similar to: Smooth line in graph

I have a problem with R. I try to compute the confidence interval for my df. When I want to create the plot I have this problem: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y' lengths differ. I try this code: library(dplR) df.rwi <- detrend(rwl = df, method = "Spline",nyrs=NULL) write.table(df.rwi,file="rwi.txt",quote=FALSE,row.names=TRUE)

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

I generally run 'make; make check' (with more settings) when building the Debian package. Running 3.2.4 rc from last night, I see a lot of package loading issues during 'make check'. Here is splines as one examples: checking package 'splines' * using log directory '/build/r-base-3.2.3.20160303/tests/splines.Rcheck' * using R version 3.2.4 RC (2016-03-02 r70270) *

Dear R-helpers, can I use a POSIXct date as the x variable in interpSpline? The help page says x and y need to be numeric... is there a workaround? example: library(splines) testdfr <- data.frame(Date=seq(as.POSIXct("2008-08-01"),as.POSIXct("2008-09-01"), length=10)) testdfr$yvar <- rnorm(10) sp <- interpSpline(yvar ~ Date, testdfr) preddfr <-

Dear list, if I do smooth.spline(tmpSec, tmpT, all.knots=T) with the attached data, I get this error-message: Error in smooth.spline(tmpSec, tmpT, all.knots = T) : smoothing parameter value too small If I do smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary number], all.knots=T) it works! I just don't see it. It works for hundrets other datasets, but not for

Greetings, A couple general questions regarding the use of splines to interpolate depth profile data. Here is an example of a set of depths, with associated attributes for a given soil profile, along with a function for calculating midpoints from a set of soil horizon boundaries: #calculate midpoints: mid <- function(x) { for( i in 1:length(x)) { if( i > 1) { a[i] = (x[i] -

Hi all, I'm trying to draw a contour plot with rounded (smoothed) contour lines instead of sharp angles; when the grid consists of only a few points in x- and y- axis, the resulting contour is in facts rather ugly since very sharp angles may appear. I did not find any way to do it, by using either "contour" or "contourplot" (from the lattice package), I wonder if there

I'm sure I'm doing something completely boneheaded here, but I've used this idiom (constructing an interpolation spline and using prediction from a backSpline to find an approximation profile confidence interval) many times before and haven't hit this particular problem: r2 <- c(1.04409027570601, 1.09953936543359, 1.15498845516117, 1.21043754488875,

Dear R-Users, I am using 'contour' to plot a graph using values x,y and a matrix z. I would like to obtain 'smooth lines' instead of no-smooth contour lines. I tried with filled.contour too. In a Post I found yy <-predict(interpSpline(x, y)) I could use that method for each range that I want to plot but I hope to find a easier method. Thank you Tommaso Letterio DIAF - UniFi

BUG IN SPLINE()? Version R-1.0.1, system i486,linux If the spline(x,y,method="natural") function is given values outside the range of the data, it does not give a warning. Moreover, the extrapolated value reported is not the ordinate of the natural spline defined by (x,y). Example. Let x <- c(2,5,8,10) and y <- c(1.2266,-1.7606,-0.5051,1.0390). Then interpolate/extrapolate with

Greetings.? My wife is teaching an introductory stat class at UC Davis.? The class emphasizes the use of simulations, rather than mathematics, to get insight into statistics, and R is the mandated tool.?? A student in the class recently inquired about different approaches to sampling from a binomial distribution.? I've appended some code that exhibits the idea, the gist of which is that using

If I have, say, five scatter points and want to connect them together into a smooth curve. I did plot(x,y,type="l"), but the graph is five segments connecting with each other, but not a smooth curve. I wonder if there is a line type that is a curve. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/connecting-points-into-a-smooth-curve-tp3021796p3021796.html Sent

Hi, I am a total beginner with this whole thing so please have patience! I am trying to run an S-plus program with a certain line: spline(1:nrow(y), y[,1],n=100); This crashes with: Error: NAs in foreign function call (arg 8) Apparently, this is caused by the last command of spline: u <- seq(xmin, xmax, length.out = n) .C("spline_eval", z$method, length(u), x = u, y =

Hi, I have a 3d array as below, I want to make this array to a matrix of p=50(rows) and n=20(columns) with the coverage values . The code before the array is: library(binom) Loading required package: lattice pi.seq<-seq(from = 0.01, to = 0.5, by = 0.01) no.seq<-seq(from = 5, to = 100, by = 5) cp.all = binom.coverage( p = pi.seq, n = no.seq , conf.level = 0.95, method = "exact")

Hello, I am having two difficulties with xspline: 1) Passing lty=2 (via ...) appears to have no effect 2) Apparently depending on the current plot axes, I may or may not get the failure: add_point - reached MAXNUMPTS (25200) This happens both on Linux/R 2.8.0 (details below) and Windows Vista/R 2.8.0 patched (2008-10-21 r46766) Simplest case: plot(c(0,1),c(0,1))

Hello everybody, I'm trying to understand how to draw a smoothed scatterplot on a geographic map with R. Have a dataframe with point locations (long, lat) and was able to simply plot these points on a shp map by using the maptools package. However, instead of having simply the raw points on the map, I would like to have a "smoothed" scatterplot of the same superimposed on the map.

Dear list, I have observed a weird behaviour from R --- apologies if I am missing something obvious! df3f826f28 df3f826f28 Say you type in R: >c.preec <- 10074 >c.gd <- 2200 >p1 <- .2 >c.neo <- p1*9451 + (1-p1)*3883 >n.preec <- 3710 >n.gd <- 2650 >n.neo <- 2120 >n.pcos <- 53000 >unit.met <- 94 >cost.met <- 94*n.pcos >effect <-

Hi, I'm trying to determine if R is useful to me. I've browsed 'The Basics of S and S-Plus' (Krause & Olson), and like the logic of the language. However, I don't see an easy way to do things like this: * given a set of observations (x,y) (x and y equal-length vectors), and a 2nd set of abscissas x2, interpolate y at the new abscissas x2. (for now, I don't really

Full_Name: Jari Oksanen Version: 2.6.2 RC (2008-02-07 r44369) OS: Linux Submission from: (NULL) (130.231.102.145) Even if function xspline() is called with argument draw=FALSE, it requires a graphics device (that it won't use since it was draw=FALSE). I run into this because I intended to use xspline within a function (that does not yet draw: there is plot method for that), and the function

Dear all, I'm trying to estimate clusters from a very large dataset using clara but the program stops with a memory error. The (very simple) code and the error: mydata<-read.dbf(file="fnorsel_4px.dbf") my.clara.7k<-clara(mydata,k=7) >Error: cannot allocate vector of size 465108 Kb The dataset contains >3,000,000 rows and 15 columns. I'm using a windows computer