英文:
Upload text document in R
问题
我正在尝试将多个文本文档上传到R中的数据框中。我的期望输出是一个具有两列的矩阵：
| DOCUMENT | CONTENT |
|---|---|
| Document A | 这是内容。 |
| Document B | 这是内容。 |
| Document C | 这是内容。 |
在“CONTENT”列中，应显示来自文本文档（10-K报告）的所有文本信息。
> setwd("C:/Users/folder")> folder <- getwd()> corpus <- Corpus(DirSource(directory = folder, pattern = "*.txt"))
这将创建一个语料库，然后可以对其进行分词。但我无法将其转换为数据框或获得我期望的输出。
有人可以帮我吗？
英文:
I am trying to upload several text document into a data frame in R. My desired output is a matrix with two colums:
| DOCUMENT | CONTENT |
|||
| Document A | This is the content.|
|: ---- |: -------:| ------:|
| Document B | This is the content.|
|: ---- |: -------:| ------:|
| Document C | This is the content.|
Within the column "CONTENT", all the text information from the text document (10-K report) shall be shown.
> setwd("C:/Users/folder")> folder <- getwd()> corpus <- Corpus(DirSource(directory = folder, pattern = "*.txt"))
This will create a corpus and I can tokenize it. But I don't achieve to convert to a data frame nor my desiret output.
Can somebody help me?
答案1
得分: 1
如果您只处理.txt文件并且最终目标是一个数据框，那么我认为您可以跳过语料库步骤，只需将所有文件读入列表中。难点在于将.txt文件的名称放入名为DOCUMENT的列中，但这可以在基本的R中完成。
# 创建一个可复现的示例a <- "this is a test"b <- "this is a second test"c <- "this is a third test"write(a, "a.txt"); write(b, "b.txt"); write(c, "c.txt")# 获取工作目录folder <- getwd()# 获取所有文件的名称/位置filelist <- list.files(path = folder, pattern = " *.txt", full.names = FALSE)# 读入文件并将它们放入列表中lst <- lapply(filelist, readLines)# 提取不带`.txt`后缀的文件名names(lst) <- filelistnamelist <- fs::path_file(filelist)namelist <- unlist(lapply(namelist, sub, pattern = ".txt", replacement = ""),use.names = FALSE)# 为列表中的每个矩阵指定其原始文件名作为名称lst <- mapply(cbind, lst, "DOCUMENT" = namelist, SIMPLIFY = FALSE)# 合并为数据框x <- do.call(rbind.data.frame, lst)# 进行一些清理rownames(x) <- NULLnames(x)[names(x) == "V1"] <- "CONTENT"x <- x[,c(2,1)]x#> DOCUMENT CONTENT#> 1 a this is a test#> 2 b this is a second test#> 3 c this is a third test
这是您提供的代码的翻译部分。
英文:
If you're only working with .txt files and your endgoal is a dataframe, then I think you can skip the corpus step and simply read in all your files as a list. The hard part is to get the names of the .txt files into a column called DOCUMENT, but this can be done in base R.
# make a reproducible examplea <- "this is a test"b <- "this is a second test"c <- "this is a third test"write(a, "a.txt"); write(b, "b.txt"); write(c, "c.txt")# get working dirfolder <- getwd()# get names/locations of all filesfilelist <- list.files(path = folder, pattern =" *.txt", full.names = FALSE)# read in the files and put them in a listlst <- lapply(filelist, readLines)# extract the names of the files without the `.txt` stuffnames(lst) <- filelistnamelist <- fs::path_file(filelist)namelist <- unlist(lapply(namelist, sub, pattern = ".txt", replacement = ""),use.names = FALSE)# give every matrix in the list its own name, which was its original file namelst <- mapply(cbind, lst, "DOCUMENT" = namelist, SIMPLIFY = FALSE)# combine into a dataframex <- do.call(rbind.data.frame, lst)# a small amount of clean-uprownames(x) <- NULLnames(x)[names(x) == "V1"] <- "CONTENT"x <- x[,c(2,1)]x#> DOCUMENT CONTENT#> 1 a this is a test#> 2 b this is a second test#> 3 c this is a third test
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