Question

我有以下数据框：

library(tidyverse)

dat <-structure(list(motif_name_binned = c("Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin1", 
"Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin2", 
"Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin3"
), motif_score = c(6.816695, 6.816695, 6.816695)), row.names = c(NA, 
-3L), class = c("tbl_df", "tbl", "data.frame"), .Names = c("motif_name_binned", 
"motif_score"))

dat

这给出了这个：

> dat
# A tibble: 3 x 2
                                                  motif_name_binned motif_score
                                                              <chr>       <dbl>
1 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin1    6.816695
2 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin2    6.816695
3 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin3    6.816695

我可以通过使用以下代码提取motif_named_binned的值来获得我想要的内容：

dat %>% 
  mutate(motif = str_match(motif_name_binned,"^(.*?)\\/.*?")[,2], 
         inst =  str_match(motif_name_binned,"^.*?\\/.*?\\/.*?\\.instid_(.*?)\\.bin\\d+")[,2],
         binno = as.integer(str_match(motif_name_binned,"^.*?\\/.*?\\/.*?\\.bin(\\d+)")[,2]))

哪个给出了

# A tibble: 3 x 5
                                                  motif_name_binned motif_score        motif                     inst binno
                                                              <chr>       <dbl>        <chr>                    <chr> <int>
1 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin1    6.816695 Ddit3::Cebpa chr1:183286845-183287245     1
2 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin2    6.816695 Ddit3::Cebpa chr1:183286845-183287245     2
3 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin3    6.816695 Ddit3::Cebpa chr1:183286845-183287245     3

但请注意，我必须执行正则表达式3次并将其逐个分配给变量。事实上，我可以使用单一的正则表达式：

str_match(motif_name_binned,"^(.*?)\\/.*?\\/.*?\\.instid_(.*?)\\.bin(\\d+)")[,c(2,3,4)]

我如何在dplyr mutate()中合并后来的一体化正则表达式？

Answer 1

您可以使用tidyr::extract将正则表达式中的捕获组转换为新列：

library(tidyr)
dat %>% 
    extract(motif_name_binned, c('motif', 'inst', 'binno'), regex = "^(.*?)\\/.*?\\/.*?\\.instid_(.*?)\\.bin(\\d+)", remove = FALSE)

# A tibble: 3 x 5
#                                                  motif_name_binned        motif                     inst binno motif_score
#*                                                             <chr>        <chr>                    <chr> <chr>       <dbl>
#1 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin1 Ddit3::Cebpa chr1:183286845-183287245     1    6.816695
#2 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin2 Ddit3::Cebpa chr1:183286845-183287245     2    6.816695
#3 Ddit3::Cebpa/MA0019.1/Jaspar.instid_chr1:183286845-183287245.bin3 Ddit3::Cebpa chr1:183286845-183287245     3    6.816695

如何在dplyr mutate中将向量分配给多个变量

1 个答案: