Reverse/backward na.approx

Question

I have a date vector with leading NAs and I'd like to generate an approximate sequence for these NAs using na.approx from package zoo.

na.approx does not work for leading NAs:

x <- as.Date(c(rep(NA,3),"1992-01-16","1992-04-16","1992-07-16",
"1992-10-16","1993-01-15","1993-04-16","1993-07-17"))
as.Date(na.approx(x,na.rm=FALSE))

[1] NA           NA           NA           "1992-01-16" "1992-04-16" 
1992-07-16" "1992-10-16" "1993-01-15" "1993-04-16" "1993-07-17"

I thought that I could reverse my vector using rev but I still get NAs

as.Date(na.approx(rev(x),na.rm=FALSE))

 [1] "1993-07-17" "1993-04-16" "1993-01-15" "1992-10-16" "1992-07-16" 
"1992-04-16" "1992-01-16" NA           NA           NA

Any ideas?

`as.Date(na.approx(x,na.rm=FALSE, rule=2))` gets a bit closer, but it seems like you want it to create a sequence outside the range? Actuallt try ` as.Date(na.spline(x,na.rm=FALSE))` — user20650, Mar 25 '17 at 15:35
Just saw your edit. Post your `na.spline` answer and I will accept it. Thank you. — Pierre Lapointe, Mar 25 '17 at 15:48

Pierre Lapointe · Accepted Answer · 2017-03-27T16:12:59.200

2

Found my answer. na.spline does a good job with lots of data. In the example above, I have few dates which causes the approximation to drift. However, in my real life example, there is no drift.

as.Date(na.spline(x,na.rm=FALSE))
 [1] "1993-07-17" "1993-04-16" "1993-01-15" "1992-10-16" "1992-07-16" 
"1992-04-16" "1992-01-16" "1991-10-15" "1991-07-13" "1991-04-06"

edited Mar 27 '17 at 16:12

answered Mar 25 '17 at 15:47

Pierre Lapointe

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I think `na.spline` is the answer, so this should get the tick (ps you dont need to rev) – user20650 Mar 25 '17 at 16:08
1

I can't tick my own answer for 48 hours. I'll leave it like that for now so other users know the question has been answered. – Pierre Lapointe Mar 25 '17 at 16:10

score 2 · Answer 2 · answered Mar 25 '17 at 16:04

na.approx requires a rule to be passed for values outside the min or max value of your vector. If rule=2 is used then the missing values are imputed with the nearest value.

as.Date(na.approx(x,na.rm=FALSE, rule=2))
# [1] "1992-01-16" "1992-01-16" "1992-01-16" "1992-01-16" "1992-04-16" "1992-07-16" "1992-10-16" "1993-01-15"
#  [9] "1993-04-16" "1993-07-17"

As an alternative you can use na.spline (as in your answer). You mention it can get a bit wild so you can write a function to impute the values based on the time difference between your measures. I use the first non-missing difference here

add_leading_seq_dates <- function(x) {
        first_non_missing = which.min(is.na(x))
        first_day_diff = na.omit(diff(x))[1]
        no_of_leadng_missing = first_non_missing - 1
        input_dates = x[first_non_missing] - cumsum(rep(first_day_diff, no_of_leadng_missing)) 
        x[is.na(x)] = rev(input_dates)
        x
}

add_leading_seq_dates(x)

# [1] "1991-04-18" "1991-07-18" "1991-10-17" "1992-01-16" "1992-04-16"
# [6] "1992-07-16" "1992-10-16" "1993-01-15" "1993-04-16" "1993-07-17"

diff(add_leading_seq_dates(x))
# Time differences in days
# [1] 91 91 91 91 91 92 91 91 92

Reverse/backward na.approx

2 Answers2