I have a data.frame df with a character column text that contains text. From that column, I would like to extract all percentage numbers (say, 1.2% and -2.3%) but not the ordinary numbers (say, 123 and 1.2) into a character vector.
A small example:
df <- data.frame(text = c("this text is 1.3% this is +1.4% and this -1.5%", "this text is 123.3% this 123.3 and this 1234.5")) Required output:
[1] "1.3%" "-1.4%" "-1.5%" "123.3%" Is that possible?
Probably not the most robust general-purpose solution, but works for your example:
unlist(stringr::str_extract_all(df$text, "[+\\-]?[0-9\\.]+%")) #[1] "1.3%" "+1.4%" "-1.5%" "123.3%" ## or using R's native forward pipe operator, since R 4.1.0 stringr::str_extract_all(df$text, "[+\\-]?[0-9\\.]+%") |> unlist() #[1] "1.3%" "+1.4%" "-1.5%" "123.3%" This meets your expected output (i.e., a character vector). But in case you are thinking about storing the results to a new data frame column, you don't really want to unlist(). Just do:
df$percentages <- stringr::str_extract_all(df$text, "[+\\-]?[0-9\\.]+%") df # text percentages #1 this text is 1.3% this is +1.4% and this -1.5% 1.3%, +1.4%, -1.5% #2 this text is 123.3% this 123.3 and this 1234.5 123.3% The new column percentages itself is a list:
str(df$percentages) #List of 2 # $ : chr [1:3] "1.3%" "+1.4%" "-1.5%" # $ : chr "123.3%" Here is an alternative tidyverse way:
First we extract the numbers with parse_number from readr package,and then within an ifelse statement we specify the combination of number and percent. Finally pull for vector output.
library(tidyverse) df %>% mutate(x = parse_number(text), x = ifelse(str_detect(text, "%"), paste0(x,"%"), NA_character_)) %>% pull(x) 1] "1.3%" "1.4%" "-1.5%" "123.3%" NA NA 