Identifying 24hour periods in GPS data - r

Question

I would like to identify sequential 24 hour periods in GPS data. I have a datetime column that is numerical (ex: 41422.29) and I know each rounded number is a day. I know how to get the day (just round), however my schedule does not specifically follow days. Instead, I would specifically like to identify all of the columns that are within 24 hours from the first column, and then go from there. I can not use a count of columns, as 24 hours is not divided into equal increments.
This is my logic so far, though it doesn't get me where I need to be:
for (i in 1:length(example)){
base<-round(example$DT_LMT[i], digits=0)
if(example$DT_LMT[i]<=base+1) {
example$DaySeq<-base
}
else {
base+1
}
}
I have a dummy data set example, with the kind of thing I would like:
structure(list(ID = 1:19, DT_LMT = c(41423.62517, 41423.79236,
41423.95868, 41424.12534, 41424.29203, 41424.45888, 41424.62535,
41424.79186, 41424.95852, 41425.12502, 41425.29185, 41425.75016,
41425.79201, 41425.83352, 41425.87534, 41425.91744, 41425.95868,
41426.00105, 41426.04257), NEED = c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L)), .Names = c("ID",
"DT_LMT", "NEED"), class = "data.frame", row.names = c(NA, -19L
))

Answer 1

Here is one approach, assuming df is the data assigned in your question. I created a new variable, need which I believe is your desired outcome.
transform(df, need = trunc(DT_LMT - DT_LMT[1]) + 1)

Answer 2

I would add 1 to the first value as the filter the data frame.
data<-data.frame(ID = 1:19, DT_LMT = c(41423.62517, 41423.79236,
41423.95868, 41424.12534, 41424.29203, 41424.45888, 41424.62535,
41424.79186, 41424.95852, 41425.12502, 41425.29185, 41425.75016,
41425.79201, 41425.83352, 41425.87534, 41425.91744, 41425.95868,
41426.00105, 41426.04257), NEED = c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L))
data[data$DT_LMT<=data$DT_LMT[1]+1,]
Output:
ID DT_LMT NEED
1 1 41423.63 1
2 2 41423.79 1
3 3 41423.96 1
4 4 41424.13 1
5 5 41424.29 1
6 6 41424.46 1
If you want to split the data into a list by 24 hour period.
split(data,unlist(lapply(data$DT_LMT,function(x){floor(x-data$DT_LMT[1])})))
Output:
$`0`
ID DT_LMT NEED
1 1 41423.63 1
2 2 41423.79 1
3 3 41423.96 1
4 4 41424.13 1
5 5 41424.29 1
6 6 41424.46 1
$`1`
ID DT_LMT NEED
7 7 41424.63 2
8 8 41424.79 2
9 9 41424.96 2
10 10 41425.13 2
11 11 41425.29 2
$`2`
ID DT_LMT NEED
12 12 41425.75 3
13 13 41425.79 3
14 14 41425.83 3
15 15 41425.88 3
16 16 41425.92 3
17 17 41425.96 3
18 18 41426.00 3
19 19 41426.04 3
To add a column with the day.
data$day<-lapply(data$DT_LMT,function(x){floor(x-data$DT_LMT[1])+1})