R 从字符串中提取字符

如何将所有字符提取到指定的字符？例如，我想提取“.”（句点）之前的所有内容：

我试过：

substr(a,1,".")

但它似乎不起作用

---

有什么想法吗？

这里有一个非常基本的方法：

sapply(strsplit(a, "\\."), `[[`, 1)
# [1] "asdasd"     "segssddfge" "se"

还有一个：

sub(".sss", "", a, fixed = TRUE)
# [1] "asdasd"     "segssddfge" "se" 
## OR sub("(.*)\\..*", "\\1", a) 
## And possibly other variations

---

使用

sub

：

# match a "." (escape with "\" to search for "." as a normal "." 
# means "any character") followed by 0 to any amount of characters
# until the end of the string and replace with nothing ("")
sub("\\..*$", "", a)

使用

subtr

和

gregexpr

（假设向量中只有1个

，并且所有字符串中都有明确的匹配）

---

这里尝试使用

gsub

gsub(pattern='(.*)[.](.*)','\\1', c("asdasd.sss","segssddfge.sss","se.sss"))
[1] "asdasd"     "segssddfge" "se"        

---

这是一个csv文件，所以应该只有一个“@Arun，忘记添加“

fixed=TRUE

”，这是我基于OP数据的假设（可能是错误的）而采取的方法。谢谢

# match a "." (escape with "\" to search for "." as a normal "." 
# means "any character") followed by 0 to any amount of characters
# until the end of the string and replace with nothing ("")
sub("\\..*$", "", a)

# get the match position of a "." for every string in "a" (returns a list)
# unlist it and get the substring of each from 1 to match.position - 1
substr(a, 1, unlist(gregexpr("\\.", a)) - 1)

gsub(pattern='(.*)[.](.*)','\\1', c("asdasd.sss","segssddfge.sss","se.sss"))
[1] "asdasd"     "segssddfge" "se"