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我用下面的代碼來創建主題模型的列表,其中主題數26〜35,按1:如何將僅有一個元素的「列表」的類更改爲R中的對象?
best.model <- lapply(seq(26,35, by=1), function(d){LDA(dtm2, d, method = "Gibbs", control = list(burnin = burnin, iter = iter, keep = keep))})
當我打電話best.model,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.
然後我嘗試每個主題模型提取到單獨的對象:
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,當我調用類各型號的,我得到:
class(Gibbs26)
[1] "list"
如何從最初的best.model列表中提取每個元素,並讓每個元素成爲一個可以輕鬆操作的對象?
見'「[」' - 一個單支架提取部分列表,'[',總是會返回另一個列表。嘗試使用雙括號:'[['。另見[這篇文章](http://stackoverflow.com/q/1169456/2572423) – JasonAizkalns
看到這個鏈接 - https://twitter.com/hadleywickham/status/643381054758363136 – jeremycg