How to reproduce exact results with LDA function in R's topicmodels package

Question

I've been unable to create reproducible results from topicmodels' LDA function. To take an example from their documentation:

library(topicmodels)
set.seed(0)
lda1 <- LDA(AssociatedPress[1:20, ], control=list(seed=0), k=2)
set.seed(0)
lda2 <- LDA(AssociatedPress[1:20, ], control=list(seed=0), k=2)
identical(lda1, lda2)
# [1] FALSE

How can I get identical results from two separate calls to LDA?

As an aside (in case the package authors are on here), I find the control=list(seed=0) snippet unfortunate and unnecessary. Behind the scenes, there's a line for if (missing(seed)) seed <- as.integer(Sys.time()). This doesn't make the process more reliably random, it only undoes a specified seed. Am I missing something?

UPDATE: As @hrbrmstr discovered below, passing a seed as a control results in effectively identical objects, with the only difference being a temp local file location. So this question is more of a misunderstanding (though still seems like it would be clearer if the function respected set.seed()).

Answer 1

Not really an "answer" but there's no other way to post code snippets :-)

I gave the following a go:

library(topicmodels)

data(AssociatedPress)

lda1 <- LDA(AssociatedPress[1:20, ], control=list(seed=0), k=2)
lda2 <- LDA(AssociatedPress[1:20, ], control=list(seed=0), k=2)

identical(lda1, lda2)
[1] FALSE

all.equal(lda1, lda2)
[1] "Attributes: < Component 5: Attributes: < Component 10: 1 string mismatch > >"

a1 <- posterior(lda1, AssociatedPress)
a2 <- posterior(lda2, AssociatedPress)

identical(a1, a2)
[1] TRUE

all.equal(a1, a2)
[1] TRUE

all.equal(lda1@alpha,lda2@alpha)
[1] TRUE
all.equal(lda1@call,lda2@call)
[1] TRUE
all.equal(lda1@Dim,lda2@Dim)
[1] TRUE
all.equal(lda1@control,lda2@control)
[1] "Attributes: < Component 10: 1 string mismatch >"
all.equal(lda1@k,lda2@k)
[1] TRUE
all.equal(lda1@terms,lda2@terms)
[1] TRUE
all.equal(lda1@documents,lda2@documents)
[1] TRUE
all.equal(lda1@beta,lda2@beta)
[1] TRUE
all.equal(lda1@gamma,lda2@gamma)
[1] TRUE
all.equal(lda1@wordassignments,lda2@wordassignments)
[1] TRUE
all.equal(lda1@loglikelihood,lda2@loglikelihood)
[1] TRUE
all.equal(lda1@iter,lda2@iter)
[1] TRUE
all.equal(lda1@logLiks,lda2@logLiks)
[1] TRUE
all.equal(lda1@n,lda2@n)
[1] TRUE

identical(lda1@alpha,lda2@alpha)
[1] TRUE
identical(lda1@call,lda2@call)
[1] TRUE
identical(lda1@Dim,lda2@Dim)
[1] TRUE
identical(lda1@control,lda2@control)
[1] FALSE
identical(lda1@k,lda2@k)
[1] TRUE
identical(lda1@terms,lda2@terms)
[1] TRUE
identical(lda1@documents,lda2@documents)
[1] TRUE
identical(lda1@beta,lda2@beta)
[1] TRUE
identical(lda1@gamma,lda2@gamma)
[1] TRUE
identical(lda1@wordassignments,lda2@wordassignments)
[1] TRUE
identical(lda1@loglikelihood,lda2@loglikelihood)
[1] TRUE
identical(lda1@iter,lda2@iter)
[1] TRUE
identical(lda1@logLiks,lda2@logLiks)
[1] TRUE
identical(lda1@n,lda2@n)
[1] TRUE

Is the "unequal" @control significant?