and read and re-read the step() and stepAIC()
documentation and I just can't figure out what it could mean. Removing
the test="F" bit also generates the same behaviour.
Any help greatly appreciated.
Chris Beeley
Institute of Mental Health, UK
_
then 23-5-2008 0 0 0 0 again, and etc.
I've had several abortive attempts and done some Googling but have got
nowhere. I'd greatly appreciate any advice.
Many thanks,
Chris Beeley
(Institute of Mental Health, UK)
structure(list(Location = structure(c(1L, 2L, 2L, 1L, 3L, 5L,
5L, 1L, 1L,
; A) sum each of the dependent variables for each of the dates (so e.g.
>>> in the example above for 1-4-2007 it would be 3+2=5, 0+1=1, 1+2=3, and
>>> 3+4=7 for each of the variables)
>>>
>>> B) do this sum, but only in each location this time (lo
r many hours of fiddling I have had to admit defeat.
I would be very grateful for any words of advice.
Many thanks,
Chris Beeley,
Institute of Mental Health, UK
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PLE
Many thanks to you both. I have now filed away for future reference the 2
factor tapply as well as the extremely useful looking plyr library. And the
code worked beautifully :-)
On 24 Aug 2010, at 19:47, "Abhijit Dasgupta, PhD"
wrote:
> The paste-y argument is my usual trick in these situat
ith the 5:8 usage in case I put it somewhere else and
don't notice the problem.
Many thanks!
Chris Beeley
Institute of Mental Health
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PLEASE do read the posting guide
anyone on any forums or anything
with the same problem.
Using R v2.13 on Windows, v 2.12 on Linux, was using RStudio but just
tested it without (just in case) and it does the same thing.
Any suggestions gratefully received.
Chris Beeley
Institute of Mental Health, UK.
Output of the operation is
ant the function to return, for the first column,
1, 1, 1, 1, 0, 0, 1, 1, 0, 0 (because those are the values of Words
which contain a 1) and for the second column return 0, 0, 0, 0, 1, 0,
0, 0, 0, 0 (because the fifth value is the only one that contains a
2).
Any suggestions gratefully received!
to accept model 2 even though they are opposite
models. Is it really just that you have to put them in the right order?
It just seems like if there were say four models you wouldn't
necessarily be able to determine the correct order.
Many thanks,
Chris Beeley, Institute of Mental Health, UK
Well that's that cleared up then. Thanks to all.
Chris B.
On 31/05/2012 17:51, Albyn Jones wrote:
No, both yield the same result: reject the null hypothesis,
which always corresponds to the restricted (smaller) model.
albyn
On Thu, May 31, 2012 at 12:47:30PM +0100, Chris Beeley wrote:
ution, which would be much more important, if say I just wanted to
replace the odd columns, or something like that.
I found this code on the internet too:
idx <- which(foo>80, arr.ind=TRUE)
foo[idx[1], idx[2]] <- NA
But I can't seem to rewrite that either, for the same reason
Many t
s.na(datavec)]]) # yields 1567
length(comments$ImproveCat[commvec[!is.na(commvec)]]) # yields 1512
I'm sorry, I did try to construct an example dataframe, but ironically
I can't make that work either! Sorry!
Any help gratefully received.
Many thanks!
Chris Be
gure out what's special about the apply
call that generates the NAs when as.numeric(mydata[,2]) doesn't and I'm
not sure what to do about it in future.
I realise I can just loop over the columns, but I would rather get to
the bottom of this if I can so I know for future.
Thanks i
Perfect, many thanks for explanation and correct line of code.
On 09/01/2012 14:29, peter dalgaard wrote:
as.data.frame(lapply(mydata, as.numeric))
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PLEASE do read the p
t of stuff about beer and not much
else (drew a blank also from RSiteSearch("Nested brew"))
Any help gratefully received.
Chris Beeley
Institute of Mental Health, UK
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e ever had this happen to them? If so did
you find a solution (other than manually removing the NAs using a final
piece of code, which admittedly is not too arduous).
Many thanks,
Chris Beeley,
Institute of Mental Health, UK
On 30/03/2012 02:27, Matt Shotwell wrote:
On Wed, 2012-03-28 at 11:40 +0
, 13L, 13L, 13L, 14L, 14L, 14L, 14L, 14L,
13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L)), .Names = c("Incident",
"Numbers"), row.names = c(NA, 50L), class = "data.frame")
Any assistance gratefully recieved.
Many thanks,
Chri
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