I have a list of bigrams that I would like to filter. Want to remove all bigrams with numbers, except for those that contain 2021 or 2020
cat <- seq(1,7)
bg <- c("2021 test", "z1 trip", "00034 job","18 mar", "2021 day","business unit", "cycle work")
df <- data.frame(cat,bg)
I have tried nested grepl, but is does not work
df %>% filter(!grepl("[[:digit:]]",!grepl("2020|2021",bg)))
The expected output is:
We could use a very simple regex pattern to check if there 4 digit number (\\b\\d{4}]]b) or (|) if there are no digits from start (^) to end ($) of the string
library(dplyr)
library(stringr)
df %>%
filter(str_detect(bg, '\\b\\d{4}\\b|^[^0-9]+$')) %>%
mutate(cat = row_number())
output
# cat bg
#1 1 2021 test
#2 2 2021 day
#3 3 business unit
#4 4 cycle work
If there are other digits such as '0004', and want only those starts with '20'
df %>%
filter(str_detect(bg, '\\b20\\d{2}\\b|^[^0-9]+$')) %>%
mutate(cat = row_number())
A simple way to do this could be:
subset(df, grepl("2021|2020", bg)|!grepl("\\d", bg))
# cat bg
# 1 1 2021 test
# 5 5 2021 day
# 6 6 business unit
# 7 7 cycle work