CODEWITHSUNDEEP
rtime-seriesaggregatextszoo
Claudia
Claudia

Reputation: 1016

R function to aggregate data

I have an irregular time series called x:

structure(c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 22, 34, 56, 25, 78, 10, 0, 
54, 55, 55, 55, 0, 0, 0, 0, 67, 0, 78, 99, 10, 10), index = structure(c(1167814140, 
1167814740, 1167815340, 1167815940, 1167816540, 1167817140, 1167817740, 
1167818340, 1167818940, 1167819540, 1167820140, 1167820740, 1167821340, 
1167821940, 1167822540, 1167823140, 1167823740, 1167824340, 1167825000, 
1167825600, 1167826200, 1167826800, 1167827400, 1167828000, 1167828600, 
1167829200, 1167829800, 1167830400, 1167831000, 1167831600, 1167832200
), class = c("POSIXct", "POSIXt")), class = "zoo")

I want to convert x into a regular time series aggregating values every 10, 15, 30, 60 minutes. The new time series should always start from 0 minutes (adding NA if necessary). Also, the aggregation should work calculating the cumulative value over the precedent period.

I have tried:

x10 <- to.minutes10(x)
x15 <- to.minutes15(x)
x30 <- to.minutes30(x)
x60 <- to.hourly(x)

However to.period does not return what I need.

EXAMPLE

to.minutes15(x)
                    x.Open x.High x.Low x.Close
2007-01-03 08:59:00      1      2     1       2
2007-01-03 09:09:00      3      3     3       3
2007-01-03 09:29:00      4      5     4       5
2007-01-03 09:39:00      6      6     6       6
2007-01-03 09:59:00      7      8     7       8
2007-01-03 10:09:00      9      9     9       9
2007-01-03 10:29:00     10     22    10      22
2007-01-03 10:39:00     34     34    34      34
...

But I was expecting:

2007-01-03 09:00:00      3
2007-01-03 09:15:00      3
2007-01-03 09:30:00      9
2007-01-03 09:45:00      6
2007-01-03 10:00:00     15
2007-01-03 10:15:00      9
2007-01-03 10:30:00     32
...

Any idea?

Upvotes: 1

Views: 305

Answers (1)

tonytonov
tonytonov

Reputation: 25608

to.period does not perform an aggregation, it just converts your time series to the desired frequency. To do aggregation, use zoo.aggregate. There's also a handy function align.time in xts package, that takes care of the by argument:

as.xts(aggregate(x, align.time(index(x), 15*60)))
                    [,1]
2007-01-03 09:00:00    3
2007-01-03 09:15:00    3
2007-01-03 09:30:00    9
2007-01-03 09:45:00    6
2007-01-03 10:00:00   15
2007-01-03 10:15:00    9
2007-01-03 10:30:00   32
2007-01-03 10:45:00   34
2007-01-03 11:00:00   81
2007-01-03 11:15:00   78
2007-01-03 11:30:00   10
2007-01-03 11:45:00   54
2007-01-03 12:00:00   55
2007-01-03 12:15:00  110
2007-01-03 12:30:00    0
2007-01-03 12:45:00    0
2007-01-03 13:00:00    0
2007-01-03 13:15:00   67
2007-01-03 13:30:00   78
2007-01-03 13:45:00  109
2007-01-03 14:00:00   10

Upvotes: 1

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