How to filter n consecutive days from a start date in a vector of dates across multiple years in R?

Question

I'm stucked trying to create a code that filters 5 consecutive days from a vector of dates starting from a given start date, across multiple years.

Example:

v = seq.Date(as.Date("2000-01-01"), as.Date("2020-12-31"), 1)

initial_date = as.Date("01-01") #(mm:dd)

Expected output:

2000-01-01 #(YYYY:mm:dd)
2000-01-02
2000-01-03
2000-01-04
2000-01-05
2001-01-01
2001-01-02
2001-01-03
2001-01-04
2001-01-05
...
2020-01-01
2020-01-02
2020-01-03
2020-01-04
2020-01-05

If the initial date is, for example, 12-31 (mm:dd), the result should be:

2000-12-31
2001-01-01
2001-01-02
2001-01-03
2001-01-04
2001-31-12
2002-01-01
2002-01-02
2002-01-03
2002-01-04
...
2019-12-31
2020-01-01
2020-01-02
2020-01-03
2020-01-04
2020-12-31

Any tips?

Answer 1

You could use str_detect() to find the starting value and use the index to find the next 4 rows.

library(dplyr)

inital_date = "12-31"

tibble(dates = v)|> 
  mutate(index = stringr::str_detect(v, "12-31")) |> 
  slice(sort(which(index) + rep(0:4, each = sum(index)))) 
      select(-index)

Output:

# A tibble: 97 × 1
   dates     
   <date>    
 1 2000-12-31
 2 2001-01-01
 3 2001-01-02
 4 2001-01-03
 5 2001-01-04
 6 2001-12-31
 7 2002-01-01
 8 2002-01-02
 9 2002-01-03
10 2002-01-04

Answer 2

n <- 5
v <- seq.Date(as.Date("2000-01-01"), as.Date("2020-12-31"), 1)
initial_date <- "12-31"
tmp <- rowSums(
  sapply(seq_len(n) - 1, function(z)
    (v-z) %in% v &
      format(v-z, format = "%m-%d") == initial_date)
  ) > 0
v[tmp]
#   [1] "2000-12-31" "2001-01-01" "2001-01-02" "2001-01-03" "2001-01-04" "2001-12-31" "2002-01-01" "2002-01-02" "2002-01-03"
#  [10] "2002-01-04" "2002-12-31" "2003-01-01" "2003-01-02" "2003-01-03" "2003-01-04" "2003-12-31" "2004-01-01" "2004-01-02"
#  [19] "2004-01-03" "2004-01-04" "2004-12-31" "2005-01-01" "2005-01-02" "2005-01-03" "2005-01-04" "2005-12-31" "2006-01-01"
#  [28] "2006-01-02" "2006-01-03" "2006-01-04" "2006-12-31" "2007-01-01" "2007-01-02" "2007-01-03" "2007-01-04" "2007-12-31"
#  [37] "2008-01-01" "2008-01-02" "2008-01-03" "2008-01-04" "2008-12-31" "2009-01-01" "2009-01-02" "2009-01-03" "2009-01-04"
#  [46] "2009-12-31" "2010-01-01" "2010-01-02" "2010-01-03" "2010-01-04" "2010-12-31" "2011-01-01" "2011-01-02" "2011-01-03"
#  [55] "2011-01-04" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04" "2012-12-31" "2013-01-01" "2013-01-02"
#  [64] "2013-01-03" "2013-01-04" "2013-12-31" "2014-01-01" "2014-01-02" "2014-01-03" "2014-01-04" "2014-12-31" "2015-01-01"
#  [73] "2015-01-02" "2015-01-03" "2015-01-04" "2015-12-31" "2016-01-01" "2016-01-02" "2016-01-03" "2016-01-04" "2016-12-31"
#  [82] "2017-01-01" "2017-01-02" "2017-01-03" "2017-01-04" "2017-12-31" "2018-01-01" "2018-01-02" "2018-01-03" "2018-01-04"
#  [91] "2018-12-31" "2019-01-01" "2019-01-02" "2019-01-03" "2019-01-04" "2019-12-31" "2020-01-01" "2020-01-02" "2020-01-03"
# [100] "2020-01-04" "2020-12-31"

• (v-z) %in% v is necessary, since otherwise it'll "match" 1999-12-31 which is not in the original v
• format(v-z, format = "%m-%d") checks to see if z days ago (for all in v) matches the initial_date
• the return value of sapply is a length(v)-row, 5-column matrix of logicals, where each row indicates if that row'th v should be retained (is within 5 days of initial_date). To determine if any on a row are true, we use rowSums(.) > 0, which will return a logical vector the same length as v.

Note: I tried doing this all at once using outer(v, 0:4, `-`), but matrices in R don't tend to keep the "Date" class, and working around that (though it can be done) is a bit onerous. The solution is to produce a matrix that isn't classes as a date, ergo doing the comparison within sapply. I tried to reduce the number of "loops" as much as possible, ergo iterating over 0:4 instead of v.