R help - Mar 2008 - Formating a zoo dataset .

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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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,
>
>
> ---------------------------------
>
>        [[alternative HTML version deleted]]
>
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