删除“。”之后的部分字符串。

问题描述 投票:53回答:4

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")  

要从biomart包中获取信息,我需要在入藏号后删除.1.2等。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?

r regex string bioinformatics biomart
4个回答
74
投票

你只需要逃避这段时间:

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" 

8
投票

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"

5
投票

你可以这样做:

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

要么

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

1
投票

如果字符串应该是固定长度,那么可以使用来自substrbase R。但是,我们可以用.获得regexpr的位置并在substr中使用它

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