如何截断字符串的特定部分(如果存在) 让我们考虑向量跟踪: x <- c("GDP_UK", "GDP_US", "GDP_UK_diff2_L2", "INC","GDP_UK_L2", "GDP_US_level", "INC_UK", "INC_L1", "INC_diff1") x <- gsub(pattern = "_L", replacement = "", x) x <- gsub(pattern = "_diff", replacement = "", x) x <- gsub(pattern = "_level", replacement = "", x)

如何截断字符串的特定部分(如果存在) 让我们考虑向量跟踪: x <- c("GDP_UK", "GDP_US", "GDP_UK_diff2_L2", "INC","GDP_UK_L2", "GDP_US_level", "INC_UK", "INC_L1", "INC_diff1") x <- gsub(pattern = "_L", replacement = "", x) x <- gsub(pattern = "_diff", replacement = "", x) x <- gsub(pattern = "_level", replacement = "", x),r,string,character,R,String,Character,但是,我们将在字符串的末尾使用剩余的数字: "GDP_UK" "GDP_US" "GDP_UK22" "INC" "GDP_UK2" "GDP_US" "INC_UK" "INC2" "INC1" 您要查找的是正则表达式“\u L\\d*”,等等。它匹配下划线、L和零或更多数

但是,我们将在字符串的末尾使用剩余的数字:

 "GDP_UK"   "GDP_US"   "GDP_UK22" "INC"      "GDP_UK2"  "GDP_US"   "INC_UK"   "INC2"     "INC1"  

您要查找的是正则表达式
“\u L\\d*”
,等等。它匹配下划线、L和零或更多数字

全部


x%
str\u replace\u all(“\u L\\d*”,”)%>%
str\u replace\u all(“\u diff\\d*”,”)%>%
str\u replace\u all(“\u level\\d*”,“”)
#>[1]“GDP_英国”“GDP_美国”“GDP_英国”“INC”“GDP_英国”“GDP_美国”“INC_英国”“INC”
#>[9]“公司”
##甚至一次性:
gsub(“|L | diff | level)\\d*”,“”,x)
#>[1]“GDP_英国”“GDP_美国”“GDP_英国”“INC”“GDP_英国”“GDP_美国”“INC_英国”“INC”
#>[9]“公司”

最后一段代码是我的最爱!非常感谢您抽出时间!