Create new column based on last 2 digits of values in another column

Question

Should be simple enough but it's become a difficult issue to solve. I have data that are grouped by their trailing decimals (a product of an upstream data source). For example, the data can be grouped for group "3" as 0.00003 while the data for group "10" is 24.00010. However, when I run both my regexpr code and my str_sub code it's as if R doesn't treat the last 0 as important.

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Example Data

df <- data.frame(a = c(0.00003, 0.00010, 24.00003, 24.00010))

print(df)
         a
1  0.00003
2  0.00010
3 24.00003
4 24.00010

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Desired Output

         a   group
1  0.00003 group03
2  0.00010 group10
3 24.00003 group03
4 24.00010 group10

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Failed Attempt 1

df %>% mutate(group = paste0("group", regmatches(a, regexpr("(\\d{2}$)", a))))         
         a   group
1  0.00003 group03
2  0.00010 group01
3 24.00003 group03
4 24.00010 group01

This failure is peculiar as this works when I check it on: https://regexr.com/, using (\d{2}$)

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Failed Attempt 2

df %>% mutate(group = paste0("group", str_sub(a, start = -2)))
         a   group
1  0.00003 group03
2  0.00010 group01
3 24.00003 group03
4 24.00010 group01

Answer 1

The key here is that when you substring or extract with regex, you are converting the number into a string. The string, however does not keep the format you are expecting.

library(tidyverse)

tibble(a = c(0.00003, 0.00010, 24.00003, 24.00010)) %>%
  mutate(group1 = paste0("group", str_extract(sprintf("%.5f", a), "\\d{2}$")),
         group2 = paste0("group", str_extract(a, "\\d{2}$")),
         sprint_char = sprintf("%.5f", a),
         char = as.character(a))
#> # A tibble: 4 x 5
#>          a group1  group2  sprint_char char    
#>      <dbl> <chr>   <chr>   <chr>       <chr>   
#> 1  0.00003 group03 group05 0.00003     3e-05   
#> 2  0.0001  group10 group04 0.00010     1e-04   
#> 3 24.0     group03 group03 24.00003    24.00003
#> 4 24.0     group10 group01 24.00010    24.0001

See here that as.character(a) does not maintain the same structure as a. You can instead set the formatting with sprintf, and then extract the text that you want.

Answer 2

We can convert to character and use str_sub. Also, make sure the options are set

options(scipen = 999)
library(stringr)
library(dplyr)
df %>% 
   mutate(group = paste0("group", str_sub(sprintf("%2.5f", a), start = -2)))
#        a   group
#1  0.00003 group03
#2  0.00010 group10
#3 24.00003 group03
#4 24.00010 group10