R: Replace last 3 zeroes with K in a column - r

Question

I am trying to replace last 3 000 with K in a column in the dataframe
eg:
data <- data.frame(abc = c(1000, 100000, 450000))
abc <- 1000
then abc <- 1K
if
abc <- 100000
then abc <- 100K
gsub or regex replaces the first 3 zeroes
I tried this:
lapply(data$abc, gsub, pattern = "000", replacement = "K", fixed = TRUE)
Also, how can I make it work on an interval like :
data <- data.frame(abc = c("150000-250000", "100000-150000", "250000K+"))

Answer 1

An option is to use %/% with 1000 and paste the "K"
library(dplyr)
library(stringr)
data %>%
mutate(abc = str_c(abc %/% 1000, "K"))
Or using sub, match the 3 zeros at the end ($) of the string and replace with "K"
options(scipen = 999)
sub("0{3}$", "K", data$abc)
#[1] "1K" "100K" "450K"
If we have a different string with interval, then change the pattern to match 3 zeros at either at the end ($) or before a - and replace with "K"
gsub("0{3}(?=-|$)", "K", "150000-250000", perl = TRUE)
#[1] "150K-250K"

Answer 2

Here is a slight modification of your code. format is to turn off the scientific notation. sapply makes the output becomes a vector. 000$ means only match those at the end.
data <- data.frame(abc = c(1000, 100000, 450000))
data$abc <- format(data$abc, scientific = FALSE)
gsub(pattern = "000$", replacement = "K", data$abc)
# [1] " 1K" "100K" "450K"