LBA in Julia / Turing: how to specify the input as vectors #1
Description
@kiante-fernandez I thought I'd first ask you as it's probably a basic issue of Turing specification.
I am testing this tutorial, and try to modify it to first make it as if I had the data stored in a dataframe. And I thought, similarly to other Turing models, to modify the input to not be a tuple of data but two vectors.
I tried the following:
using Turing
using SequentialSamplingModels
using Random
using DataFrames
using LinearAlgebra
# Generate some data
Random.seed!(254)
data1 = DataFrame(rand(LBA(ν=[3.0, 2.0], A=0.8, k=0.2, τ=0.3), 500))
data1[!, :condition] = repeat(["A"], nrow(data1))
data2 = DataFrame(rand(LBA(ν=[1.0, 1.5], A=0.7, k=0.2, τ=0.3), 500))
data2[!, :condition] = repeat(["B"], nrow(data2))
data = vcat(data1, data2)
# Specify model
@model model(choice, rt) = begin
min_rt = minimum(rt) # Get min RT
ν ~ MvNormal(zeros(2), I * 2)
A ~ truncated(Normal(0.8, 0.4), 0.0, Inf)
k ~ truncated(Normal(0.2, 0.2), 0.0, Inf)
τ ~ Uniform(0.0, min_rt)
data ~ LBA(; ν, A, k, τ)
end
sample(model(data[!, :choice], data[!, :rt]), Prior(), 5000)ERROR: MethodError: no method matching iterate(::LBA{Vector{Float64}, Float64, Float64, Float64})
I suspect this is because LBA() returns a tuple. But I'm not sure what to do here, I tried things like having:
(choice, rt) ~ LBA(; ν, A, k, τ)
but it doesn't seem to be the solution. Any thoughts?