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Formating a zoo dataset .
2 messages · Megh Dal, Gabor Grothendieck
Your code below changes the class of the time index from "Date" to "character". Perhaps you want "yearmon" class. The following averages all values in each month producing a series with time class "yearmon". See ?yearmon library(zoo) z <- zoo(c(0.1423065, -0.03276228, -0.0652789, -0.04999873, -0.01447902, 0.22265729), as.Date(c(11047, 11075, 11108, 11138, 11169, 11200))) z aggregate(z, as.yearmon, mean) # Its also possible, though normally undesirable, to use "character" as your # index class but then for it to make sense you will need to choose a # representation which sorts appropriately under the usual rules for sorting: aggregate(z, format(time(z), "%Y-%m"), mean)
On Mon, Mar 3, 2008 at 2:13 AM, Megh Dal <megh700004 at yahoo.com> wrote:
Suppose I have following dataset :
> head(data1)
Date Return 1 03/31/00 0.14230650 2 04/28/00 -0.03276228 3 05/31/00 -0.06527890 4 06/30/00 -0.04999873 5 07/31/00 -0.01447902 6 08/31/00 0.22265729 Now I convert it to zoo object :
> data11 = zoo(data1[,2], as.Date(data1[,1], format="%m/%d/%y")) > head(data11)
2000-03-31 2000-04-28 2000-05-31 2000-06-30 2000-07-31 2000-08-31 0.14230650 -0.03276228 -0.06527890 -0.04999873 -0.01447902 0.22265729 Clearly those are monthly data. Therefore I want to convert it to mm-yy format. I used following code : data111 = zoo(coredata(data11), format(index(data11), "%m/%y")) However what I got is that :
> head(data111)
01/01 01/02 01/03 01/04 01/05 01/06 -0.001388889 -0.016274826 -0.047707664 0.001104362 -0.077961541 0.017637141
tail(data111)
12/02 12/03 12/04 12/05 12/06 12/07
0.058660676 -0.018067833 -0.055569851 0.007142888 0.051162052 0.052643733
It is evident that month order has been changed. Can anyone here tell me how to get correct order like :
01/01, 02/01, 03/01..................
Your help is highly appreciated
Regards,
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