R str_replace / remove access zeros in string after the decimal seperator, (zero following dot[1-9])-CodePudding

I have vector of strings containing numbers. Within these strings, I would like to remove all access zeros behind the decimal separator. So I tried mutate_all(funs(str_replace(., ".00", ""))).

This works for the first number in my vector: v <- c("bla 500.00", "bla 1.20", "bla 1.10", "bla 2.34").

For the rest I would not like to hard code mutate_all(funs(str_replace(., ".10", ".1")))%>%mutate_all(funs(str_replace(., ".20", ".2")))%>% ..., but use some kind of smart regex, which automatically does the job. So, removing every zero which is behind ".non-zero-integer" (dot non-zero-integer), while keeping the ".non-zero-integer" the same.

CodePudding user response：

You can do this. Note that mutate_all has been superseded by mutate(across()) (see the docs).

v <- c("bla 500.00", "bla 1.20", "bla 1.10", "bla 2.34")
dat  <- data.frame(v = v)

dat  |>
    mutate(
        across(v, \(x) stringr::str_replace(x, "\\.\\d{2}", ""))
    )

#         v
# 1 bla 500
# 2   bla 1
# 3   bla 1
# 4   bla 2

Base R

gsub("\\.\\d{2}", "", dat$v)
# "bla 500" "bla 1"   "bla 1"   "bla 2"

CodePudding user response：

You could try to find:

\b(\d )(?:\.0 \b|(\.\d ?))0*\b

And replace with \1\2. See an online demo

\b - Word-boundary;
(\d ) - Capture trailing digits upto;
(?: - Open a non-capture group;
- \.0 \b - 1 zero's upto word-boundary;
- | - Or;
- (\.\d ?) - A nested 2nd capture group to match a dot followed by a digit and 0 (lazy) digits;
- ) - Close non-capture group;
0* - 0 (greedy) digits;
\b - Trailing word-boundary.

CodePudding user response：

How about this base R solution:

v <- c("bla 500.00", "bla 1.20", "bla 1.10", "bla 2.34")
v <- gsub("(.*[\\.1-9])0*$", "\\1", v)
v <- gsub("\\.$", "", v)
v
#> [1] "bla 500"  "bla 1.2"  "bla 1.1"  "bla 2.34"

^{Created on 2022-07-13 by the reprex package (v2.0.1)} The first gsub() function removes trailing zeros and the second gsub() removes the period if it is at the end of the number, so it changes 500. into 500