similar to: Extremum index from a sampling time series data

Dear All, I was looking for the color index in?image function,?such as from topo.colors(n) and etc. but still never found it. For instance, from the help menu. ########################################### # Volcano data visualized as matrix. Need to transpose and flip # matrix horizontally. image(t(volcano)[ncol(volcano):1,]) # A prettier display of the volcano x <- 10*(1:nrow(volcano)) y

Dear All, Let: dp: depth of the?river tp: temperature with respect to depth We can?have a simple scatter plot, between depth as y-axis and temperature as x-axis,?by using a?plot function as shown?below. #####################? dp <- c(1,4,3,2,5,7,9,8,9,2) tp <- 1:10 plot(tp,dp, type= 'l') ##################### ? Could someone advice me on?the way to?plot the same pair of

Dear All, I have a series of data in which the first column consist of a combination of date and time, for instance 17 April 2008 at 4.01pm, such data is recorded as: 4/17/2008 16:01 I'd like to seperate it into four different columns which consist of Day, Month,Year and Time, respectively. Could someone please advice me on this mater? Thank you, Fir

Dear All, Let a, b and c are?three matrices with same no. of column but different no. of row. a <- matrix(1, 1, 2) b <- matrix(2, 2, 2) c <- matrix(3, 3, 2) Could someone help me to combine these matrices together as a single matrix? Thank you Fir

Dear All, I've been?searching for?appropriate codes to compute the rate of change and the curvature?of ?nonparametric regression model whish was denoted by a smooth function?but?unfortunately?don't manage to?do?it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a

> On 26 Aug 2017, at 16:39, niharika singhal <niharikasinghal1990 at gmail.com> wrote: > > Hi, > > Thanks for your mail, and time > > It is not working for some arguments, when mean value is like >6. > > > case > > mc0 <- c(0.08844446,0.1744455,0.1379778,0.1209769,0.1573065,0. > 1134463,0.2074027) > > rv

Greetings, R-Wizards: I'm trying to find an extremum subject to a nonlinear constraint. (Yes, I have perused the archives but have found nothing positive.) The details of the problem are these: In a paper published some years ago in Technometrics, ("Confidence bands for cumulative distribution functions of continuous random variables" Technometrics, 25, 77-86. 1983), Cheng and

Dear All, Could anyone?give me a hand?to suggest few packages in R to running Kalman prediction and filtration ? Thanks Fir

Dear Niharika, As I said before, the problem is basically an optimization issue. You should isolate the problematic part from the rest of your study. Sometimes, more information does not help to solution. All the answers from us (Ulrik, David, me) are more or less are correct to find a maximum point. Newton?s method is also correct. But after answers, you only say, it didn?t find the right

---------- Forwarded message ---------- From: niharika singhal <niharikasinghal1990 at gmail.com> Date: Sun, Aug 27, 2017 at 11:57 AM Subject: Re: Find maxima of a function To: "David Winsemius [via R]" <ml+s789695n4745009h56 at n4.nabble.com>, "Ismail SEZEN [via R]" <ml+s789695n4744993h60 at n4.nabble.com>, Ulrik Stervbo <ulrik.stervbo at gmail.com>

Dear All,   Could someone please give some advice the way to do linear modelling via best subset regression in R? I'd really appreciate for your kindness.   Thanks, Kagba [[alternative HTML version deleted]]

Dear All, I'm sorry if my question does not suit with this R group. I have recently installed?R software with version 2.9.2, but i found the program took almost 1 minute as soon as it was opened, before it can be used. However,?the previous version 2.9.1 only take few seconds?after the menu bar was clicked. This circumstance has caused me to wait for couple of minutes as several R windows

I have not followed the history of this thread, but I am quite flummoxed as to why the OP is rewriting code to estimate parameters from an univariate Gaussian mixture model when alternatives such as EMCluster (which generally appears to handle initialization better than MClust) exist. Or perhaps there is more to it in which case I apologize. But I thought that I would make the OP aware of the

Dear All, ? Let: dp: depth of the river tp: temperature with respect to depth These pair of observations are in 3 different groups i.e: Obs. 1,3,5,7?from the first group Obs. 2,4 and?10?from second group Obs 6,8 and 9?from third group. We can have a simple scatter plot, between depth as y-axis and temperature as x-axis, with each pairs are denoted by a red dot, by using a plot function shown

Hi,   I have an original data frame with 8 columns of variables, which are stored in 'data1' frame.   data1 <- read.csv("E:\\PHD GLASGOW UNIVERSITY\\Data\\R\\Colin\\Cailness21.csv") attach(data1) names(data1)   [1] "Date"       "d"          "m"          "y"          "Time"      [6] "Depth"      "Temp"      

Dear All, I'm trying to do the estimation in a changepoint regression problem via R, but never found any suitable function which might help me to do this. Could someone give me a hand?on this matter? Thank you.

Hi,   Could any one please help how I can transform data based on Box-Cox Transformations in R.   Any helps will be much appreciated.   thanks, Kagba [[alternative HTML version deleted]]

Dear R-Users, I need to find a smooth function f() and coefficients a_i that give the best fit to y ~ a_0 + a_1*f(x_1) + a_2*f(x_2) Note that it is the same non-linear transformation f() that is applied to both x_1 and x_2. So my first question is how can I do it in R? A more general question is this: suppose I have a utility function U(a_i, f()), where f() is say a spline. Is there a general

Dear All, Let? 499 piece-wise lines were buit up by 500 pair of observations, via R code below. x <- 1:500 y <- rnorm(500) plot(x, y, type = 'b') I was trying to compute all the slopes?for the ?lines which were connected between two adjacent points. For instance,?slopes?of lines between first and second points, second and third points, and so on, until?between??points 499 and 500,

Dear All, Let have 10 pair of observations, as shown below. ###################### x <- 1:10 y <- c(1,3,2,4,5,10,13,15,19,22) plot(x,y) ###################### Two fitted? models, with ranges of [1,5] and [5,10],?can be easily fitted separately by lm function as shown below: ####################### mod1 <- lm(y[1:5] ~ x[1:5]) mod2 <- lm(y[5:10] ~ x[5:10]) #######################