On 18 Oct 2001 at 16:21, Robin Hankin wrote:
> Hello everybody.
>
> I have a question that has stumped me and the usual "apply"
tricks> don't seem to work. I run a course where each student's
performance> is marked by one or more assessors.
>
> I have a data frame containing students' names, assessors' names
and> their marks, arranged as follows:
>
>
> ID student assessor Q1A Q1B Q1C Q2A Q2B
Q3> 1 2152833 Fred.Smith Robin 15.0 17.0 13.0 14.0 13.0
8.0 2 > 2152833 Fred.Smith Steph 14.0 13.0 11.0 16.0 13.0 8.0 3
> 2152833 Fred.Smith Julie 15.0 17.0 13.0 17.0 13.0 8.0 4
> 9821091 Joe.Bloggs John 13.0 12.0 12.0 15.0 12.0 8.0 5
> 9821091 Joe.Bloggs Julie NA NA NA 18.0 14.0 8.0 6
> 9821091 Joe.Bloggs Robin 12.0 12.5 NA 16.0 13.5 8.0
7 > 9821091 Joe.Bloggs Steph NA NA 12.0 17.0 12.5 NA
8 > 9791734 Bob.Jones Julie 13.0 12.0 8.0 16.0 13.0 8.0 9
> 9791734 Bob.Jones Robin 11.0 12.0 7.0 14.0 11.0 8.0 ...
>
>
> note that some students are marked by more assessors than
others.>
>
> What I want is each student's mean mark for each question,
averaged> over the appropriate assessors. Is there a neat vectorized
method I> could use?
>
If I understood it correctly you need something like that:
ID student assessor Q1A Q1B Q1C Q2A Q2B Q3
1 2152833 Fred Robin 15 17.0 13 14 13.0 8
2 2152833 Fred Steph 14 13.0 11 16 13.0 8
3 2152833 Fred Julie 15 17.0 13 17 13.0 8
4 9821091 Joe John 13 12.0 12 15 12.0 8
5 9821091 Joe Julie NA NA NA 18 14.0 8
6 9821091 Joe Robin 12 12.5 NA 16 13.5 8
7 9821091 Joe Steph NA NA 12 17 12.5 NA
8 9791734 Bob Julie 13 12.0 8 16 13.0 8
9 9791734 Bob Robin 11 12.0 7 14 11.0 8
***10 9821091 Joe John 14 15.0 14 15 11.0 9***
I added one line because each student was marked by the assessor
only once in your example...
> aggregate(ppp[,4:9],list(ppp$student,ppp$assessor),mean)
Group.1 Group.2 Q1A Q1B Q1C Q2A Q2B Q3
***1 Joe John 13.5 13.5 13 15 11.5 8.5***
2 Bob Julie 13.0 12.0 8 16 13.0 8.0
3 Fred Julie 15.0 17.0 13 17 13.0 8.0
4 Joe Julie NA NA NA 18 14.0 8.0
5 Bob Robin 11.0 12.0 7 14 11.0 8.0
6 Fred Robin 15.0 17.0 13 14 13.0 8.0
7 Joe Robin 12.0 12.5 NA 16 13.5 8.0
8 Fred Steph 14.0 13.0 11 16 13.0 8.0
9 Joe Steph NA NA 12 17 12.5 NA
> The best I can come up with is
"tapply(x$Q1A,x$name,mean,na.rm=T)" but> this works question-by-question and I would like a reduced data
frame> with the same column headings.
>
>
> this is driving me crazy; thanks in advance!
>
>
>
> --
>
> Robin Hankin, Lecturer,
> School of Environmental and Marine Science
> Private Bag 92019 Auckland
> New Zealand
>
> r.hankin at auckland.ac.nz
> tel 0064-9-373-7599 x6820; FAX 0064-9-373-7042
>
>
>
> --f9295XO03655.1002013533/r.hankin.sems.auckland.ac.nz--
>
>
>
> --
>
> Robin Hankin, Lecturer,
> School of Environmental and Marine Science
> Private Bag 92019 Auckland
> New Zealand
>
> r.hankin at auckland.ac.nz
> tel 0064-9-373-7599 x6820; FAX 0064-9-373-7042
>
>
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