正则表达式提取“--!!”之后的所有文本在 R dplyr

Question

我正在尝试在 R 中使用 dplyr 在以下示例中由变量 name 的某些实例过滤的数据帧中的变量字符串之后提取子字符串。我试图将所需的结果传递给一个名为income_rent 的新变量。

我是正则表达式的新手。我的尝试是：

income_cashrent <- v18 %>% 
filter(str_detect(name, "B25122")) %>% 
mutate(income_rent = str_extract(label, "[^--!!]*$"))

但是，我得到了结果： Error in stri_extract_first_regex(string, pattern, opts_regex = opts(pattern)) : Syntax error in regexp pattern. (U_REGEX_RULE_SYNTAX)

name的前四行是：

Estimate!!Total
Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000
Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent
Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent!!Less than $100

期望的结果是：

[not sure how to indicate an empty result here]
Less than $10,000
Less than $10,000!!With cash rent
Less than $10,000!!With cash rent!!Less than $100

到目前为止，我一直无法调试这个，参考堆栈上的其他正则表达式示例。任何指导都将受到欢迎。提前谢谢大家！

Answer 1

regmatches(vec, gregexpr("(?<=--!!).*", vec, perl = TRUE))
# [[1]]
# character(0)
# [[2]]
# [1] "Less than $10,000"
# [[3]]
# [1] "Less than $10,000!!With cash rent"
# [[4]]
# [1] "Less than $10,000!!With cash rent!!Less than $100"

如果您从这里unlist，您会注意到您“丢失”了第一个条目，不确定这是否有问题。

unlist(regmatches(vec, gregexpr("(?<=--!!).*", vec, perl = TRUE)))
# [1] "Less than $10,000"                                
# [2] "Less than $10,000!!With cash rent"                
# [3] "Less than $10,000!!With cash rent!!Less than $100"

如果这是个问题，那么

vecout <- regmatches(vec, gregexpr("(?<=--!!).*", vec, perl = TRUE))
unlist(replace(vecout, lengths(vecout) < 1, NA))
# [1] NA                                                 
# [2] "Less than $10,000"                                
# [3] "Less than $10,000!!With cash rent"                
# [4] "Less than $10,000!!With cash rent!!Less than $100"

（或者您也可以替换为""。）

---

在dplyr 管道中：

tibble(vec = c("Estimate!!Total",
# "Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000",
# "Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent",
# "Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent!!Less than $100")) %>%
  mutate(out = regmatches(vec, gregexpr("(?<=--!!).*", vec, perl = TRUE)), out = replace(out, lengths(vecout) < 1, NA), out = unlist(out))
+ + # A tibble: 4 x 2
#   vec                                             out                           
#   <chr>                                           <chr>                         
# 1 Estimate!!Total                                 <NA>                          
# 2 Estimate!!Total!!Household income in the past ~ Less than $10,000             
# 3 Estimate!!Total!!Household income in the past ~ Less than $10,000!!With cash ~
# 4 Estimate!!Total!!Household income in the past ~ Less than $10,000!!With cash ~

---

数据：

vec <- c("Estimate!!Total",
"Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000",
"Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent",
"Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent!!Less than $100")

Answer 2

我们可以使用str_extract to extract the characters after the pattern--!!` 使用正则表达式查找

library(stringr)
library(dplyr)
 v18 %>%        
     mutate(income_rent = str_extract(label, "(?<=--!!).*"))                                                                                                                                                label
#1                                                                                                                                    Estimate!!Total
#2                                 Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000
#3                 Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent
#4 Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent!!Less than $100
 #                                       income_rent
#1                                              <NA>
#2                                 Less than $10,000
#3                 Less than $10,000!!With cash rent
#4 Less than $10,000!!With cash rent!!Less than $100

或者另一个选项是str_match

v18$income_rent <-  str_match(v18$label, ".*--!!(.*)")[,2]

数据

v18 <- structure(list(label = c("Estimate!!Total", "Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000", 
"Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent", 
"Estimate!!Total!!Household income in the past 12 months (in 2018 inflation-adjusted dollars) --!!Less than $10,000!!With cash rent!!Less than $100"
)), class = "data.frame", row.names = c(NA, -4L))