similar to: getting values from arrays using which()

Frede Aakmann T?gersen wrote: > Perhaps > > http://stinet.dtic.mil/cgi-bin/GetTRDoc?AD=ADA266969&Location=U2&doc=GetTRDoc.pdf > > is something that you can use? Thanks a lot - that might help. Rainer > > > > Best regards > > Frede Aakmann T?gersen > Scientist > > > UNIVERSITY OF AARHUS > Faculty of Agricultural Sciences > Dept.

Hello R, According to: g <- expand.grid(x = 1:10, y = 5:15, gr = 1:2) g$z <- log((g$x^g$g + g$y^2) * g$gr) wireframe(z ~ x * y, data = g, groups = gr, scales = list(arrows = FALSE), drape = TRUE, colorkey = TRUE, screen = list(z = 30, x = -60)) i have two wireframes in one plot. How could i change the color of the top - one to transparent (or only the grid).

R-listers, In using the adapt function, I am getting the following warning: Ifail=2, lenwrk was too small. -- fix adapt() ! Check the returned relerr! in: adapt(ndim = 2, lower = lower.limit, upper = upper.limit, functn = pr.set, Would someone explain what the 'lenwrk' value indicates in order to help diagnose this issue. Also, what are the possible codes for Ifail, so I can set

Does anyone know if it is possible to incorporate image files (e.g., jpgs or gifs) in an R graphic? I would like to make a scatterplot in which each point is a small picture of the animal represented by each value. Each picture would be a graphics file. Is there some way to use such files as plot symbols in R? Cheers, Steve -- -------------------------------------------------------- Steve C.

The compositional data xi=(x_i1,x_i2,...,x_in), for each fixed i , xij>0, and sum(xij)=1; I want to compare the mean( u_i) of several groups i.e. H0: u_1=u_2=...=u_N or H0: u_11=u_21=...=u_N1 Are there any ANOVA tpye tools to do this work in R? Thanks, WEN S Q [[alternative HTML version deleted]]

Hello, I am using library fda and I can not run a lot of functions because I receive the error: Error in bsplineS(evalarg, breaks, norder, nderiv) : couldn't find function "spline.des" do you know how I can fix that? Thnaks. Liliana

Hello everybody, I'm trying to predict a linear regression model but it does not work. My Model: y = Worktime + Vacation + Illnes + Bankholidays My modelmatrix is of dimension 28x4 Then I want to make use of the function predict because there confidence.intervals are include. My idea was: mod <- lm(y~Worktime+Vacation+Illnes+Bankholidays) newdate=data.frame(x=c(324,123,0.9,0.1))

Hello list I have a theoretical differential equation model (two coupled differential equations) describing the acidification of dairy cultures in milk: dX/dt = f(H, param)*X dH/dt = g(param)*dX/dt pH = -log10(H) I also have actual data of the same. I would like to estimate the parameters of the theoretical model from real data, but don't know how to go about it in R (I'm fairly new

Dear R gurus, I have got stuck on how to customise the colorkey generated by levelplot and contourplot, in the lattice package. This best illustrated by an example: library(lattice) levelplot(volcano^20/1e45,at=c(0,0.001,0.01,0.1,1,10)) The reason for raising the volcano dataset to the 20th power is to create a dataset with a large vertical exaggeration - this is similar in nature to the data

I have a vector whose length is nearly 70 thousand. I need randomize it for 1000 times . for randomizing , I mean ,the elements of the vector remain intact while their order in the vector get changed randomly. I have written a function which seems to be able to solve short vectors , but waste a lot of time when dealing with the the long vector mentioned above. -- View this message in context:

Hi, why don't you try try ks.test(VeriSeti1, VeriSeti2)$p.value All the best Jenny >How can i print only the P-Value of the kolmogorov smirnov test? > > >> ks.test(VeriSeti1, VeriSeti2) > > Two-sample Kolmogorov-Smirnov test > >data: VeriSeti1 and VeriSeti2 >D = 0.5, p-value = 0.4413 >alternative hypothesis: two-sided > > >This expression

R-help, I'm trying to work out the density function by doing: > df(5.22245, 3, 22) [1] 0.005896862 > df(15.20675, 6, 4) [1] 0.001223825 In excell the result is : 0.0071060464 <*> FDIST(5.22245,3,22) 0.0100001406 <--> FDIST(15.20675,6,4) >From my point of view the differences in the second case are substantial. Can anyone give me a hint on this? Thanks in

Hello, I am new to R exploratory data analysis and plotting. Is anyone aware of a way to overlay a set of conditional histograms with conditional PDFs? Below, I generate a lattice plot of precipitation histograms based on different months and stations, given a subset of the dataset: histogram(~ data | month * station, data = sta.stack[sta.stack[,"type"]=="precip" &

I am wanting to generate panels showing scatterplots with the linear fitted line for two groups within each panel superimposed. I have two conditioning factors, "variable" and "Group" and I want separate panels for each level of "variable" with different symbols and "lmline"s for each level of "Group". However all observations for a group are

Hello all, I'm hoping to find a way to evaluate the following sort of integral in R. \int_a^b \int_{g(y)}^Inf f(x,y) dx dy. The integral has no closed form and so must be evaluated numerically. The "adapt" package provides for multidimensional integration but does not appear to allow the limits of integration to be a function. I need to evaluate a number of integrals of this

If I have a contourplot (in the lattice package) and I want to add straight lines to it, how do I do this? I see that there are llines() and lsegement() functions for lattice plots, but they don't seem to do anything in this case: library(lattice) library(KernSmooth) x=rnorm(10000) y=x+rnorm(x,0,.5) a=bkde2D(cbind(x,y),.7) z=as.vector(a$fhat) grid=expand.grid(x=a$x1,y=a$x2) grid$z=z

I have following syntax for putting a legend : legend("bottom", fill=c("red","blue"), legend=expression(p==0.30, p==0.50), bty="n") However what I want is that : the value "0.30" should be a value of a variable instead of a constant, so that I can put the name of this variable and in legend it's value will be displayed. Can anyone tell me how

R-help, I have a function which is exporting the output to a file via write.table(df, file = "file name.xls" ) What I want is to paste the file name (above) by taking the argument to the function as a file name something like this: MY.function<- function(df) { ... ... write.table(df,"argument.xls") } MY.function(argument) Thank you

Hi! I had a database with some variables in sequence. Let me say: TX01, TX02, TX03 and TX04. But I need to run some regressions changing the variables... so: variable <- paste("TX0", 1:4, sep="") for(i in 1:4){ test[i] <- lm(variable[i] ~ INCOME, data=database) } But doesn't work... lm tries to find a variable inside database named variable[i] ... Suggestions?

What is the most efficient alternative to x[order(x)][1:n] where length(x)>>n? I also need the positions of the mins/maxs perhaps by preserving names. Thanks for any suggestions. -- View this message in context: http://www.nabble.com/Largest-N-Values-Efficiently--tf4788033.html#a13697535 Sent from the R help mailing list archive at Nabble.com.