R 删除“"；之后的部分字符串&引用；

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")  

但正如您所看到的，这不是该变量的正确方法。有人能帮我吗？

你可以：

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

或

---

你只需要逃离这段时间：

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" 

---

a我们可以假装它们是文件名并删除扩展名：

如果字符串应为固定长度，则可以使用
base R
中的
substr
。但是，我们可以使用
regexpr
获得
的位置，并将其用于
substr

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

备选方案：
stru-replace（a，“\\.[0-9]”，“）
和
stru-replace（a，“\\..*”，“）
说明：对于基本包中的函数（即没有其他包，如
string r
），选项如下所示：b1您不会将其与fixed as TRUE一起使用，因为我们在这里使用正则表达式。
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" 

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

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