How can I get the middle date of three dates in R? - r

Question

I have a datatable with three date columns x, y and z and I am trying to create a new column (new_col) that is the middle date of the three dates in each row once ranked from earliest to latest, i.e., I want the date between the min and max date – please see table below:
x
y
z
new_col
1st Jan 2005
4th May 1998
2nd Mar 2009
1st Jan 2005
9th May 2010
14th Feb 2003
9th Jan 2008
9th Jan 2008
7th Sept 2002
8th Dec 2010
23rd May 2012
8th Dec 2010
So, for rows 1, 2, and 3 I would like the dates from column x, z, and y, respectively. How can I go about this in R? I have used pmin and pmax but I can't isolate the date in the middle
Thanks in advance!

Answer 1

The approach below
coerces the character date strings to numeric type Date as there is no arithmetic with character dates,
finds the position of the "middle" date in each row
and returns the corresponding character string
which eventually becomes new_col.
This can be implemented using apply() on each row using an appropriate function:
df$new_col <- apply(df, 1L, function(x) x[order(lubridate::dmy(x))][2L])
df
x y z new_col
1 1st Jan 2005 4th May 1998 2nd Mar 2009 1st Jan 2005
2 9th May 2010 14th Feb 2003 9th Jan 2008 9th Jan 2008
3 7th Sept 2002 8th Dec 2010 23rd May 2012 8th Dec 2010
Note
This returns the expected result. new_col is a character date string.
However, if the OP intends to continue working with type Date, e.g. doing more arithmetic, I recommend to follow Ben's example and to coerce the whole data.frame to type Date and to stick to it.

Answer 2

First make sure all your dates are "Date" type, you can use dmy from lubridate for this (assumes your data frame is called df):
library(lubridate)
df[] <- lapply(df, dmy)
Next, sort each row in chronological order, and take the middle column (column 2) to be the new_col:
df$new_col <- as.Date(t(apply(df, 1, sort))[,2])
Finally, if you want the result to be displayed in same text format (e.g., "1st Jan 2005" instead of "2005-01-01") then you can use a custom function based on this answer:
library(dplyr)
date_to_text <- function(dates){
dayy <- day(dates)
suff <- case_when(dayy %in% c(11,12,13) ~ "th",
dayy %% 10 == 1 ~ 'st',
dayy %% 10 == 2 ~ 'nd',
dayy %% 10 == 3 ~'rd',
TRUE ~ "th")
paste0(dayy, suff, " ", format(dates, "%b %Y"))
}
df[] <- lapply(df, date_to_text)
Output
x y z new_col
1 1st Jan 2005 4th May 1998 2nd Mar 2009 1st Jan 2005
2 9th May 2010 14th Feb 2003 9th Jan 2008 9th Jan 2008
3 7th Sep 2002 8th Dec 2010 23rd May 2012 8th Dec 2010
Data
df <- structure(list(x = c("1st Jan 2005", "9th May 2010", "7th Sept 2002"
), y = c("4th May 1998", "14th Feb 2003", "8th Dec 2010"), z = c("2nd Mar 2009",
"9th Jan 2008", "23rd May 2012")), class = "data.frame", row.names = c(NA,
-3L))