R: Replace last 3 zeroes with K in a column

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+"))

akrun · Accepted Answer · 2019-05-31T18:32:38.517

4

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"

edited May 31 '19 at 18:32

answered May 31 '19 at 18:23

akrun

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100000 becomes 1e+05 – Bruce Wayne May 31 '19 at 18:25
Also, how can I make it work on an interval like : 150000-250000 – Bruce Wayne May 31 '19 at 18:27
@BruceWayne The scientific format can be changed with `options` (updated). regardng the second option we need `gsub` and pattern should be slightly changed – akrun May 31 '19 at 18:28
Could you help me with the interval part please? – Bruce Wayne May 31 '19 at 18:29
1

@BruceWayne Updated the post – akrun May 31 '19 at 18:30
@akrun this will mean if you had 1000:2000, then you will need to change the regex, i would rather prefer `0{3}(?!0)` instead – Onyambu May 31 '19 at 18:32
@Onyambu Sure, I thought the OP had only a single pattern or `gsub("0{3}(?=[:-]|$)", "K", "1000:2000", perl = TRUE) [1] "1K:2K"` – akrun May 31 '19 at 18:34
1

@BruceWayne Please include all your patterns in the question. This was based on the initial pattern showed – akrun May 31 '19 at 18:37
https://stackoverflow.com/questions/56400344/r-fuzzy-merge-two-data-frame can you help with this question too please? – Bruce Wayne May 31 '19 at 18:55

www · Answer 2 · 2019-05-31T19:00:24.317

1

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"

edited May 31 '19 at 19:00

answered May 31 '19 at 18:27

www

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1

`gsub` is vectorized, whats the purpose of `sapply here? – Onyambu May 31 '19 at 18:29
1

@Onyambu Good point. I was affected by the `lapply` OP used. I have updated my solution now. Thanks. – www May 31 '19 at 19:00

R: Replace last 3 zeroes with K in a column

2 Answers2