How to remove text both sides of character?

Question

How can I transform a column of characters written as

c("0 y", "0 m", "23 d", "0 y",  "0 m", "8 d")

into number values

c(0, 0, 23, 0, 0, 0)

example of what I'm talking about

another example that has some single-digit dates

Try `gsub("^.*(\\d+) d.*$", "\\1", x)` where x is you vector of strings — Allan Cameron, Mar 07 '20 at 21:56
It is almost always preferable to add code and data as formatted text to the question than to use pictures of code and data. Also a [minimal reproducible example](https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example/5963610#5963610) is always weclome! Although in this really simple case it might not be really necessary. — dario, Mar 07 '20 at 22:42
the answers never include months or years, its a short term follow up value, hope that clears it up — bdg67, Mar 08 '20 at 02:00

score 0 · Answer 1 · edited Mar 07 '20 at 22:32

0

Assuming y and m are always 0

Oy.date.diff <- c("0 y, 0 m, 12 d", "0 y, 0 m, 13 d", "0 y, 0 m, 12 d", "0 y, 0 m, 15 d") 
as.numeric(gsub(" d", "", gsub("0 y, 0 m, ", "", Oy.date.diff)))
# [1] 12 13 12 15

Note that R does not allow variables (or columns) to begin with a digit so the first character is uppercase letter O.

edited Mar 07 '20 at 22:32

akrun

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answered Mar 07 '20 at 21:57

dcarlson

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akrun · Accepted Answer · 2020-03-07T22:25:41.960

We can use sub to capture the digits before the space followed by 'd'

as.integer(sub(".*\\s(\\d+) d", "\\1", v1))
#[[1] 12 13 12 15 12

Or with regmatches/regexpr

regmatches(v1, regexpr("(\\d+)(?= d$)", v1, perl = TRUE))
#[1] "12" "13" "12" "15" "12"

If we need to convert to all days, then

library(dplyr)
library(tidyr)
tibble(col1 = v1) %>% 
  tidyr::extract(col1, into = c('year', 'month', 'day'),
       "^(\\d+) y, (\\d+) m, (\\d+) d$",  convert = TRUE) %>% 
  transmute(days = year * 365 + month * 30 + day)

data

v1 <- c("0 y, 0 m, 12 d", "0 y, 0 m, 13 d", "0 y, 0 m, 12 d",
        "0 y, 0 m, 15 d", "1 y, 2 m, 12 d")

score 0 · Answer 3 · answered Mar 07 '20 at 22:03

You can try this capturing regex with gsub, which captures any numbers before a " d" and doesn't make any assumptions about the rest of the string:

x <- c("0 y, 0 m, 12 d", "0 y, 0 m, 13 d", "0 y, 0 m, 12 d", "0 y, 0 m, 15 d") 
gsub("^.*(\\d+) d.*$", "\\1", x)
#> [1] "2" "3" "2" "5"

How to remove text both sides of character?

3 Answers3

data