如何在字符串中查找模式并将其提取为数据框的新列

Question

我有一个如下图的数据框：

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% 1ST $100000 1.1% BALANCE")

[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% 1ST $100000 1.1% BALANCE"   

一般来说，我可以说所有的 cmets 都是这种形式： x.y% 1ST abcdef AND a.b% BALANCE （它可以有第四个数字作为奖励。这个数据框有 52000 行，所以绝对不可能捕获各种 cmets，但总的来说，我上面提到的格式是一个很好的起始格式）。 我想知道如何提取这些 cmets 中的每个数字并将它们保存为数据框格式。例如，我想要一个这样的数据框：

    First%   cut-off second%  Bonus 
1    3.2     100000    1.1      NA
2    3.3     100000    1.2      3000
3    NA        NA      NA       NA
4    3.3     100000    1.2      NA
5    3.3     100000    1.2      NA
6    3.2     100000    1.1      NA

我大概可以使用str_subset 和这样的模式：

str_subset(data$comment, "(\\d\\.\\d)% 1ST \\$(\\d{3,8}) AND (\\d\\.\\d)% BALANCE \\w+")

但是str_subset 完全返回字符串，我不确定如何单独保存每个部分。我也查了这个链接Extract a regular expression match，看来str_extract是个不错的选择；但是，仅用于检查一种模式。就我而言，如果我提取\\d\\.\\d%，则不清楚是提取第一个数字还是提取第二个数字。

谢谢

Answer 1

您可以尝试以下方法：

library(tidyverse)

df %>%
  extract(col, c('First', 'cut-off', 'Second'), 
               '(\\d+.*?)% 1ST\\s*\\$(\\d+).*?(\\d+.*?)%.*?', remove = FALSE) %>%
  mutate(Bonus = str_extract(col, '\\d+(?=\\sBONUS)')) %>%
  select(-col)

#   First cut-off Second Bonus
#1   3.2  100000    1.1  <NA>
#2   3.3  100000    1.2  3000
#3  <NA>    <NA>   <NA>  <NA>
#4   3.3  100000    1.2  <NA>
#5   3.3  100000    1.2  <NA>
#6   3.2  100000    1.1  <NA>

数据

df <- data.frame(col = 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% 1ST $100000 1.1% BALANCE"))

Answer 2

这行得通吗：

st <- 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% 1ST $100000 1.1% BALANCE")
dat <- plyr::ldply(str_extract_all(st, '[0-9]\\.[0-9]|[0-9]{4,6}'), rbind)
names(dat) <- c('First%', 'cut-off', 'second%',  'Bonus')
dat <- dat %>% type.convert(as.is = T)
dat$`First%`[dat$`First%`>10] <- NA
dat
  First% cut-off second% Bonus
1    3.2  100000     1.1    NA
2    3.3  100000     1.2  3000
3     NA      NA      NA    NA
4    3.3  100000     1.2    NA
5    3.3  100000     1.2    NA
6    3.2  100000     1.1    NA

Answer 3

我选择了 dplyr 管道。结果并不完美，但它是一种更流畅的方式来逐个处理正则表达式匹配而不是长字符串。

library(tidyverse)

data_ <- data.frame(dat = 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% 1ST $100000 1.1% BALANCE"))

data_ %>%
  mutate(first_perc = str_extract(dat, "\\d+\\.\\d%?"),
         cut_off = str_extract(dat, "[^\\d+\\.\\d% 1ST.]\\d+"),
         second_perc = str_extract(dat, "[^\\d+\\.\\d% 1ST?\\d+]?\\d+\\.\\d%"),
         bonus = str_extract(dat, "[^\\d+\\.\\d% 1ST?\\d+\\d+\\.\\d%\\D+]?\\d\\d+"))

我可能会建议使用条件返回不完整的行，例如 case_when 参数，具体取决于各种变异行捕获或未捕获的内容。