subtract pairs of rows for long dataset - r

Question

I would like to find a way to subtract in a long dataset, both time and meters values every two rows (two measures per day) and therefore create a new table that store those values.
(03:21-09:37 and 3.2-0.9, so on...). Is there a function that can do it automatically. how is possible to set it? I am completly new using R and I need to figure out those things only with R
time <- c("03:21","09:37","15:41","21:46","03:54","10:12")
day <- c(1,1,1,1,2,2)
meters <- c(3.2,0.9,3.2,0.9,3.2,0.9)
df <- data.frame(day,time,meters)
day time meters
1 1 03:21 3.2
2 1 09:37 0.9
3 1 15:41 3.2
4 1 21:46 0.9
5 2 03:54 3.2
6 2 10:12 0.9

Answer 1

Here are a couple of options that quickly come to mind to consider:
Option 1: Subset with TRUE and FALSE to calculate the difference:
Time <- strptime(df$time, format="%H:%M")
TimeD <- Time[c(TRUE, FALSE)] - Time[c(FALSE, TRUE)]
MetersD <- df$meters[c(TRUE, FALSE)] - df$meters[c(FALSE, TRUE)]
cbind(meters = MetersD, time = TimeD)
# meters time
# [1,] 2.3 -6.266667
# [2,] 2.3 -6.083333
# [3,] 2.3 -6.300000
Option 2: Use %/% to create a grouping variable and use aggregate
df$pairs <- c(0, 1:(nrow(df)-1) %/% 2)
df$time2 <- strptime(df$time, format="%H:%M")
aggregate(list(meters = df$meters, time = df$time2),
by = list(pairs = df$pairs), FUN=function(y) diff(rev(y)))
# pairs meters time
# 1 0 2.3 -6.266667
# 2 1 2.3 -6.083333
# 3 2 2.3 -6.300000
Update
It's not too difficult to extend the idea to get your "day" column back tooL
with(df, {
time <- strptime(time, format="%H:%M")
time <- time[c(TRUE, FALSE)] - time[c(FALSE, TRUE)]
meters <- meters[c(TRUE, FALSE)] - meters[c(FALSE, TRUE)]
day <- day[c(TRUE, FALSE)]
data.frame(day, time, meters)
})
# day time meters
# 1 1 -6.266667 hours 2.3
# 2 1 -6.083333 hours 2.3
# 3 2 -6.300000 hours 2.3

Answer 2

Using diff
# Create a proper date
df$date <- strptime(paste(df$day,df$time),format="%d %H:%M")
new_df <- data.frame(
diff_meters = abs(diff(df$meters)),
diff_time = diff(df$date))
new_df
diff_meters diff_time
1 2.3 6.266667 hours
2 2.3 6.066667 hours
3 2.3 6.083333 hours
4 2.3 6.133333 hours
5 2.3 6.300000 hours
It's pretty easy to get every other row, if that's what you're actually looking for (not really clear from the question nor your comment:
new_df[seq(1,nrow(new_df),2),]
diff_meters diff_time
1 2.3 6.266667 hours
3 2.3 6.083333 hours
5 2.3 6.300000 hours