Question

我有一个像这样的数据框：

structure(list(mut = c("Q184H/CAA-CAT", "I219V/ATC-GTC", "A314T/GCG-ACG, P373Q/CCG-CAG, A653E/GCG-GAA","0")), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))

我想做的是，对于每行中每个逗号分隔的值，在新列中的“ /”之后获取所有内容，而不管每行中有多少个条目。

我最后想要得到的东西：

    mut                    nt
1   Q184H                  CAA-CAT
2   I219V                  ATC-GTC
3   A314T, P373Q, A653E    GCG-ACG, CCG-CAG, CGC-GAA
4   0                      0

我尝试为此使用正则表达式，但似乎无法匹配每个用逗号分隔的条目。

library(dplyr)
df %>%
    mutate(nt = gsub(".+/(.*?)", "\\1", mut))

如何使每个条目都匹配？我必须将它们分开然后进行匹配吗？

Answer 1

您只需要稍微调整一下正则表达式即可；请注意，我是如何将您的.更改为[^,]的。在正则表达式中，如果将字符放在方括号中并在^之前，则表示匹配所有但字符。因此，[^,]+意味着要匹配尽可能多的非逗号连续字符。

df = structure(list(mut = c("Q184H/CAA-CAT", "I219V/ATC-GTC",
                            "A314T/GCG-ACG, P373Q/CCG-CAG, A653E/GCG-GAA","0")),
               row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))
library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union
df %>%
    mutate(nt = gsub("[^,]+?/([^,]+?)", "\\1", mut),
           mut = gsub("([^/]+)/[^,]+", "\\1", mut))
#> # A tibble: 4 x 2
#>   mut                 nt                     
#>   <chr>               <chr>                  
#> 1 Q184H               CAA-CAT                
#> 2 I219V               ATC-GTC                
#> 3 A314T, P373Q, A653E GCG-ACG,CCG-CAG,GCG-GAA
#> 4 0                   0

^{由reprex package（v0.2.1）于2018-10-10创建}

Answer 2

不将此作为答案（@duckmayr进行了正则表达式调试）。发布此soley来向人们展示，通过使用stringi，我们可以获得自我记录的正则表达式，因此我们将来的自己不会讨厌我们过去的自己：

library(stringi) # it's what stringr uses
library(tidyverse)

xdf <- structure(list(mut = c("Q184H/CAA-CAT", "I219V/ATC-GTC", "A314T/GCG-ACG, P373Q/CCG-CAG, A653E/GCG-GAA","0")), row.names = c(NA, -4L), class = c("tbl_df", "tbl", "data.frame"))

mutate(
  xdf, 
  nt = stri_replace_all_regex(
    str = mut,
    pattern = "
[^,]+?  # match anything but a comma and if there is one, match at most once
/       # followed by a forward slash
(       # start of match group
 [^,]+? # same as above
)       # end of match group
",
    replacement = "$1", # take the match group value as the value
    opts_regex = stri_opts_regex(comments=TRUE)
  ),
  mut = stri_replace_all_regex(
    str = mut,
    pattern = "
(      # start of match group
 [^/]+ # match anything but a forward slash
)      # end of match group
/      # followed by a forward slash
[^,]+  # match anything but a comma
",
    replacement = "$1", # take the match group value as the value
    opts_regex = stri_opts_regex(comments=TRUE)
  )
)

从具有逗号分隔值的列中提取多个字符串

2 个答案: