從字符串中提取字符

Question

如何提取指定字符的所有字符？舉例來說，我想提取「。」之前的所有內容。 （期）：

a<-c("asdasd.sss","segssddfge.sss","se.sss") 

我想回去：

asdasd segssddfge se 

我想：

substr(a,1,".") 

，但它似乎並沒有工作。

有什麼想法？

Answer 1

這裏是一個非常基本的方法：

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

而另：

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

Answer 2

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（假設只有1 .有，有一個在所有條紋確定匹配向量中的ngs）。

# 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) 

Answer 3

這裏使用gsub

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