Question about apply()

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