[R] is that possible to graph 4 dimention plot

[joris meys]

I'm basically put off by the question itself. Plotting a 4-dimensional
graph is rather "complicated" if the world has only 3 dimensions. A
4-dimensional representation is typically a movie (with time as the
4th dimension). You could try to project a heatmap on a 3D surface
graph, but I doubt this will make things much more clear.

So the standard (and correct) way of solving this problem, is to think
about a clever way to represent the needed information in less
dimensions. Ryan gives some nice examples and tips of how to do that,
but those are dismissed as "not helpful".

People on the list answer voluntarily. They do not like to be told
that "you don't think it will really help". Did you actually try it
out? You certainly don't give that impression. Maybe you should have
another look at it, and question your own approach to the problem as
well.

Keep it in mind for next time, you're not making yourself popular this way.

Kind regards
Joris



On Sat, Oct 10, 2009 at 10:01 PM, Duncan Murdoch <murdoch at stats.uwo.ca> wrote:
> On 07/10/2009 5:50 PM, gcheer3 wrote:
>>
>> Thanks for your reply.
>>
>> But I don't think it will really help. My problem is as follows:
>>
>> I have 20 observations
>> y <- rnorm(N,mean= rep(th[1:2],N/2),sd=th[3])
>>
>> I have a loglikelihood function for 3 variables mu<-(mu1,mu2) and sig
>>        loglike <- function(mu,sig){
>>        temp<-rep(0,length(y))
>>        for (i in 1:(length(y)))
>>        {
>>
>> temp[i]<-log((1/2)*dnorm(y[i],mu[1],sig)+(1/2)*dnorm(y[i],mu[2],sig))}
>>        return(sum(temp))
>>         }
>>
>> for example
>>>
>>> mu<-c(1,1.5)
>>> sig<-2
>>> loglike(mu,sig)
>>
>> [1] -34.1811
>>
>> I am interested how mu[1], mu[2], and sig changes, will effect the
>> loglikelihood surface. At what values of mu and sig will make
>> loglikelihood
>> the maximum and at what values of mu and sig will make loglikelihood has
>> local max (smaller hills) and at what values of mu and sig the
>> loglikelihood
>> is flat , etc.
>> I tried contour3d also, seems doesn't work
>
> I haven't seen any replies to this.  One explanation would be that everyone
> was turned off (as I was) by the rude remark above.
>
> On this list, before saying that something "doesn't work", it's polite to
> give a simple, nicely formatted, self-contained reproducible example of what
> went wrong, and to ask whether it is your error or an error in the package.
>  Taking that approach will usually result in someone pointing out your error
> (and fixing your code); sometimes it will result in a package author
> agreeing it's a bug, and fixing it.
>
> Duncan Murdoch
>
>>
>> Thanks for any advice
>>
>>
>> Ryan-50 wrote:
>>>>
>>>> Suppose there are 4 variables
>>>> d is a function of a , b and c
>>>> I want to know how a, b and c change will make d change
>>>> It will be straightforward to see it if we can graph the d surface
>>>>
>>>> if d is only a function of a and b, I can use 'persp' to see the surface
>>>> of
>>>> d. I can easily see at what values of a and b, d will get the maxium or
>>>> minium or multiple modes, etc
>>>>
>>>> But for 4 dimention graph, is there a way to show the surface of d
>>>> Will use color help
>>>>
>>>> Thanks a lot
>>>
>>> Not sure what your data looks like, but you might also consider looking
>>> at a 2 dimensional version.  See ?coplot
>>> for example:
>>>
>>> coplot(lat ~ long | depth * mag, data = quakes)
>>>
>>> Or you can make 2 or 3-dimensional plots using the lattice package
>>> conditioning on some of the variables - e.g. d ~ a | b * c,
>>> etc.
>>> If a, b, and c are "continuous", you can use equal.count.  Here is
>>> an uninteresting example, considering a, b, and c as points along
>>> a grid:
>>>
>>> a <- b <- c <- seq(1:10)
>>> dat <- data.frame(expand.grid(a, b, c))
>>> names(dat) <- letters[1:3]
>>>
>>> dat$d <- with(dat, -(a-5)^2 - (b-5)^2 - (c-5)^2)
>>>
>>> library(lattice)
>>> # 2-d:
>>> xyplot(d ~ a | equal.count(b)*equal.count(c), data=dat, type="l")
>>> # etc.
>>>
>>> # 3-d:
>>> contourplot(d ~ a * b | equal.count(c), data=dat)
>>> wireframe(d ~ a * b | equal.count(c), data=dat)
>>>
>>> ______________________________________________
>>> R-help at 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.
>>>
>>>
>>
>
> ______________________________________________
> R-help at 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.
>