anything random with mrgsovle in order for the results to be reproducible.

mrgsolve: mrgsolve.github.io | metrum research group: metrumrg.com