删除"."之后的部分字符串.

Question

我正在使用NCBI参考序列登录号,如变量a:

a <- c("NM_020506.1","NM_020519.1","NM_001030297.2","NM_010281.2","NM_011419.3", "NM_053155.2")  

要获得从biomart包我需要删除的信息.1,.2登录号等设备中后.我通常使用以下代码执行此操作:

b <- sub("..*", "", a)

# [1] "" "" "" "" "" ""

但正如您所看到的,这不是这个变量的正确方法.谁能帮我这个？

Answer 1

你只需要逃避这个时期:

a <- c("NM_020506.1","NM_020519.1","NM_001030297.2","NM_010281.2","NM_011419.3", "NM_053155.2")

gsub("\\..*","",a)
[1] "NM_020506"    "NM_020519"    "NM_001030297" "NM_010281"    "NM_011419"    "NM_053155" 

Answer 2

我们可以假装它们是文件名并删除扩展名:

tools::file_path_sans_ext(a)
# [1] "NM_020506"    "NM_020519"    "NM_001030297" "NM_010281"    "NM_011419"    "NM_053155"

Answer 3

你可以这样做:

sub("*\\.[0-9]", "", a)

要么

library(stringr)
str_sub(a, start=1, end=-3)

Answer 4

如果字符串应该是固定长度的，那么可以使用substrfrom base R。但是，我们可以得到.with的位置regexpr并在substr

substr(a, 1, regexpr("\\.", a)-1)
#[1] "NM_020506"    "NM_020519"    "NM_001030297" "NM_010281"    "NM_011419"    "NM_053155"   

Answer 5

我们可以使用前瞻正则表达式来提取之前的字符串.。

library(stringr)

str_extract(a, ".*(?=\\.)")
[1] "NM_020506"    "NM_020519"    "NM_001030297" "NM_010281"   
[5] "NM_011419"    "NM_053155"