Difference between nearest times in time two time series

Question

I have a time series on which two events (x and y) occur irregularly. I'm trying to write a bit of script that describes the time from x to the closest y event.

For example:

Time         x          y
11:01:00     1          0
11:03:04     0          1
11:03:34     0          0
11:06:12     1          1
11:12:00     0          0

I'm trying to create a vector where each row is the time from each '1' in y to the nearest '1' in the x vector.

So the above would return:

diff
02:04   (closest point is the previous row here)
0       (occurred on the same row so time difference is 0)
05:48   (occurred on the previous row)

reproducible example:

time<-c("11:01:00","11:03:04","11:03:34","11:06:12","11:12:00")
x<-c(1,0,0,1,0)
y<-c(0,1,0,1,0)
df<-data.frame(time,x,y)

I'm not really sure how to go about this and any help would be appreciated!

Answer 1

We can do in a couple of steps:

First by converting your data into POSIXct format so we can use arithmatic on the time column

df$time <- as.POSIXct(df$time, format = "%H:%M:%S")

Then we create two new columns, with the index of x where x is 1. I'm assuming your data is in time order here.

df$nextx <- ifelse(df$x == 1, which(df$x == 1), NA)
df$prevx <- rev(ifelse(df$x == 1, which(df$x == 1), NA))

By using tidyr::fill, we fill in the xx, to get the next and previous x for each y:

library(tidyr)
df <- df %>% fill(nextx, rev(prevx))

Then we use pmin to find the minimum distance at each row:

x = pmin(abs(df$time - df$time[df$nextx]), abs(df$time - df$time[df$prevx]))

and subset by the rows that have ys:

x[df$y == 1]
Time differences in secs
[1] 124   0

(presumably, you wanted your data to have a 1 for y in the last place, in which case we would get your desired answer):

Time differences in secs
[1] 124   0 348