Question

我编写了一个获得所需结果的代码。我想要一些帮助来缩短我的代码。代码的作用：

检索包含文件的所有目录（和子目录）的路径
将这些行拆分为两列 - a）一列是路径，b）另一列是带扩展名的文件名

我确信可以有更短的版本。期待着帮助。

这是我的代码：

    library(stringr)
    setwd("/Users/Guest/Desktop/Project") #set Working Directory
    path <-"/Users/Guest/Desktop/Project"  #set path to retrieve files
    a <- list.files(path,recursive = TRUE) #retrieve files in variable a
    last <- str_locate(a,"(.*)/") #locate the last "/"
    sub <- str_sub(a,last[,2:2] + 1) #split from the last "/"
    adf <- as.data.frame(a,stringsAsFactors= FALSE) #convert to DF
    colnames(adf) <- "FPath" #ColumnName
    subdf <- as.data.frame(sub, stringsAsFactors = FALSE) #Convert to DF
    colnames(subdf) <- "FileName" #ColumnName
    Final <- cbind(adf,subdf) #Join both DF's
    Final <- within(Final, FileName <- ifelse(is.na(FileName), FPath, FileName)) #If there are files directly in root folder (Project), then FileName is NULL so replace it with FPath.
    Final
    write.table(Final, file = "Final_Import2.txt", quote = FALSE, row.names = FALSE, sep ="\t") #WritetoFile

Answer 1

除了链接中提供的答案之外，这是一种可能有帮助的方法：

['Mon Sep 1 16:40:20 2015', 'Mon Sep 1 16:45:20 2015',
 'Mon Sep 1 16:50:20 2015', 'Mon Sep 1 16:55:20 2015']

将文件路径拆分为R

1 个答案: