如何拼合列表列表? [英] How to flatten a list of lists?

问题描述

tm包扩展了c，因此，如果给定了一组PlainTextDocument，它将自动创建一个Corpus.不幸的是，似乎每个PlainTextDocument必须分别指定.

The tm package extends c so that, if given a set of PlainTextDocuments it automatically creates a Corpus. Unfortunately, it appears that each PlainTextDocument must be specified separately.

例如如果我有:

foolist <- list(a, b, c); # where a,b,c are PlainTextDocument objects

我这样做是为了获得Corpus:

foocorpus <- c(foolist[[1]], foolist[[2]], foolist[[3]]);

我有一个'PlainTextDocument列表，看起来像这样:

I have a list of lists of 'PlainTextDocuments that looks like this:

> 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

因此，要将我所有的PlainTextDocument放入Corpus中，这将起作用:

So, to get all my PlainTextDocuments into a Corpus, this would work:

sectioned.Corpus <- c(sectioned[[1]][[1]], sectioned[[1]][[2]], ..., sectioned[[154]][[6]])

请问有人可以建议一种更简单的方法吗?

Can anyone suggest an easier way, please?

ETA:foo<-unlist(foolist, recursive=FALSE)生成一个PlainTextDocuments的平面列表，这仍然使我面临着将一个列表元素逐个馈送到c

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

推荐答案

我希望unlist(foolist)会为您提供帮助.它具有选项recursive，默认情况下为TRUE.

I expect that unlist(foolist) will help you. It has an option recursive which is TRUE by default.

因此unlist(foolist, recursive = FALSE)将返回文档列表，然后可以通过以下方式将它们组合在一起:

So unlist(foolist, recursive = FALSE) will return the list of the documents, and then you can combine them by:

do.call(c, unlist(foolist, recursive=FALSE))

do.call只是将函数c应用于获得的列表的元素

do.call just applies the function c to the elements of the obtained list

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