问题
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()).
回答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?
来源：https://stackoverflow.com/questions/22623309/how-to-reproduce-exact-results-with-lda-function-in-rs-topicmodels-package