Kevin,

Notice the subtle difference between Hadley's and your code:

Hadley
m2008$DayOfYear <- factor(m2008$DayOfYear, levels = 1:365)

Kevin
m2007$DayOfYear <- factor(m2008$DayOfYear, levels = 1:365)

Your are using the m2007 object instead of the suggested m2008 object!

HTH,

Thierry



------------------------------------------------------------------------
----
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-----Oorspronkelijk bericht-----
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens [EMAIL PROTECTED]
Verzonden: donderdag 28 augustus 2008 3:14
Aan: r-help@r-project.org
Onderwerp: Re: [R] Updating a list.

Since this didn't work:

> m2007$DayOfYear <- factor(m2008$DayOfYear, levels = 1:365)
Error in `$<-.data.frame`(`*tmp*`, "DayOfYear", value = c(1L, 1L, 1L,  :

  replacement has 432267 rows, data has 1592009

Perhaps I need to clarify how the m2007 object was generated.

t2007 <- read.csv("Total2007.dat", header = TRUE)
m2007 <- melt(t2007,
id.var=c("DayOfYear","Category","SubCategory","Sku"),
measure.var=c("Quantity"))

Kevin


---- hadley wickham <[EMAIL PROTECTED]> wrote:
>
> Try this:
>
> m2008$DayOfYear <- factor(m2008$DayOfYear, levels = 1:365)
> r2007 <- cast(m2008, DayOfYear ~ variable | Sku, sum, fill = 0)
>
> Hadley
>
> --
> http://had.co.nz/

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