search for: zmat

Hello R users... how to align this Zmat (triangle image) in X axis? I would like that the triangle's base become in the X axis and the triangle's height become in the Y axis Is there some trick for make this? Thanks. Cleber ######################## my test and try f <- function(x,y){ z=1-x-y z[ z < (-1e-15) ]...

Please I am new to R. I got the following code from a friend:   gmat <- cbind(gmat,p*(1-p)*wxb) for (j in seq(1:ncol(gmat))) {   gmat[,j] <- fitted(lm(gmat[,j]~zmat)) }   It is for spatial logit.   After defining all the matrices in it such as P, wxb, gmat, and zmat, I tried to run it in R and got the following error message: Error in model.frame.default(formula = gmat[, j] ~ zmat, drop.unused.levels = TRUE) :   invalid type (list) for variable 'zmat'...

...-c(Re(f(z),Im(f(z)) wireframe(func~x*y,data=plotz,groups=groups) But that seems like a clunky way to go, especially if I happen to have started out with a nice matrix of the f(z) values. So, is there some simpler way to write the formula in wireframe? I envision, for a matrix of complex values zmat, pseudocode: wireframe(c(Re(zmat),Im(zmat), groups=1:2) -- and yes, I'm fully aware that without a connection between the 'groups' variable and zmat, this won't work as written. All suggestions (including "read the help file for {some lattice func I didn't know abou...

...<- c(1, mu.target) cUp <- c(1, Inf) bLo <- rep(0, n) bUp <- rep(1, n) #I convert [cLo <= A <= cUp] to Amat >= bvec and [bLo=0 <= w #<=1=bUp] to Amat <- rbind(-1, 1, -mu, mu) dim(bLo) <- c(n,1) dim(bUp) <- c(n,1) bvec <- rbind(-1, 1, mu.target, Inf, bLo, -bUp) zMat <- matrix(rep(0,2*n*n),ncol=n, nrow=n*2) zMat[,1] <- c(rep(1,n), rep(-1,n)) Amat <- t(rbind(Amat, zMat)) #So I set Dmat=Cov and set dvec=0 Dmat=Cov dvec=rep(0, nrow(Amat)) #The first two rows of Amat should be equality constraints (so weights sum to 1) meq <- 2 sol <- solve.Q...

...;- c("x","y","o") xor.nn <- nnet(o ~ x + y, data=xor.data, linout=FALSE, size=1) # Create an (x.y) surface and predict over all points d <- data.frame(expand.grid(seq(0,1,.1), seq(0,1,.1))) names(d) <- c("x","y") p <- predict(xor.nn, d) zmat <- as.matrix(cbind(d,p)) Now my z matrix consists of x and y points, and the corresponding prediction value for each (x,y) tuple. What would be the best way to plot these? I tried persp(), but it didnt like the z matrix. Is there an alternative plot function that I could use (I am presuming I n...

Hi, All: Is there a simple way to detect complex conjugates in the roots returned by 'polyroot'? The obvious comparison of each root with the complex conjugate of the next sometimes produces roundoff error, and I don't know how to bound its magnitude: (tst <- polyroot(c(1, -.6, .4))) tst[-1]-Conj(tst[-2]) [1] 3.108624e-15+2.22045e-16i

Hi, I am plotting values of log(hazards ratio) as a function of two predictors, using the plotting function filled.contour(). Here is a simple simulated example of this: x <- seq(0,1, length=20) y <- seq(0,1, length=20) z <- outer(x,y, function(x,y) x^2 + y^2 ) zmat <- matrix( rexp(n=400, rate = z+0.001), 20, 20) filled.contour(x, y, log(zmat), xlab="X", ylab="Y", color=heat.colors, main="log-Hazard ratios for different X & Y cutoffs") Here lower hazard-ratios are reddish and higher values are lighter. I would like...

...it appears only with the default color palette despite various efforts to supply a col argument. I even took a crack at hacking the plot.Design function, adding a col=col parameter to be passed in the function call to image(), but failed to get the desired effect: # else image(xseqn, y, zmat, xlab = xlab, ylab = laby , col = col) library(Hmisc); library(Design) lr.fit6 <- lrm(death ~ rcs(BL_CHOLEST.A,c(180,220,280))*rcs(BL_HDL.A,c(40,55,70))*Sex, data = pref900) # str(pref900[,c("BL_HDL.A","BL_CHOLEST.A","death")]) $'data.frame': 910659 ob...

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90