Question about apply()

Dear Peter,

I tried with the comma already, it did not work.

rets

            TY1.lev SP1.lev GC1.lev CL1.lev
2000-01-03 66.60938  1702.7   453.7   18.34
2000-01-04 67.09375  1647.7   447.8   18.29
2000-01-05 66.40625  1649.4   446.2   17.65
2000-01-06 66.73438  1639.9   446.5   17.52
2000-01-07 67.10938  1696.4   447.0   16.96
2000-01-10 66.87500  1710.9   446.8   17.41
2000-01-11 66.09375  1690.2   448.5   18.51
2000-01-12 65.92188  1677.9   447.8   19.02
2000-01-13 66.54688  1694.4   449.2   19.43
2000-01-14 66.14062  1713.9   449.0   20.76
2000-01-17 66.14062  1713.9   449.0   20.76
2000-01-18 65.65625  1705.4   453.7   21.59
2000-01-19 65.87500  1708.4   454.4   22.28
2000-01-20 65.59375  1692.9   453.4   22.40
2000-01-21 65.51562  1689.6   453.8   22.63
2000-01-24 66.09375  1647.7   452.2   22.26
2000-01-25 66.01562  1654.5   450.7   22.71
2000-01-26 66.14062  1651.3   450.6   22.27
2000-01-27 66.10938  1645.7   451.4   21.75
2000-01-28 66.42188  1602.4   447.5   21.65

func <- function(x,z){

+
+ a <- embed(x[,z, drop=FALSE], 2)[,1]/embed(x[,z, drop=FALSE], 2)[,2] - 1
+
+ return(a)
+ }

 merge(rets, Bond.ret=zoo(apply(rets,2,func,z="TY1.lev"), order.by=time(rets)[-1]))

Error in x[, z, drop = FALSE] : incorrect number of dimensions


Nothing was wrong with your suggestion, except that I did not come up
with it. :-)
I've been using a lot of embed and apply(, 1, FUNC) functions lately
and out of inertia I tried the same.

I specify a new variable (with suffix ."ret") within merge() function.

Kind Regards,
Sergey



On Wed, Feb 11, 2009 at 13:47, Peter Dalgaard <P.Dalgaard at biostat.ku.dk> wrote:

Sergey Goriatchev 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.

The immediate problem is that you are missing a comma in x[, z, drop =
FALSE]. But what was wrong with

cbind(data,data/lag(data,-1)-1, suffixes=c("","r"))

(except that it adds a "." to the orginal names, I see no way of NOT
adding a suffix?)

--
  O__  ---- Peter Dalgaard             ?ster Farimagsgade 5, Entr.B
 c/ /'_ --- Dept. of Biostatistics     PO Box 2099, 1014 Cph. K
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