如何使用 dplyr 重命名存储为 tibble/dataframe 的文件名

Question

我有以下包含文件列表的数据框。

library(tidyverse)
dat <- structure(list(source_file = structure(c("data/monroe_20180214/180131 WT PB d5/PB x10_01.tif", 
"data/monroe_20180214/180131 WT PB d5/PB x10_02.tif", "data/monroe_20180214/180131 WT PB d5/PB x10_03.tif", 
"data/monroe_20180214/180131 WT PB d5/PB x10_04.tif", "data/monroe_20180214/180131 WT PB d5/PB x10_05.tif", 
"data/monroe_20180214/180131 WT PB d5/PB x10_06.tif"), class = c("fs_path", 
"character"))), .Names = "source_file", row.names = c(NA, -6L
), class = c("tbl_df", "tbl", "data.frame"))


dat
#> # A tibble: 6 x 1
#>   source_file                                       
#>   <chr>                                             
#> 1 data/monroe_20180214/180131 WT PB d5/PB x10_01.tif
#> 2 data/monroe_20180214/180131 WT PB d5/PB x10_02.tif
#> 3 data/monroe_20180214/180131 WT PB d5/PB x10_03.tif
#> 4 data/monroe_20180214/180131 WT PB d5/PB x10_04.tif
#> 5 data/monroe_20180214/180131 WT PB d5/PB x10_05.tif
#> 6 data/monroe_20180214/180131 WT PB d5/PB x10_06.tif

我想要做的是创建第二列new_filename，方法是将前两个目录路径替换为新路径pooled/，并将空格替换为.，反斜杠替换为__。我怎样才能做到这一点？

想要的结果是

  source_file                                         new_filename                                   
1 data/monroe_20180214/180131 WT PB d5/PB x10_01.tif  pooled/180131.WT.PB.d5__PB.x10_01.tif 
2 data/monroe_20180214/180131 WT PB d5/PB x10_02.tif  ...
3 data/monroe_20180214/180131 WT PB d5/PB x10_03.tif  .etc.
4 data/monroe_20180214/180131 WT PB d5/PB x10_04.tif  
5 data/monroe_20180214/180131 WT PB d5/PB x10_05.tif  
6 data/monroe_20180214/180131 WT PB d5/PB x10_06.tif  

Answer 1

使用来自string 的gsub()，您也可以做到这一点

     dat %>% mutate(new_var = gsub("data/monroe_20180214", "pooled", source_file),
+                new_var = gsub(" ", ".", new_var), 
+                new_var = gsub("/", "_", new_var), 
+                new_var = gsub("pooled_", "pooled/", new_var))
# A tibble: 6 x 2
                                         source_file                              new_var
                                               <chr>                                <chr>
1 data/monroe_20180214/180131 WT PB d5/PB x10_01.tif pooled/180131.WT.PB.d5_PB.x10_01.tif
2 data/monroe_20180214/180131 WT PB d5/PB x10_02.tif pooled/180131.WT.PB.d5_PB.x10_02.tif
3 data/monroe_20180214/180131 WT PB d5/PB x10_03.tif pooled/180131.WT.PB.d5_PB.x10_03.tif
4 data/monroe_20180214/180131 WT PB d5/PB x10_04.tif pooled/180131.WT.PB.d5_PB.x10_04.tif
5 data/monroe_20180214/180131 WT PB d5/PB x10_05.tif pooled/180131.WT.PB.d5_PB.x10_05.tif
6 data/monroe_20180214/180131 WT PB d5/PB x10_06.tif pooled/180131.WT.PB.d5_PB.x10_06.tif

Answer 2

一个班轮：

paste0("pooled/",chartr(" /", "._",(sub("^(?:[^\\/]*\\/){2}","",dat$source_file))))


#[1] "pooled/180131.WT.PB.d5_PB.x10_01.tif"
#[2] "pooled/180131.WT.PB.d5_PB.x10_02.tif"
#[3] "pooled/180131.WT.PB.d5_PB.x10_03.tif"
#[4] "pooled/180131.WT.PB.d5_PB.x10_04.tif"
#[5] "pooled/180131.WT.PB.d5_PB.x10_05.tif"
#[6] "pooled/180131.WT.PB.d5_PB.x10_06.tif"

这里我们首先用空字符串（""）替换前两次出现/的部分，然后使用base R中的chartr函数将空格替换为点（.）和正斜杠（@ 987654327@) 带下划线 (_) 和paste 带pooled/ 的字符串。

sub 部分的正则表达式取自 here。

在dplyr 调用中添加这个：

dat %>%
 mutate(new_filename =paste0("pooled/", chartr(" /", "._", 
                            (sub("^(?:[^\\/]*\\/){2}", "", source_file))))) %>%
 select(new_filename)


#new_filename                        
#  <chr>                               
#1 pooled/180131.WT.PB.d5_PB.x10_01.tif
#2 pooled/180131.WT.PB.d5_PB.x10_02.tif
#3 pooled/180131.WT.PB.d5_PB.x10_03.tif
#4 pooled/180131.WT.PB.d5_PB.x10_04.tif
#5 pooled/180131.WT.PB.d5_PB.x10_05.tif
#6 pooled/180131.WT.PB.d5_PB.x10_06.tif