Binary data example using data from Remiro-Azocar et al. (2020)

Introduction

This is a simpler version of the Binary Data Example with Simulated Data example. See this document for more details and exposition.

General problem

Consider one AB trial, for which the company has IPD, and one AC trial, for which only published aggregate data are available. We wish to estimate a comparison of the effects of treatments B and C on an appropriate scale in some target population P, denoted by the parameter $d_{BC(P)}$. We make use of bracketed subscripts to denote a specific population. Within the AB population there are parameters $\mu_{A( AB)}$, $\mu_{B(AB)}$ and $\mu_{C(AB)}$ representing the expected outcome on each treatment (including parameters for treatments not studied in the AB trial, e.g. treatment C). The AB trial provides estimators $\bar{Y}_{A(AB)}$ and $\bar{Y}_{B(AB)}$ of $\mu_{A( AB)}$, $\mu_{B(AB)}$, respectively, which are the summary outcomes. It is the same situation for the AC trial.

For a suitable scale, for example a logit, or risk difference, we form estimators $\Delta_{AB(AB)}$ and $\Delta_{AC(AC)}$ of the trial level (or marginal) relative treatment effects.

$$\Delta_{AB(AB)} = g(\bar{Y}_{B{(AB)}}) - g(\bar{Y}_{A{(AB)}})$$

Example analysis

First, let us load necessary packages.

library(boot)      # non-parametric bootstrap in MAIC and ML G-computation
library(copula)    # simulating BC covariates from Gaussian copula
library(rstanarm)  # fit outcome regression, draw outcomes in Bayesian G-computation
library(outstandR)

Data

Next, we load the data to use in the analysis. The data comes from a simulation study in Remiro‐Azócar A, Heath A, Baio G (2020). We consider binary outcomes using the log-odds ratio as the measure of effect. The binary outcome may be response to treatment or the occurrence of an adverse event. For trials AC and BC, outcome $y_n$ for subject $n$ is simulated from a Bernoulli distribution with probabilities of success generated from logistic regression.

For the BC trial, the individual-level covariates and outcomes are aggregated to obtain summaries. The continuous covariates are summarized as means and standard deviations, which would be available to the analyst in the published study in a table of baseline characteristics in the RCT publication. The binary outcomes are summarized in an overall event table. Typically, the published study only provides aggregate information to the analyst.

set.seed(555)

ipd_trial <- read.csv(here::here("raw-data", "AC_IPD.csv"))  # AC patient-level data
ald_trial <- read.csv(here::here("raw-data", "BC_ALD.csv"))  # BC aggregate-level data

This general format of data sets consist of the following.

ipd_trial: Individual patient data

• X*: patient measurements
• trt: treatment ID (integer)
• y: (logical) indicator of whether event was observed

ald_trial: Aggregate-level data

• mean.X*: mean patient measurement
• sd.X*: standard deviation of patient measurement
• y.*.sum: total number of events
• y.*.bar: proportion of events
• N.*: total number of individuals

Note that the wildcard * here is usually an integer from 1 or the trial identifier B, C.

Let us label the treatment levels

ipd_trial$trt <- factor(ipd_trial$trt, labels = c("C", "A"))

Our data look like the following.

head(ipd_trial)
#>            X1        X2          X3          X4 trt y
#> 1  0.43734111 0.6747901  0.93001035  0.09165363   A 0
#> 2  0.05643081 0.5987971  0.03557646  0.59954129   A 1
#> 3 -0.08048882 0.6843784  0.93147222 -0.11419716   A 0
#> 4 -0.38580926 0.5716644 -0.32252212  0.02551808   A 0
#> 5  1.00755116 0.8220826  0.92735892  0.84414221   A 1
#> 6  0.19443956 0.2031329  0.34990179  0.15633009   A 0

There are 4 correlated continuous covariates generated per subject, simulated from a multivariate normal distribution.

ald_trial
#>     mean.X1   mean.X2   mean.X3   mean.X4     sd.X1     sd.X2     sd.X3
#> 1 0.5908996 0.6414179 0.5856529 0.6023671 0.3863145 0.4033615 0.4076097
#>      sd.X4 y.B.sum y.B.bar N.B y.C.sum y.C.bar N.C
#> 1 0.395132     182   0.455 400     149   0.745 200

In this case, we have 4 covariate mean and standard deviation values; and the event total, average and sample size for each treatment B, and C.

Model fitting in R

The outstandR package has been written to be easy to use and essential consists of a single function, outstandR(). This can be used to run all of the different types of model, which we will call strategies. The first two arguments of outstandR() are the individual and aggregate-level data, respectively.

A strategy argument of outstandR takes functions called strategy_*(), where the wildcard * is replaced by the name of the particular method required, e.g. strategy_maic() for MAIC. Each specific example is provided below.

MAIC

Using the individual level data for AC firstly we perform non-parametric bootstrap of the maic.boot function with R = 1000 replicates. This function fits treatment coefficient for the marginal effect for A vs C. The returned value is an object of class boot from the {boot} package. We then calculate the bootstrap mean and variance in the wrapper function maic_boot_stats.

The formula used in this model has all covariates as prognostic variable and is

$$y = X_1 + X_2 + X_3 + X_4 + (\beta_t + \beta_x X_1 + \beta_x X_2) t$$

which corresponds to the following R formula object passed as an argument to the strategy function.

lin_form <- as.formula("y ~ X3 + X4 + trt*X1 + trt*X2")

outstandR_maic <- outstandR(ipd_trial, ald_trial,
                            strategy = strategy_maic(formula = lin_form,
                                                     family = binomial(link = "logit")))

The returned object is of class outstandR.

outstandR_maic
#> Object of class 'outstandR' 
#> Model: binomial 
#> Scale: log_odds 
#> Common treatment: C 
#> Individual patient data study: AC 
#> Aggregate level data study: BC 
#> Confidence interval level: 0.95 
#> 
#> Contrasts:
#> 
#> # A tibble: 3 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl>      <dbl>      <dbl>
#> 1 AB            0.101    0.173      -0.715      0.917
#> 2 AC           -1.15     0.137      -1.88      -0.427
#> 3 BC           -1.25     0.0364     -1.63      -0.879
#> 
#> Absolute:
#> 
#> # A tibble: 2 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl> <lgl>      <lgl>     
#> 1 A             0.434   0.00255 NA         NA        
#> 2 C             0.704   0.00329 NA         NA

STC

STC is the conventional outcome regression method. It involves fitting a regression model of outcome on treatment and covariates to the IPD plugging-in covariate mean values.

outstandR_stc <- outstandR(ipd_trial, ald_trial,
                           strategy = strategy_stc(formula = lin_form,
                                                   family = binomial(link = "logit")))
outstandR_stc
#> Object of class 'outstandR' 
#> Model: binomial 
#> Scale: log_odds 
#> Common treatment: C 
#> Individual patient data study: AC 
#> Aggregate level data study: BC 
#> Confidence interval level: 0.95 
#> 
#> Contrasts:
#> 
#> # A tibble: 3 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl>      <dbl>      <dbl>
#> 1 AB           -0.176   NA           NA        NA    
#> 2 AC           -1.43    NA           NA        NA    
#> 3 BC           -1.25     0.0364      -1.63     -0.879
#> 
#> Absolute:
#> 
#> # A tibble: 2 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl> <lgl>      <lgl>     
#> 1 A             0.117        NA NA         NA        
#> 2 C             0.356        NA NA         NA

For the last two approaches, we perform G-computation firstly with a frequentist MLE approach and then a Bayesian approach.

Parametric G-computation with maximum-likelihood estimation

G-computation marginalizes the conditional estimates by separating the regression modelling from the estimation of the marginal treatment effect for A versus C.

outstandR_gcomp_ml <- outstandR(ipd_trial, ald_trial,
                                strategy = strategy_gcomp_ml(formula = lin_form,
                                                             family = binomial(link = "logit")))
outstandR_gcomp_ml
#> Object of class 'outstandR' 
#> Model: binomial 
#> Scale: log_odds 
#> Common treatment: C 
#> Individual patient data study: AC 
#> Aggregate level data study: BC 
#> Confidence interval level: 0.95 
#> 
#> Contrasts:
#> 
#> # A tibble: 3 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl>      <dbl>      <dbl>
#> 1 AB            0.164    0.147      -0.586      0.915
#> 2 AC           -1.09     0.110      -1.74      -0.438
#> 3 BC           -1.25     0.0364     -1.63      -0.879
#> 
#> Absolute:
#> 
#> # A tibble: 2 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl> <lgl>      <lgl>     
#> 1 A             0.485   0.00235 NA         NA        
#> 2 C             0.734   0.00251 NA         NA

Bayesian G-computation with MCMC

The difference between Bayesian G-computation and its maximum-likelihood counterpart is in the estimated distribution of the predicted outcomes. The Bayesian approach also marginalizes, integrates or standardizes over the joint posterior distribution of the conditional nuisance parameters of the outcome regression, as well as the joint covariate distribution.

