如何根据正则表达式分隔 dplyr 中的列

Question

我有以下数据框：

df <- structure(list(X2 = c("BB_137.HVMSC", "BB_138.combined.HVMSC", 
"BB_139.combined.HVMSC", "BB_140.combined.HVMSC", "BB_141.HVMSC", 
"BB_142.combined.HMSC-bm")), .Names = "X2", row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))

看起来像这样

> df
# A tibble: 6 x 1
                       X2
                    <chr>
1            BB_137.HVMSC
2   BB_138.combined.HVMSC
3   BB_139.combined.HVMSC
4   BB_140.combined.HVMSC
5            BB_141.HVMSC
6 BB_142.combined.HMSC-bm

我想做的是将最后一个字段保留为第二列，分成两列（使用 . 作为分隔符）

              col1 col2
            BB_137 HVMSC
   BB_138.combined HVMSC
   BB_139.combined HVMSC
   BB_140.combined HVMSC
            BB_141 HVMSC
   BB_142.combined HMSC-bm

正确的做法是什么？

我的尝试是这样的：

> df %>% separate(X2, into = c("sid","status", "tiss"), sep = "[.]") 
# A tibble: 6 x 3
     sid   status    tiss
*  <chr>    <chr>   <chr>
1 BB_137    HVMSC    <NA>
2 BB_138 combined   HVMSC
3 BB_139 combined   HVMSC
4 BB_140 combined   HVMSC
5 BB_141    HVMSC    <NA>
6 BB_142 combined HMSC-bm

Warning message: Too few values at 2 locations: 1, 5

Answer 1

我们可以在单独的函数中使用负前瞻作为分隔符。

library(tidyr)
separate(data = df, col = X2, into = c("col1", "col2"), sep = "(\.)(?!.*\.)")

#            col1    col2
#           <chr>   <chr>
#1          BB_137   HVMSC
#2 BB_138.combined   HVMSC
#3 BB_139.combined   HVMSC
#4 BB_140.combined   HVMSC
#5          BB_141   HVMSC
#6 BB_142.combined HMSC-bm

正则表达式取自 this 答案。

Answer 2

我们也可以使用tidyr::extract()

extract(df, X2, c("col1","col2"), "(.*)\.(H.*)")

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