[R] Question about apply()

[Sergey Goriatchev]

Understood.

Peter, Gabor, thank you for your time and help.

Regards,
Sergey

On Wed, Feb 11, 2009 at 14:24, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> There is no zoo method for embed.  embed is implicit in
> rollapplly so its not normally needed in zoo and rollapply,
> lag or diff would normally be used instead.
>
> If you did want to use embed then unzoo data first but
> that will lose the time index so add it back in later:
>
> e <- embed(coredata(data),2)
> dif <- zoo(e[, 1:4] / e[, 5:8] - 1, time(data)[-1])
> cbind(data, dif)
>
> On Wed, Feb 11, 2009 at 7:45 AM, Sergey Goriatchev <sergeyg at gmail.com> wrote:
>> Dear Gabor,
>>
>> Thank you as always. I will follow your suggestion.
>> But, I still want to know what I do wrong in the above code (with
>> embed(), apply(), and merge() functions). Why do I get that error
>> message and how can I rewrite the code to make it run.
>>
>> Best Regards,
>> Sergey
>>
>> On Wed, Feb 11, 2009 at 13:34, Gabor Grothendieck
>> <ggrothendieck at gmail.com> wrote:
>>> See ?diff.zoo
>>>
>>> dif <- diff(data, arithmetic = FALSE) - 1
>>> cbind(data, dif)
>>>
>>> On Wed, Feb 11, 2009 at 6:21 AM, Sergey Goriatchev <sergeyg at gmail.com> wrote:
>>>> Hello, everyone!
>>>>
>>>> Assume you have this data:
>>>> data <- structure(c(66.609375, 67.09375, 66.40625, 66.734375, 67.109375,
>>>> 66.875, 66.09375, 65.921875, 66.546875, 66.140625, 66.140625,
>>>> 65.65625, 65.875, 65.59375, 65.515625, 66.09375, 66.015625, 66.140625,
>>>> 66.109375, 66.421875, 1702.7, 1647.7, 1649.4, 1639.9, 1696.4,
>>>> 1710.9, 1690.2, 1677.9, 1694.4, 1713.9, 1713.9, 1705.4, 1708.4,
>>>> 1692.9, 1689.6, 1647.7, 1654.5, 1651.3, 1645.7, 1602.4, 453.7,
>>>> 447.8, 446.2, 446.5, 447, 446.8, 448.5, 447.8, 449.2, 449, 449,
>>>> 453.7, 454.4, 453.4, 453.8, 452.2, 450.7, 450.6, 451.4, 447.5,
>>>> 18.34, 18.29, 17.65, 17.52, 16.96, 17.41, 18.51, 19.02, 19.43,
>>>> 20.76, 20.76, 21.59, 22.28, 22.4, 22.63, 22.26, 22.71, 22.27,
>>>> 21.75, 21.65), .Dim = c(20L, 4L), .Dimnames = list(NULL, c("TY1.lev",
>>>> "SP1.lev", "GC1.lev", "CL1.lev")), index = structure(c(10959,
>>>> 10960, 10961, 10962, 10963, 10966, 10967, 10968, 10969, 10970,
>>>> 10973, 10974, 10975, 10976, 10977, 10980, 10981, 10982, 10983,
>>>> 10984), class = "Date"), class = "zoo")
>>>>
>>>> What I want to do is to calculate simple return on each column (return=P1/P0-1)
>>>> and put it in new columns.
>>>>
>>>> I've tried like this:
>>>>
>>>> #convenience function
>>>> func <- function(x,z){
>>>>        a <- embed(x[z, drop=FALSE], 2)[,1]/embed(x[z, drop=FALSE], 2)[,2] - 1
>>>>        return(a)
>>>> }
>>>>
>>>> data <- merge(data, Bond.ret=zoo(apply(data, 2, func, z="TY1.lev"),
>>>> order.by=time(data)[-1]))
>>>>
>>>> and I get this:
>>>> Error in embed(x[z, drop = FALSE], 2) : wrong embedding dimension
>>>>
>>>> What am I doing wrong?
>>>> (Somehow I cannot understand how to work with columns in apply(), with
>>>> rows, that is apply(,1,FUN)  I have no problem.
>>>>
>>>> Thank you for help,
>>>> Sergey
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>
>>
>>
>>
>> --
>> I'm not young enough to know everything. /Oscar Wilde
>> Experience is one thing you can't get for nothing. /Oscar Wilde
>> When you are finished changing, you're finished. /Benjamin Franklin
>> Tell me and I forget, teach me and I remember, involve me and I learn.
>> /Benjamin Franklin
>> Luck is where preparation meets opportunity. /George Patten
>>
>



-- 
I'm not young enough to know everything. /Oscar Wilde
Experience is one thing you can't get for nothing. /Oscar Wilde
When you are finished changing, you're finished. /Benjamin Franklin
Tell me and I forget, teach me and I remember, involve me and I learn.
/Benjamin Franklin
Luck is where preparation meets opportunity. /George Patten