similar to: Persp plotting of kernel density estimate.

Displaying 20 results from an estimated 6000 matches similar to: "Persp plotting of kernel density estimate."

2004 Mar 17
1
(no subject)
Dear All, I am trying to visualize the surface of a bivariate kernel density estimate. I have a vector of bivariate observations(x,y), and a function which computes the kernel density estimate z corresponding to each observation. I cannot generate the (x,y) data in the ascending order needed by persp(x,y,z). I was wondering whether there is an R version of the S function interp. Would anybody
2006 Jun 28
1
persp/ scatterplot3d
Dear WizaRds, I would like to create a 3d-plot with persp(). I sampled 17 points with xyz-coordinates (real-life example!), representing the peaks of the whole plane with "zero coordinates" x=3,y=3,z=3. My intention is to show which entries are above or below the "zero" level with persp() on a nicely created grid. I also tried scatterplot3d(), but, alas, I am unable
2005 Mar 07
1
plot gam object
Dear R help, I am fitting a gam model with package GAM ( I use lo() to fit 2 non-parametric additive components). When I plot the GAM object, along with the graphics window displaying plots of each components, I read the following warnings at the prompt: 3: the condition has length 1 and only the first element will be used in: if (scale2 < scale) rep(mean(ylim), 2) + ((ylim - mean(ylim))
2003 Jan 14
4
density plot - beginner's question
Hi, I am trying to plot densities given on a two dimensional grid. My data is in the an external file, and is arranged in three columns: x, y, density how may i get a plot of this? i would like to get (1) a three dimensional plot and (2) a color coded two dimensional plot. I have tried using image(x, y, density) but i am asked to put the data in ascending order. i am not sure how i may
2009 Nov 09
1
persp function question
Hello, I have a question about persp function: I made my classical matrix with x, y and z variables and I dont know why I obtain a 3D image with overestimate heights. Here is my script: microtopo = read.csv2("topo2.csv", dec=".") attach(microtopo) require(akima) # for persp or zz <-interp (x, y, z) persp(zz) See also my attach file Thanks, Best regards JS
2012 Feb 29
2
Contour plot
Hi, I would like to make a contour plot with the data below, x-axis= Day (increasing order, day number is out of 365 days of a year), y-axis= Depth (varies), and z=pH. I tried to use filled.contour function, but not sure the proper way to ordinate my z results into a matrix with different x and y length. After that, if I attempted to plot the data, the x and y values are not in ascending
2006 Feb 13
2
transforming data frame for use with persp
Hi, This is probably documented, but I cannot find the right words or expression for a search. My attempts failed. I have a data frame of 3 vectors (x, y and z) and would like to transform this so that I could use persp. Presently I have y-level copies of each x level, and a z value for each x-y pair. I need 2 columns giving the possible levels of x and y, and then a transformation of
2003 Sep 20
4
persp graphs
Hi, I am doing my way on R, with much experimenting. So, I am trying to plot a 3d graphic ussing "persp" and entering a data set (attached) of UTM coordinates as x,y and a pH values as z. However when I try an error message comes out telling that increasing x and y values are expected. Jus ordering the first vector does not help, and, of course, order the first two independently
2007 Apr 11
2
persp Error. x,y,z with the same length
Hello, I have a problem in drawing a 3d graphic. I simplified the program to show you the problem. I have a text file with hundreds of entries in 3 columns. I scaned these columns with matrix(scan). Then, I transformed some datas. Now I have the datas in h,x,y. I created a new matrix to use persp . But I got an error. The reason is probably the same length of x,y,z. But what can I do the draw
2003 May 30
1
bootstrapping data.frame and matrix
Dear All, When bootstrapping a statistics based on more than one vector, from a data.frame or a matrix object, it looks like I am not able to pass the data to R. What am I doing wrong? I use the library "bootstrap". Here is an example with a data.frame called "data" "boot2_bootstrap(data, theta, nboot) I get the following error message: Error in inherits(x,
