stringr:从字符串中删除圆括号和方括号 [英] stringr: Removing Parentheses and Brackets from string

问题描述

我想从字符串中删除圆括号和方括号.我得到了所需的结果，但正在寻找更紧凑的方法.

I want to remove parentheses and brackets from a string. I got the required result, however looking for more compact method.

Test <- c("-0.158)", "0.426)", "1.01)", "1.6)", "2.18)", "2.77]")

stringr::str_replace(
  string = Test
  , pattern = "\\)"
  , replacement = ""
)

# [1] "-0.158" "0.426"  "1.01"   "1.6"    "2.18"   "2.77]" 
stringr::str_replace(
    string = Test
  , pattern = "\\]"
  , replacement = ""
)

# [1] "-0.158)" "0.426)"  "1.01)"   "1.6)"    "2.18)"   "2.77"  
stringr::str_replace(
  string = stringr::str_replace(
            string = Test
            , pattern = "\\]"
            , replacement = ""
          )
  , pattern = "\\)"
  , replacement = ""
)

# [1] "-0.158" "0.426"  "1.01"   "1.6"    "2.18"   "2.77" 

想知道是否可以通过这样的单个命令获得它

Wonder if it can be obtained with a single command something like this

stringr::str_replace(
  string = Test
  , pattern = "\\)^]"
  , replacement = ""
)

已编辑

找到了一个非常简单的解决方案

Found a very simple solution

readr::parse_number(Test) 
[1] -0.158 0.426 1.010 1.600 2.180 2.770.

推荐答案

我们可以使用|

gsub("\\)|\\]", "", Test)
#[1] "-0.158" "0.426"  "1.01"   "1.6"    "2.18"   "2.77"  

或者不是转义，而是将括号放在 []

or instead of escaping place the brackets inside the []

gsub("[][()]", "", Test)
#[1] "-0.158" "0.426"  "1.01"   "1.6"    "2.18"   "2.77"  

<小时>

如果我们想要进行提取而不是删除，请使用 base R 中的 gregexpr/regmatches 或 stringr<中的 str_extract/code> 检查数字可以以 - 开头并包含 的模式.

---

If we want to do the extract instead of removing use either gregexpr/regmatches from base R or str_extract from stringr to check for patterns where a number could start with - and include .

library(stringr)
str_extract(Test, "-?[0-9.]+")
#[1] "-0.158" "0.426"  "1.01"   "1.6"    "2.18"   "2.77"  

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