Question

也许是一个幼稚的问题，我下载了一堆具有以下结构的不同生物的程序集；

-Parent_folder
--Genus_species_1
---genome_filename_1
---genome_filename_2
---genome_filename_n
--Genus_species_2
---genome_filename_1
---genome_filename_2
---genome_filename_n
--Genus_species_N
---genome_filename_1
---genome_filename_2
---genome_filename_n

我想创建一个表，其中一列具有种类名称，第二列具有程序集的文件名。像这样的东西

    colum1      |     column2
Genus_species_1 | genome_filename_1
Genus_species_1 | genome_filename_2
Genus_species_1 | genome_filename_n
Genus_species_2 | genome_filename_1
Genus_species_2 | genome_filename_2
Genus_species_2 | genome_filename_n
Genus_species_N | genome_filename_1
Genus_species_N | genome_filename_2
Genus_species_N | genome_filename_n

我尝试了很多事情，但不知道这段代码有什么问题；

#listing the folders containing different number of genomes;
folder_list<- list.dirs(".", full.names = FALSE)

#Remove the parent folder;
folder_list<- folder_list[-1]

#Creating two vectors to populate with the genome filename and another with the species name(same as folder name);
genomes<-NULL
species<- NULL

#Generate a loop to populate;
for (dir in 1:length(folder_list)){
  files<- as.vector(list.files(file.path(WD, dir))) #Vector containing all the genome filenames
  genomes<- c(genomes, files) #add the one before to the genomes vector

  #next, create a vector with the number of the folder(which is the species) and repeat it as much as the number of genomes;
  directories<-rep(dir, length(list.files(file.path(WD, dir))))
  species<- append(species, directories) #add it to species vector

} #end of the loop

希望有人可以提供帮助！

提前谢谢！

Answer 1

我对您的文件结构进行了简化。

-Parent_Folder
--Genus_Species_1
---genome_filename1.txt
---genome_filename2.txt
---genome_filename3.txt
---genome_filename4.txt
--Genus_Species_2
---genome_filename1.txt
---genome_filename2.txt
---genome_filename3.txt

下面的代码使用here包。有关软件包here的更多信息。从本质上讲，它使构造路径更容易（imo）。

library(here)

# save the name of genus directories
GenusDirs <- list.files(here("Parent_Folder"))
GenusDirs

# initialize list to save name of genome files
GenomeFiles <- vector("list", length = length(GenusDirs))
# name elements in the list with Genus names
names(GenomeFiles) <- GenusDirs

# create list of genome files associated with each genus
for (i in 1:length(upperDir)){
  GenomeFiles[[i]] <- list.files(here("Parent_Folder", upperDir[i]))
}

# initialize vectors
GenusNames <- c()
GenomeNames <- c()

# in this loop, construct two vectors by appending Genus and Genome names
for (i in 1:length(GenomeFiles)){
  # repeat Genus Name by how many Genome files in that Genus directory
  GenusNames <- append(GenusNames, rep(names(GenomeFiles[i]), length(GenomeFiles[[i]])))

  GenomeNames <- append(GenomeNames, GenomeFiles[[i]])
}

# create data frame
GenusGenomeData <- data.frame(
  Genus = GenusNames,
  Genome = GenomeNames
)

Answer 2

我重新创建了目录结构，然后创建了一个data.frame，其中第1列列出了所有子目录，第2列列出了每个子目录中的文件。

首先，这是用于复制目录结构的代码。

# Create parent folder
dir.create("Parent_folder")

# Create child directories and files in one go
sample_dirnames <- seq_len(3)
sapply(sample_dirnames, function(index){

    # build dir path and create
    dirname <- paste0("Genus_species_", sample_dirnames[index])
    dir.create(paste0("Parent_folder/",dirname))

    # generate three files in the current directory
    sapply(seq_len(3), function(n) {
        file.create(paste0("Parent_folder/", dirname, "/genome_filename_", n, ".R"))      
    })
})

以list.files函数为起点，使用recursive = TRUE。然后将输出通过管道传递到data.frame并将路径分为目录和文件名。此示例使用dplyr包中的函数和基本函数（substring，gregexpr）。

# pkg
library(tidyverse)

# build object
files <- list.files("Parent_folder/", recursive = TRUE) %>%
    as.data.frame(.) %>%
    rename(., "path" = .) %>%
    mutate(
        column1 = substring(
            text = path,
            first = 1,
            last = as.numeric(
                gregexpr(
                    pattern = "/",
                    text = path
                )[1]
            ) - 1
        ),
        column2 = substring(
            text = path,
            first = as.numeric(
                gregexpr(
                    pattern = "/",
                    text = path
                )[1]
            ) + 1
        )
    ) %>%
    select(-path)

这将打印出以下对象

files
#          column1             column2
# 1 Genus_species_1 genome_filename_1.R
# 2 Genus_species_1 genome_filename_2.R
# 3 Genus_species_1 genome_filename_3.R
# 4 Genus_species_2 genome_filename_1.R
# 5 Genus_species_2 genome_filename_2.R
# 6 Genus_species_2 genome_filename_3.R
# 7 Genus_species_3 genome_filename_1.R
# 8 Genus_species_3 genome_filename_2.R
# 9 Genus_species_3 genome_filename_3.R

Answer 3

使用stringr和dplyr软件包尝试

library(stringr)
library(dplyr)
file_list<- list.files(".",recursive=T)
> file_list

[1] "Genus_species_1/genome_filename_1" "Genus_species_1/genome_filename_2" "Genus_species_1/genome_filename_n"
[4] "Genus_species_2/genome_filename_1" "Genus_species_2/genome_filename_2" "Genus_species_2/genome_filename_n"
[7] "Genus_species_n/genome_filename_1" "Genus_species_n/genome_filename_2" "Genus_species_n/genome_filename_n"

使用stringr包中的str_split_fixed函数将file_list变量拆分成矩阵，然后使用%>%中的管道dplyr将其保存为数据帧{{1} }

df

从list.files（）函数填充两个向量；

3 个答案: