me additional packages (R.oo, R.methodsS3
before you can use it.
On Fri, Aug 2, 2013 at 12:03 AM, Mike Rennie wrote:
> Indeed- thanks for the tips to get me going. This just kicks things up a
> notch for me, having happily used packages without wanting to tinker with
> them till now.
>
>
Try typing this into google search bar:
[R] install packages
The majority of the results on the first page will help you out.
On Thu, Aug 1, 2013 at 8:43 PM, Said Filahi wrote:
> hello,
> i am new and I want to know how to install a packare on R
>
> thank you
>
>
> said filahi
>
> [[a
Hi folks,
I've not before had to edit code right in a function, but I think I need
to. I am using lmer() and want to use a model that uses zero weights. I
found this thread from 2009:
https://stat.ethz.ch/pipermail/r-sig-mixed-models/2009q1/001995.html
but I'm unsure of how I would actually go a
Doh- I'm a moron. I get it now. The last line is the confirmation that
function "realroots" is working. Sorry- late in the day on a friday.
Thanks everyone for your help with this- Uber-useful, and much appreciated.
Mike
On 3/1/13, Mike Rennie wrote:
> Hi Peter,
>
> Wit
else
+ r <- polyroot(c(-b, coef(model)))
+ Re(r[is.zero(Im(r))])
+ }
>
> r <- realroots(po.lm, 5)
> predict(po.lm, newdata = data.frame(b = r))
1 2
5 5
This function just returns what I feed it as written.
Mike
On 3/1/13, Peter Ehlers wrote:
>
>
> if(names(model)[1] == "(Intercept)")
>
> A.K.
>
>
>
> - Original Message -
> From: Rui Barradas
> To: Mike Rennie
> Cc: r-help Mailing List
> Sent: Friday, March 1, 2013 3:18 PM
> Subject: Re: [R] solving x in a polynomial function
Hi there,
Does anyone know how I solve for x from a given y in a polynomial
function? Here's some example code:
##example file
a<-1:10
b<-c(1,2,2.5,3,3.5,4,6,7,7.5,8)
po.lm<-lm(a~b+I(b^2)+I(b^3)+I(b^4)); summary(po.lm)
(please ignore that the model is severely overfit- that's not the point).
try tapply() if you want the values for all levels of x, or calculate your
mean after a subset()
Check the documentation on both of these.
Mike
On Wed, Dec 8, 2010 at 9:54 AM, madr wrote:
>
> for example I have matrix with two columns
>
> x,y
> 1,0.56
> 2,9.55
> 2,7.56
> 5,2.55
> 5,0.56
> 3,0.
There's a pretty good section in the R book by Crawley on contrast
statements in R, including some discussion of the contrasts being
orthogonal.
I would say you should just make your own table and sort it out there- if
you have equal sample sizes, then the contrast coefficients along the row
shoul
Try looking at subset()... I typically use this when I'm trying to isolate
specific cases from a data frame.
Also, Dalgaard's introductory R book has some great examples for parsing
data in the manner you're attempting you might want to look at that.
Mike
On Thu, Dec 2, 2010 at 2:08 PM, Dioge
see also tapply()
e.g.
a<-c(1,1,1,2,2,2,3,3,3)
b<-c(10,10,10,15,15,15,20,20,20)
c.dat<-data.frame(a,b)
tapply(c.dat[,2],c.dat[,1],sum)
Mike
On Thu, Dec 2, 2010 at 10:33 AM, Ivan Calandra wrote:
> see ?aggregate, and ?summaryBy (in package doBy)
> I think ddply (in package plyr) could also do
You could try writing a loop
a<-data.frame(c(1:10),c(21:30))
M<-10 #number of iterations- scale up to 1000 once you get your sampling
function working
res<-NULL #place to store your results
for i in (1:M)
{
ares<-sample(a[,2],1)
res<-c(res, ares)
}
res
It's up to you how to
Hi Wallace,
Have you tried playing with sample()? Note that you can apply this function
both to whole dataframes, as well as specific items within a vector. If you
play with applying the function to different ways of indexing your sample
data, you will likely arrive at your solution.
for example:
Hi Mauluda,
Next time, please read the posting guide- helping you is made alot easier if
you provide the code that didn't work.
It sounds like you might want something like this?
#make a data frame, with some column names assigned...
aa<-data.frame(c(rep("a",5), rep("c",3)),c(rep(7,5), rep(2,3))
(apologies for any double hits; forgot to reply all...)
Or, you could just go back to basics, and write yourself a general loop that
goes through whatever levels of a variable and gives you back whatever
statistics you want... below is an example where you estimate means for each
level, but you co
Hi Anand,
Try creating a variable where you can store your data, and append it in your
loop. See added lines of code to include below...
On Thu, Nov 4, 2010 at 9:43 AM, Anand Bambhania wrote:
> Hi all,
>
> I am processing 24 samples data and combine them in single table called
> CombinedSamples
By range on the y-axis, do you mean distance? It would have to be if time is
on your x? Or am I misreading this?
You could just plot() with the data for your first individual, and then add
additional individuals after that using lines(), specifying a different
colour and/or line type for each indi
First, you might want to start by generating a new column to identify your
'pdf" and "bdf" or whatever once it's merged.
For the merging, see
?merge
But as someone's already pointed out, it's not clear what you are trying to
merge by.
Also, as your example calculations show, you don't need to m
In your call to polygon(), include lty="dashed" or "dotted" or whatever you
want your line type to look like.
take a good look at all the options in
?par
For everything you can customize in plots. Alternatively, Paul Murrel's R
Graphics book is the best reference I know for this sort of stuff.
Hi Ralf
try hist()
obl<-hist(x1, plot=FALSE)
it returns midpoints and their respective frequencies. You can specify the
breakpoints as well.
?hist
for details.
Mike
On Wed, Sep 22, 2010 at 1:44 PM, Ralf B wrote:
> Dear R users,
>
> I would like to great a frequency table from raw data and
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