如何将只有一个元素的"list"类更改为 R 中的对象？

Question

我使用以下代码创建主题模型列表，其中主题的数量从26到35不等，按1表示:

best.model <- lapply(seq(26,35, by=1), function(d){LDA(dtm2, d, method = "Gibbs", control = list(burnin = burnin, iter = iter, keep = keep))})

当我呼唤最好。模型，我得到:

> best.model
[[1]]
A LDA_Gibbs topic model with 26 topics.
[[2]]
A LDA_Gibbs topic model with 27 topics.
[[3]]
A LDA_Gibbs topic model with 28 topics.
[[4]]
A LDA_Gibbs topic model with 29 topics.
[[5]]
A LDA_Gibbs topic model with 30 topics.
[[6]]
A LDA_Gibbs topic model with 31 topics.
[[7]]
A LDA_Gibbs topic model with 32 topics.
[[8]]
A LDA_Gibbs topic model with 33 topics.
[[9]]
A LDA_Gibbs topic model with 34 topics.
[[10]]
A LDA_Gibbs topic model with 35 topics.

然后我尝试将每个主题模型提取到单独的对象中:

Gibbs26 <- best.model[1]
Gibbs27 <- best.model[2]
Gibbs28 <- best.model[3]
Gibbs29 <- best.model[4]
Gibbs30 <- best.model[5]
Gibbs31 <- best.model[6]
Gibbs32 <- best.model[7]
Gibbs33 <- best.model[8]
Gibbs34 <- best.model[9]
Gibbs35 <- best.model[10]

然而，当我调用每个模型的类时，我得到:

class(Gibbs26)
[1] "list"

如何从初始最优中提取每个元素。模型列表，并让每个元素都是我可以轻松操作的对象?

Answer 1

你有两个问题。首先，正如@JasonAizkalns在评论中提到的，当你想要两个括号时，你只使用一个括号:

Gibbs26 <- best.model[[1]]

其次，你不会真的想要输入那么多东西，因为你不可避免地会搞砸一个。相反，您可以使用lapply和assign来分配所有对象:

lapply(1:length(bestmodel), function(x){assign(paste0("Gibbs", x + 25), bestmodel[[x]], envir = .GlobalEnv)})