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) <- filelist
namelist <- 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) <- NULL
names(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 example
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")
# get working dir
folder <- getwd()
# get names/locations of all files
filelist <- list.files(path = folder, pattern =" *.txt", full.names = FALSE)
# read in the files and put them in a list
lst <- lapply(filelist, readLines)
# extract the names of the files without the `.txt` stuff
names(lst) <- filelist
namelist <- 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 name
lst <- mapply(cbind, lst, "DOCUMENT" = namelist, SIMPLIFY = FALSE)
# combine into a dataframe
x <- do.call(rbind.data.frame, lst)
# a small amount of clean-up
rownames(x) <- NULL
names(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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