I think this is what you want to do. I did not have 'shaprio.test' so
I just returned a value:
> x <- structure(list(RM215alk = c(15, 13, 12, 14, NA, 13, 16, 13, 16,
+ #subset of data
+ 13, 17, 13, 19, 13, 14, 14, 15, 15, 14, 16, 14, 15, 13, 14),
+ NSCl = c(NA, NA, 2.9, NA, NA, NA, NA, NA, NA, NA, NA, NA,
+ NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("RM215alk",
+ "NSCl"), class = "data.frame", row.names = c(NA, -24L))
>
> apply(x, 2, function(.col){
+ if (sum(!is.na(.col)) >= 3) 3 # I don't have this function:
shaprio.test(.col)
+ else NULL
+ })
$RM215alk
[1] 3
$NSCl
NULL
On Fri, Apr 4, 2008 at 10:44 AM, stephen sefick <[EMAIL PROTECTED]> wrote:
> structure(list(RM215alk = c(15, 13, 12, 14, NA, 13, 16, 13, 16,
> #subset of data
> 13, 17, 13, 19, 13, 14, 14, 15, 15, 14, 16, 14, 15, 13, 14),
> NSCl = c(NA, NA, 2.9, NA, NA, NA, NA, NA, NA, NA, NA, NA,
> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("RM215alk",
> "NSCl"), class = "data.frame", row.names = c(NA, -24L))
>
> apply(if n>=3(x.f, 2, shapiro.test)else(NULL)) #crappy code showing my
> lack of knowledge on the subject
>
> I would like to apply this to a 208 column data matrix, and this is
> the first time I have tried a repetitive task in R- so please bear
> with me. the shapiro.test requires 3 to 5000 observations and there
> are some columns that don't have three observation making it
> unnecessary to preform a test for normality. How do I do this?
> thanks
>
> stephen
> --
> Let's not spend our time and resources thinking about things that are
> so little or so large that all they really do for us is puff us up and
> make us feel like gods. We are mammals, and have not exhausted the
> annoying little problems of being mammals.
>
> -K. Mullis
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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