Question

我有随附的数据框。

数据

structure(list(associated_gene = c(NA, NA, "A4GALT", NA, NA, 
   "NOT FOUND"), chr_name = c("22", "22", "22", "22", "22", "NOT FOUND"
   ), chrom_start = c(42693910L, 42693843L, 42693321L, 42693665L, 
   42693653L, 0L), allele = c("G/A/T", "T/C", "G/C", "C/T", "G/A/T", 
   "NOT FOUND"), refsnp_id = c("rs778598915", "rs11541159", "rs397514502", 
   "rs762949801", "rs776304817", "NOT FOUND")), row.names = c("s3a", 
   "s3b", "s3c", "s3d", "s3e", "s3f"), class = "data.frame")

    associated_gene  chr_name chrom_start    allele   refsnp_id
s3a            <NA>        22    42693910     G/A/T rs778598915
s3b            <NA>        22    42693843       T/C  rs11541159
s3c          A4GALT        22    42693321       G/C rs397514502
s3d            <NA>        22    42693665       C/T rs762949801
s3e            <NA>        22    42693653     G/A/T rs776304817
s3f       NOT FOUND NOT FOUND           0 NOT FOUND   NOT FOUND

我想将等位基因列的第一个“ /”分为两部分（参考和变量），然后将它们插入到$ chrom_start和$ refsnp_id之间

理想的输出是：

     associated_gene  chr_name chrom_start   Ref   Var   refsnp_id
s3a            <NA>         22    42693910     G   A/T rs778598915
s3b            <NA>         22    42693843     T     C  rs11541159

我不知道我是否可以加载awk，但是在bash中我会这样做：

猫等位基因| awk -F“ /”'{print $ 1“ \ t” $ 2}'

Answer 1

我们可以使用Activity中的extract从字符串的开头（tidyr）捕获不是/（[^/]+）的字符然后是^，然后捕获其余字符

或者另一个选择是library(tidyr) library(dplyr) df1 %>% extract(allele, into = c("Ref", "Var"), "^([^/]+)/(.*)") # associated_gene chr_name chrom_start Ref Var refsnp_id #s3a <NA> 22 42693910 G A/T rs778598915 #s3b <NA> 22 42693843 T C rs11541159 #s3c A4GALT 22 42693321 G C rs397514502 #s3d <NA> 22 42693665 C T rs762949801 #s3e <NA> 22 42693653 G A/T rs776304817 #s3f NOT FOUND NOT FOUND 0 <NA> <NA> NOT FOUND

str_split

或使用library(stringr) do.call(rbind, str_split(df$allele, "/", 2))创建定界符并在sub中用read.table/read.csv读取

base R

Answer 2

另一种解决方案是使用“字符串”包：

install.packages("stringr")
library(stringr)

数据：

df <- structure(list(associated_gene = c(NA, NA, "A4GALT", NA, NA, 
                               "NOT FOUND"), chr_name = c("22", "22", "22", "22", "22", "NOT FOUND"
                               ), chrom_start = c(42693910L, 42693843L, 42693321L, 42693665L, 
                                                  42693653L, 0L), allele = c("G/A/T", "T/C", "G/C", "C/T", "G/A/T", 
                                                                             "NOT FOUND"), refsnp_id = c("rs778598915", "rs11541159", "rs397514502", 
                                                                                                         "rs762949801", "rs776304817", "NOT FOUND")), row.names = c("s3a", "s3b", "s3c", "s3d", "s3e", "s3f"), class = "data.frame")

创建一个包含两个新变量的新df：

new_df <- data.frame(
Ref = str_extract(df$allele, "\\w(?=/)"), 
Var = str_extract(df$allele, "(?<=/)\\w.*")
  )
new_df
   Ref  Var
1    G  A/T
2    T    C
3    G    C
4    C    T
5    G  A/T
6 <NA> <NA>

然后将new_df与df进行列绑定（减去现在已经过时的allele列）：

cbind(df[,-4], new_df)
    associated_gene  chr_name chrom_start   refsnp_id  Ref  Var
s3a            <NA>        22    42693910 rs778598915    G  A/T
s3b            <NA>        22    42693843  rs11541159    T    C
s3c          A4GALT        22    42693321 rs397514502    G    C
s3d            <NA>        22    42693665 rs762949801    C    T
s3e            <NA>        22    42693653 rs776304817    G  A/T
s3f       NOT FOUND NOT FOUND           0   NOT FOUND <NA> <NA>

Answer 3

根据最后一行的输出方式，可以使用以下一种：

使用tidyr::separate

library(tidyr)
separate(df, allele,into = c("Ref", "Var"),sep = "/",extra = "merge",fill = "right")

#   associated_gene  chr_name chrom_start        Ref  Var   refsnp_id
#s3a            <NA>        22    42693910         G  A/T rs778598915
#s3b            <NA>        22    42693843         T    C  rs11541159
#s3c          A4GALT        22    42693321         G    C rs397514502
#s3d            <NA>        22    42693665         C    T rs762949801
#s3e            <NA>        22    42693653         G  A/T rs776304817
#s3f       NOT FOUND NOT FOUND           0 NOT FOUND <NA>   NOT FOUND

或与stringr::str_match

stringr::str_match(df$allele, "(.*?)/(.*)")[,c(2, 3)]

#     [,1] [,2] 
#[1,] "G"  "A/T"
#[2,] "T"  "C"  
#[3,] "G"  "C"  
#[4,] "C"  "T"  
#[5,] "G"  "A/T"
#[6,] NA   NA

在分隔符“ /”的首次出现之前和之后将一列分为两部分

3 个答案: