如何将字符串中的所有数字提取为向量

Question

有没有办法将字符串中的所有数字提取为向量？我有一个不遵循任何特定模式的大型数据集，因此使用 extract + regex 模式不一定会提取所有数字。因此，例如对于如下所示的每一行数据框：

c("3.2% 1ST $100000 AND 1.1% BALANCE", "3.3% 1ST $100000 AND 1.2% BALANCE AND $3000 BONUS FULL PRICE ONLY", 
"$4000", "3.3% 1ST $100000 AND 1.2% BALANCE", "3.3% 1ST $100000 AND 1.2% BALANCE", 
"3.2 - $100000")

[1] "3.2% 1ST $100000 AND 1.1% BALANCE"                                
[2] "3.3% 1ST $100000 AND 1.2% BALANCE AND $3000 BONUS FULL PRICE ONLY"
[3] "$4000"                                                            
[4] "3.3% 1ST $100000 AND 1.2% BALANCE"                                
[5] "3.3% 1ST $100000 AND 1.2% BALANCE"                                
[6] "3.2 - $100000"   

我想要这样的输出：

[1] "3.2 100000 1.1"                                
[2] "3.3 100000 1.2 3000"
[3] "4000"                                                            
[4] "3.3 100000 1.2 "                                
[5] "3.3 100000 1.2 "                                
[6] "3.2 100000 "   

我查看了资源，找到了这个链接：https://statisticsglobe.com/extract-numbers-from-character-string-vector-in-r

regmatches(x, gregexpr("[[:digit:]]+", x))

上面的函数似乎有效，但它不能同时对各种数字执行此任务。我知道"[[:digit:]]+" 只查找整数，但我们如何更改它以使其涵盖各种数字？

Answer 1

我们还需要在匹配模式中添加.

sapply(regmatches(x, gregexpr("\\b[[:digit:].]+\\b", x)), paste, collapse= ' ')
#[1] "3.2 100000 1.1"    
#[2] "3.3 100000 1.2 3000" 
#[3] "4000"              
#[4] "3.3 100000 1.2"   
#[5] "3.3 100000 1.2"     
#[6] "3.2 100000"   

Answer 2

Akrun 的答案是完美的，但只是为了添加另一个解决方案，使用一个包来创建我最近发现的正则表达式模式。

library(stringr)
library(rebus)
library(magrittr)

pattern = one_or_more(DIGIT) %R% optional(DOT) %R% optional(one_or_more(DIGIT))

str_remove(x, "1ST") %>% 
str_match_all( pattern = pattern) %>% 
  lapply( function(x) paste(as.vector(x), collapse = " ")) %>% 
  unlist()

Answer 3

您可以使用负前瞻正则表达式：

stringr::str_extract_all(x, '\\d+(\\.\\d+)?(?![A-Z])')

#[[1]]
#[1] "3.2"    "100000" "1.1"   

#[[2]]
#[1] "3.3"    "100000" "1.2"    "3000"  

#[[3]]
#[1] "4000"

#[[4]]
#[1] "3.3"    "100000" "1.2"   

#[[5]]
#[1] "3.3"    "100000" "1.2"   

#[[6]]
#[1] "3.2"    "100000"

如果您希望输出为一个字符串：

sapply(stringr::str_extract_all(x, '\\d+(\\.\\d+)?(?![A-Z])'), paste, collapse = ' ')
#[1] "3.2 100000 1.1"      "3.3 100000 1.2 3000" "4000"               
#[4] "3.3 100000 1.2"      "3.3 100000 1.2"      "3.2 100000"