outstandR_gcomp_stan <-
  outstandR(ipd_trial, ald_trial,
            strategy = strategy_gcomp_stan(formula = lin_form,
                                           family = binomial(link = "logit")))
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 1).
#> Chain 1: 
#> Chain 1: Gradient evaluation took 7.4e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.74 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1: 
#> Chain 1: 
#> Chain 1: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 1: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 1: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 1: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 1: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 1: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 1: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 1: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 1: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 1: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 1: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 1: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 1: 
#> Chain 1:  Elapsed Time: 0.285 seconds (Warm-up)
#> Chain 1:                0.327 seconds (Sampling)
#> Chain 1:                0.612 seconds (Total)
#> Chain 1: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 2).
#> Chain 2: 
#> Chain 2: Gradient evaluation took 1.6e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.16 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2: 
#> Chain 2: 
#> Chain 2: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 2: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 2: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 2: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 2: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 2: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 2: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 2: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 2: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 2: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 2: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 2: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 2: 
#> Chain 2:  Elapsed Time: 0.301 seconds (Warm-up)
#> Chain 2:                0.385 seconds (Sampling)
#> Chain 2:                0.686 seconds (Total)
#> Chain 2: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 3).
#> Chain 3: 
#> Chain 3: Gradient evaluation took 4.4e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.44 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3: 
#> Chain 3: 
#> Chain 3: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 3: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 3: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 3: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 3: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 3: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 3: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 3: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 3: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 3: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 3: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 3: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 3: 
#> Chain 3:  Elapsed Time: 0.32 seconds (Warm-up)
#> Chain 3:                0.328 seconds (Sampling)
#> Chain 3:                0.648 seconds (Total)
#> Chain 3: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 4).
#> Chain 4: 
#> Chain 4: Gradient evaluation took 1.9e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.19 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4: 
#> Chain 4: 
#> Chain 4: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 4: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 4: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 4: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 4: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 4: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 4: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 4: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 4: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 4: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 4: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 4: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 4: 
#> Chain 4:  Elapsed Time: 0.298 seconds (Warm-up)
#> Chain 4:                0.262 seconds (Sampling)
#> Chain 4:                0.56 seconds (Total)
#> Chain 4:
outstandR_gcomp_stan
#> Object of class 'outstandR' 
#> Model: binomial 
#> Scale: log_odds 
#> Common treatment: C 
#> Individual patient data study: AC 
#> Aggregate level data study: BC 
#> Confidence interval level: 0.95 
#> 
#> Contrasts:
#> 
#> # A tibble: 3 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl>      <dbl>      <dbl>
#> 1 AB            0.151    0.141      -0.585      0.887
#> 2 AC           -1.10     0.105      -1.74      -0.467
#> 3 BC           -1.25     0.0364     -1.63      -0.879
#> 
#> Absolute:
#> 
#> # A tibble: 2 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl> <lgl>      <lgl>     
#> 1 A             0.485   0.00218 NA         NA        
#> 2 C             0.736   0.00258 NA         NA

Multiple imputation marginalisation

Fit the model as before.