2011 Aug 04
3
persp()
Hello I am trying to draw a basic black and white map of two European countries. After searching some key words in google and reading many pages I arrived to the conclusion that persp() could be used to draw that map. I have prepared three small example files, which are supposed to be the files required for running that function. xvector is a vector with the longitudes yvector is a vector with
2004 Mar 29
2
Problems with "optimize"
Dear All, I'm trying to maximize a likelihood with respect one parameter using "optimize" on simulated data (without error component). I've iterated the maximization procedure 1000 times and I should always obtain the same estimate of the parameter (equal to the simulated one) but, instead, i obtain different results (the likelihood function shouldn't be flat). Does
2010 Jan 01
2
How to calculate density function of Bivariate binomial distribution
Am trying to do some study on bivariate binomial distribution. Anyone knows if there is package in R that I can use to calculate the density function of bivariate binomial distribution and to generate random samples of it. Thanks, -- View this message in context: http://n4.nabble.com/How-to-calculate-density-function-of-Bivariate-binomial-distribution-tp992002p992002.html Sent from the R help
1999 Nov 16
1
spikes in contour and persp (PR#327)
The following matrix of normed likelihoods should give a smooth surface but instead gives a series of spikes in both persp and contour (the dim labels are the axes values). I know an algorithm cannot be infallible but it would be nice to have some parameters to control the smoothing. (It takes close to an hour to produce this matrix on a Pentium II 300mh. It was after that that it crashed with the
2008 Jan 24
1
surface plot
Hi, there is following issue, which I encounter when doing data analysis. Say we have data in form of the matrix x: > x [,1] [,2] [,3] [1,] 0.1 4 10 [2,] 0.2 5 11 [3,] 0.1 5 12 [4,] 0.1 6 13 [5,] 0.2 4 14 [6,] 0.3 5 16 now I want to do surface plot of x[,3] with respect to axes OX:x[,1] and OY: x[,2] in ascending orders of course. Thanks in advance,
2012 Apr 30
3
R2 in multilevel modelling
Goodmorning everybody, i'm an italian statistician and i'm using R for research. Could someone tell me some indices to see the goodness of fit in multilevel modelling? I'm using the lmer function, and I want to know if my model fit well my data. I actually want to justify the use of multilevel model instead the classical one. Hope someone can help me. Thank you. Greetings
2004 Dec 08
2
Surface graph.
What package/code could I use to create a 3-d surface graph of the predicted values over two of the explanatory variables' coefficients? Jim James Bang Department of Economics University of Illinois Well I AM missing the back of my head.you COULD cut me a little slack! -Homer Simpson
2003 Jul 12
2
help with bivariate density plot question
Dear R users: I have a dataset with two variables (>20000 observations, two samples from same subject) and I used "kernSur" from library(Genkern) to get a estimated bivariate density and corresponding plots as follows: new.data.normal<-data.normal[!is.na(data.normal[,2]),] x<-new.data.normal[,2] y<-new.data.normal[,3] op <- KernSur(x,y, xgridsize=50, ygridsize=50,
2023 Jan 26
1
akima interp results to zero with less than 10 values
The akima package has a problematic license (it doesn't allow commercial use), so it's been recommended that people use the interp package instead. When I use interp::interp instead of akima::interp, I get reasonable output from your example. So that's another reason to drop akima... Duncan Murdoch On 26/01/2023 9:35 a.m., PIKAL Petr wrote: > Dear all > > I have this
2023 Jan 26
1
akima interp results to zero with less than 10 values
Dear all I have this table > dput(mat) mat <- structure(c(2, 16, 9, 2, 16, 1, 1, 4, 7, 7, 44.52, 42.8, 43.54, 40.26, 40.09), dim = c(5L, 3L)) And I want to calculate result for contour or image plots as I did few years ago. However interp does not compute the z values and gives me zeros in z matrix. library(akima) > interp(mat[,1], mat[,2], mat[, 3], nx=5, ny=5) $x [1] 2.0 5.5