问题描述
我一直无法从 topicmodels 的 LDA 函数中创建可重现的结果.以他们的文档为例:
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
如何从对 LDA 的两次单独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(以防软件包作者在这里),我发现 control=list(seed=0)
片段很不幸且不必要.在幕后,有一行表示 if (missing(seed)) seed .这不会使过程更可靠地随机,它只会撤消指定的种子.我错过了什么吗?
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有效相同的对象,唯一的区别是临时本地文件位置.所以这个问题更像是一个误解(尽管如果函数尊重
set.seed()
似乎仍然会更清楚).
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
并不是真正的答案",但没有其他方法可以发布代码片段 :-)
Not really an "answer" but there's no other way to post code snippets :-)
我试了一下:
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
不平等"
@control
重要吗?
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