list 如何展平列表列表？
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How to flatten a list of lists?
提问by dnagirl
The tmpackage extends cso that, if given a set of PlainTextDocuments it automatically creates a Corpus. Unfortunately, it appears that each PlainTextDocumentmust be specified separately.
该tm包进行扩展，c因此，如果给定一组PlainTextDocuments，它会自动创建一个Corpus. 不幸的是，似乎每个都PlainTextDocument必须单独指定。
e.g. if I had:
例如，如果我有：
foolist <- list(a, b, c); # where a,b,c are PlainTextDocument objects
I'd do this to get a Corpus:
我这样做是为了得到一个Corpus：
foocorpus <- c(foolist[[1]], foolist[[2]], foolist[[3]]);
I have a list of lists of 'PlainTextDocuments that looks like this:
我有一个'PlainTextDocument看起来像这样的s列表：
> str(sectioned)
List of 154
$ :List of 6
..$ :Classes 'PlainTextDocument', 'TextDocument', 'character' atomic [1:1] Developing assessment models Developing models
.. .. ..- attr(*, "Author")= chr "John Smith"
.. .. ..- attr(*, "DateTimeStamp")= POSIXlt[1:1], format: "2013-04-30 12:03:49"
.. .. ..- attr(*, "Description")= chr(0)
.. .. ..- attr(*, "Heading")= chr "Research Focus"
.. .. ..- attr(*, "ID")= chr(0)
.. .. ..- attr(*, "Language")= chr(0)
.. .. ..- attr(*, "LocalMetaData")=List of 4
.. .. .. ..$ foo : chr "bar"
.. .. .. ..$ classification: chr "Technician"
.. .. .. ..$ team : chr ""
.. .. .. ..$ supervisor : chr "Bill Jones"
.. .. ..- attr(*, "Origin")= chr "Smith-John_e.txt"
#etc., all sublists have 6 elements
So, to get all my PlainTextDocuments into a Corpus, this would work:
因此，要将我所有的PlainTextDocuments 放入 a Corpus，这将起作用：
sectioned.Corpus <- c(sectioned[[1]][[1]], sectioned[[1]][[2]], ..., sectioned[[154]][[6]])
Can anyone suggest an easier way, please?
任何人都可以建议更简单的方法吗？
ETA: foo<-unlist(foolist, recursive=FALSE)produces a flat list of PlainTextDocuments, which still leaves me with the problem of feeding a list element by element to c
ETA：foo<-unlist(foolist, recursive=FALSE)生成一个纯文本文档的平面列表，这仍然给我留下了一个逐个元素提供列表的问题c
采纳答案by DrDom
I expect that unlist(foolist)will help you. It has an option recursivewhich is TRUEby default.
我希望这unlist(foolist)会帮助你。它有一个选项recursive是TRUE默认。
So unlist(foolist, recursive = FALSE)will return the list of the documents, and then you can combine them by:
Sounlist(foolist, recursive = FALSE)将返回文档列表，然后您可以通过以下方式组合它们：
do.call(c, unlist(foolist, recursive=FALSE))
do.calljust applies the function cto the elements of the obtained list
do.call只是将函数应用于c获得的列表的元素
回答by Michael
Here's a more general solution for when lists are nested multiple times and the amount of nesting differs between elements of the lists:
当列表多次嵌套并且列表元素之间的嵌套量不同时，这是一个更通用的解决方案：
flattenlist <- function(x){
morelists <- sapply(x, function(xprime) class(xprime)[1]=="list")
out <- c(x[!morelists], unlist(x[morelists], recursive=FALSE))
if(sum(morelists)){
Recall(out)
}else{
return(out)
}
}
