Finding average of values in the past 2 minutes in

Question

I am trying to find average of values that are within a certain time frame within the same data.table and save it to a new column.

Below is a sample data set

Updated the dataset to represent the discontinuous timeline in my original dataset.

> x
                     ts value avg
 1: 2015-01-01 00:00:23     9   0
 2: 2015-01-01 00:01:56    11   0
 3: 2015-01-01 00:02:03    18   0
 4: 2015-01-01 00:03:16     1   0
 5: 2015-01-01 00:05:19     6   0
 6: 2015-01-01 00:05:54    16   0
 7: 2015-01-01 00:06:27    13   0
 8: 2015-01-01 00:06:50     7   0
 9: 2015-01-01 00:08:41    12   0
10: 2015-01-01 00:09:08    17   0
11: 2015-01-01 00:09:28     8   0
12: 2015-01-01 00:10:56     5   0
13: 2015-01-01 00:11:44    10   0
14: 2015-01-01 00:12:23    20   0
15: 2015-01-01 00:12:28     2   0
16: 2015-01-01 00:12:37    15   0
17: 2015-01-01 00:12:42     4   0
18: 2015-01-01 00:12:48    19   0
19: 2015-01-01 00:13:41     3   0
20: 2015-01-01 00:16:04    14   0

My code assigns value 10.5 to all the rows and I donot get the expected results. Here is my code.

require(lubridate)
x[, avg :=  x[ts>=ts-minutes(2) & ts<=ts , mean(value)], verbose=TRUE  ]

Updated

I want the results to be as below

                   ts   value   avg
1   01-01-2015 00:00:23 9   0
2   01-01-2015 00:01:56 11  9
3   01-01-2015 00:02:03 18  10
4   01-01-2015 00:03:16 1   14.5
5   01-01-2015 00:05:19 6   0
6   01-01-2015 00:05:54 16  6
7   01-01-2015 00:06:27 13  11
8   01-01-2015 00:06:50 7   11.66666667
9   01-01-2015 00:08:41 12  7
10  01-01-2015 00:09:08 17  12
11  01-01-2015 00:09:28 8   14.5
12  01-01-2015 00:10:56 5   12.5
13  01-01-2015 00:11:44 10  5
14  01-01-2015 00:12:23 20  7.5
15  01-01-2015 00:12:28 2   11.66666667
16  01-01-2015 00:12:37 15  9.25
17  01-01-2015 00:12:42 4   10.4
18  01-01-2015 00:12:48 19  9.333333333
19  01-01-2015 00:13:41 3   11.666667
20  01-01-2015 00:16:04 14  0

I want to do this to a data with a larger data set, also with min and max values in separate columns separately( here I have shown only the average function). Any help would be great.

Updated

Below is the reproducible code.

#reproducible code
ts<-  seq(from=ISOdatetime(2015,1,1,0,0,0,tz="GMT"),to=ISOdatetime(2015,1,1,0,0,19,tz="GMT"), by="sec")
set.seed(2)
ts <-ts + seconds(round(runif(20,0,1000),0))
value <- 1:20
avg <- 0
x <- data.table(ts,value,avg)  
setkey(x,ts)
x

Solution

Thanks to @Saksham for poining me towards apply functions. Here is the solution that I have come up with.

find <- function(y){
  mean(x[ts>=y-minutes(2) & ts<y,value])
}
x$avg <- mapply(find,x[,ts])
> x
                     ts value       avg
 1: 2015-01-01 00:00:23     9       NaN
 2: 2015-01-01 00:01:56    11  9.000000
 3: 2015-01-01 00:02:03    18 10.000000
 4: 2015-01-01 00:03:16     1 14.500000
 5: 2015-01-01 00:05:19     6       NaN
 6: 2015-01-01 00:05:54    16  6.000000
 7: 2015-01-01 00:06:27    13 11.000000
 8: 2015-01-01 00:06:50     7 11.666667
 9: 2015-01-01 00:08:41    12  7.000000
10: 2015-01-01 00:09:08    17 12.000000
11: 2015-01-01 00:09:28     8 14.500000
12: 2015-01-01 00:10:56     5 12.500000
13: 2015-01-01 00:11:44    10  5.000000
14: 2015-01-01 00:12:23    20  7.500000
15: 2015-01-01 00:12:28     2 11.666667
16: 2015-01-01 00:12:37    15  9.250000
17: 2015-01-01 00:12:42     4 10.400000
18: 2015-01-01 00:12:48    19  9.333333
19: 2015-01-01 00:13:41     3 11.666667
20: 2015-01-01 00:16:04    14       NaN

Answer 1

Will this do

ts[,avg] <- ts[,val] - 0.5

As logically and seeing your expected result, it is doing the same thing. You can edit you expected result to make it more flexible if I interpreted it wrong.

EDIT:

This base R approach should do the trick. As I an not familiar with manipulating time, I am assuming that arithmetic works in the same way as it does in most of the languages

interval <- minutes(2) #Assuming this is how we define 5 minutes

x$avg <- apply( x, 1, function(y){
               mean(x$value[x$time > ( y["time"]) - interval ) && x$time < y["time"]])
               })