HI Ista,

Thanks for the help.  The 'cut' function seems to do the trick .

I'm not sure why you suggested this line of code:
> ddply(dat, .(image.group), transform, measure.median = median(Measurement))

I think I might have confused the issue by putting a 'Measurement' column in my 
example in the body of the e-mail, while there is no such column in the actual 
data.

The second ddply function on the cut data file seems to do the trick for taking 
the median of the relevant data. However, I still have one more question.  
Would it be possible to assign the median data back to the original 
a.ImageNumber number.  In this situation, the same data would be associated 
with images 1 through 3 and another set associated with images 4 through 6 and 
so on.

For example (again, I use 'Measurement' just as a generic column):

ImageNumber     Measurement
1                               1
1                               2
1                               3
2                               2
2                               2
3                               4
3                               3
3                               3
3                               4

where the median of all the 'Measurement' data is 3 and the output would be:

ImageNumber     Measurement
1                               3
1                               3
1                               3
2                               3
2                               3
3                               3
3                               3
3                               3
3                               3

or

ImageNumber     Measurement
1                               3
2                               3
3                               3

I really appreciate your help with this.

JT

Johnny Tkach, PhD
Donnelly CCBR, Rm. 1230 
Department of Biochemistry
University of Toronto
160 College Street
M5S 3E1

phone - 416 946 5774
fax - 416 978 8548
johnny.tk...@utoronto.ca

"Beauty's just another word I'm never certain how to spell"




On Aug 27, 2010, at 2:01 PM, Ista Zahn wrote:

> Hi Johnny,
> 
> If I understand correctly, I think you can use cut() to create a grouping 
> variable, and then calculate your summaries based on that. Something like
> 
> dat <- read.csv("~/Downloads/exampledata.csv")
> 
> dat$image.group <- cut(dat$a.ImageNumber, breaks = seq(0, 
> max(dat$a.ImageNumber), by = 3))
> library(plyr)
> ddply(dat, .(image.group), transform, measure.median = median(Measurement))
> 
> dat.med <- ddply(dat, .(image.group), summarize,
>       a.AreaShape_Area.median = median(a.AreaShape_Area),
>       a.Intensity_IntegratedIntensity_OrigRFP.median = 
> median(a.Intensity_IntegratedIntensity_OrigRFP),
>       a.Intensity_IntegratedIntensity_OrigGFP.median = 
> median(a.Intensity_IntegratedIntensity_OrigGFP),
>       b.Intensity_MeanIntensity_OrigGFP.median = 
> median(b.Intensity_MeanIntensity_OrigGFP),
>       EstCytoIntensity.median = median(EstCytoIntensity),
>       TotalIntensity.median = median(TotalIntensity),
>       NucToCytoRatio.median = median(NucToCytoRatio)
>       )
> 
> Best,
> Ista
> On Fri, Aug 27, 2010 at 5:28 PM, Johnny Tkach <johnny.tk...@utoronto.ca> 
> wrote:
> Hi all,
> 
> 
> Since I could not attach a file to my original e-mail request, for those who 
> want to look at an example of a data file I am working with, please use this 
> link:
> 
> http://dl.dropbox.com/u/4637975/exampledata.csv
> 
> Thanks again,
> 
> Johnny.
> 
> ______________________________________________
> R-help@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.
> 
> 
> 
> -- 
> Ista Zahn
> Graduate student
> University of Rochester
> Department of Clinical and Social Psychology
> http://yourpsyche.org


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