outstandR_mim <-
  outstandR(ipd_trial, ald_trial,
            strategy = strategy_mim(formula = lin_form,
                                    family = binomial(link = "logit")))
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 1).
#> Chain 1: 
#> Chain 1: Gradient evaluation took 2.2e-05 seconds
#> Chain 1: 1000 transitions using 10 leapfrog steps per transition would take 0.22 seconds.
#> Chain 1: Adjust your expectations accordingly!
#> Chain 1: 
#> Chain 1: 
#> Chain 1: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 1: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 1: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 1: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 1: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 1: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 1: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 1: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 1: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 1: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 1: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 1: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 1: 
#> Chain 1:  Elapsed Time: 0.309 seconds (Warm-up)
#> Chain 1:                0.336 seconds (Sampling)
#> Chain 1:                0.645 seconds (Total)
#> Chain 1: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 2).
#> Chain 2: 
#> Chain 2: Gradient evaluation took 1.9e-05 seconds
#> Chain 2: 1000 transitions using 10 leapfrog steps per transition would take 0.19 seconds.
#> Chain 2: Adjust your expectations accordingly!
#> Chain 2: 
#> Chain 2: 
#> Chain 2: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 2: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 2: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 2: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 2: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 2: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 2: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 2: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 2: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 2: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 2: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 2: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 2: 
#> Chain 2:  Elapsed Time: 0.309 seconds (Warm-up)
#> Chain 2:                0.335 seconds (Sampling)
#> Chain 2:                0.644 seconds (Total)
#> Chain 2: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 3).
#> Chain 3: 
#> Chain 3: Gradient evaluation took 2.2e-05 seconds
#> Chain 3: 1000 transitions using 10 leapfrog steps per transition would take 0.22 seconds.
#> Chain 3: Adjust your expectations accordingly!
#> Chain 3: 
#> Chain 3: 
#> Chain 3: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 3: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 3: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 3: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 3: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 3: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 3: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 3: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 3: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 3: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 3: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 3: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 3: 
#> Chain 3:  Elapsed Time: 0.323 seconds (Warm-up)
#> Chain 3:                0.322 seconds (Sampling)
#> Chain 3:                0.645 seconds (Total)
#> Chain 3: 
#> 
#> SAMPLING FOR MODEL 'bernoulli' NOW (CHAIN 4).
#> Chain 4: 
#> Chain 4: Gradient evaluation took 1.8e-05 seconds
#> Chain 4: 1000 transitions using 10 leapfrog steps per transition would take 0.18 seconds.
#> Chain 4: Adjust your expectations accordingly!
#> Chain 4: 
#> Chain 4: 
#> Chain 4: Iteration:    1 / 2000 [  0%]  (Warmup)
#> Chain 4: Iteration:  200 / 2000 [ 10%]  (Warmup)
#> Chain 4: Iteration:  400 / 2000 [ 20%]  (Warmup)
#> Chain 4: Iteration:  600 / 2000 [ 30%]  (Warmup)
#> Chain 4: Iteration:  800 / 2000 [ 40%]  (Warmup)
#> Chain 4: Iteration: 1000 / 2000 [ 50%]  (Warmup)
#> Chain 4: Iteration: 1001 / 2000 [ 50%]  (Sampling)
#> Chain 4: Iteration: 1200 / 2000 [ 60%]  (Sampling)
#> Chain 4: Iteration: 1400 / 2000 [ 70%]  (Sampling)
#> Chain 4: Iteration: 1600 / 2000 [ 80%]  (Sampling)
#> Chain 4: Iteration: 1800 / 2000 [ 90%]  (Sampling)
#> Chain 4: Iteration: 2000 / 2000 [100%]  (Sampling)
#> Chain 4: 
#> Chain 4:  Elapsed Time: 0.322 seconds (Warm-up)
#> Chain 4:                0.279 seconds (Sampling)
#> Chain 4:                0.601 seconds (Total)
#> Chain 4:
outstandR_mim
#> Object of class 'outstandR' 
#> Model: binomial 
#> Scale: log_odds 
#> Common treatment: C 
#> Individual patient data study: AC 
#> Aggregate level data study: BC 
#> Confidence interval level: 0.95 
#> 
#> Contrasts:
#> 
#> # A tibble: 3 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>         <dbl>     <dbl>      <dbl>      <dbl>
#> 1 AB            0.153    0.131      -0.556      0.861
#> 2 AC           -1.10     0.0942     -1.70      -0.499
#> 3 BC           -1.25     0.0364     -1.63      -0.879
#> 
#> Absolute:
#> 
#> # A tibble: 2 × 5
#>   Treatments Estimate Std.Error lower.0.95 upper.0.95
#>   <chr>      <lgl>    <lgl>     <lgl>      <lgl>     
#> 1 A          NA       NA        NA         NA        
#> 2 C          NA       NA        NA         NA

Model comparison

Combine all outputs for log-odds ratio table of all contrasts and methods.

knitr::kable(
  data.frame(
  `MAIC` = unlist(outstandR_maic$contrasts),
  `STC` = unlist(outstandR_stc$contrasts),
  `Gcomp ML` = unlist(outstandR_gcomp_ml$contrasts),
  `Gcomp Bayes` = unlist(outstandR_gcomp_stan$contrasts),
  `MIM` = unlist(outstandR_mim$contrasts))
)

	MAIC	STC	Gcomp.ML	Gcomp.Bayes	MIM
means.AB	0.1007707	-0.1758708	0.1641674	0.1513437	0.1525650
means.AC	-1.1518383	-1.4284798	-1.0884416	-1.1012654	-1.1000440
means.BC	-1.2526090	-1.2526090	-1.2526090	-1.2526090	-1.2526090
variances.AB	0.1732497	NA	0.1465980	0.1410717	0.1305757
variances.AC	0.1368488	NA	0.1101971	0.1046708	0.0941748
variances.BC	0.0364009	0.0364009	0.0364009	0.0364009	0.0364009
CI.AB1	-0.7150305	NA	-0.5862659	-0.5848092	-0.5556730
CI.AB2	0.9165719	NA	0.9146007	0.8874966	0.8608031
CI.AC1	-1.8768893	NA	-1.7390702	-1.7353698	-1.7015160
CI.AC2	-0.4267873	NA	-0.4378130	-0.4671609	-0.4985720
CI.BC1	-1.6265510	-1.6265510	-1.6265510	-1.6265510	-1.6265510
CI.BC2	-0.8786671	-0.8786671	-0.8786671	-0.8786671	-0.8786671