如何在R的topicmodels包中使用LDA函数重现准确的结果 [英] How to reproduce exact results with LDA function in R's topicmodels package
问题描述
我无法从topicmodels的LDA函数创建可重复的结果.从他们的文档中举个例子:
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
如何从两个单独的LDA调用中获得相同的结果?
How can I get identical results from two separate calls to LDA?
顺便说一句(如果软件包作者在这里),我发现control=list(seed=0)
片段是不幸的,也是不必要的.在幕后,有一行if (missing(seed)) seed <- as.integer(Sys.time())
.这不会使过程更可靠地随机化,只会撤消指定的种子.我想念什么吗?
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?
更新:正如@hrbrmstr在下面发现的那样,将种子作为控件传递会产生实际上相同的对象,唯一的区别是临时本地文件位置.因此,这个问题更像是一个误解(尽管如果该函数遵守set.seed()
的话,看起来仍然会更清楚).
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()
).
推荐答案
不是真正的答案",但是没有其他方法可以发布代码段了:-)
Not really an "answer" but there's no other way to post code snippets :-)
我做了以下尝试:
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
不相等" @control
重要吗?
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