similar to: ordinary polynomial coefficients from orthogonal polynomials?

I am trying to use the SeasonalMannKendall function in the Kendall package. My dataset (alb_data) is in the same format as the example dataset (manaus) in the package. > class(manaus) [1] "ts" > is.ts(manaus) [1] TRUE > typeof(manaus) [1] "double" > alb_data=read.table("R:/albemarle_manken.txt", header=T) > head(alb_data) year Jan Feb

Dear list, I have a dataframe with two column as fellow. > head(dat) V1 V2 0.15624 0.94567 0.26039 0.66442 0.16629 0.97822 0.23474 0.72079 0.11037 0.83760 0.14969 0.91312 I want to get the column V2 mean value based on the bin of column of V1. I write the code as fellow. It works, but I think this is not the elegant way. Any suggestions?

I am getting the following erro rmessage in ordistep. I have a number of similarly structured datasets using ordistep in a loop, and the message only occurs for some of the datasets. I cannot include a reproducible sample - the specific datasets where this is occur ing are fairly large and there are several pcnm's in the rhs of the formula. thanks for any pointers that may allow me to

Good Morning. Below I like statement like j<-grep(".r\\b",colnames(mydata),value=TRUE); j with the \\b option which I read long time ago which Ive found useful. Are there more or these options, other than ? grep? Thanks. dstat is just my own descriptive routine. > x ?[1] "age"????????? "sleep"??????? "primary"????? "middle" ?[5]

Hello everybody! I have an ARIMA model for a time series. This model was obtained through an auto.arima function. The resulting model is a ARIMA(2,1,4)(2,0,1)[12] with drift (my time series has monthly data). Then I perform a 12-step ahead forecast to the cited model... so far so good... but when I look the plot of my forecast I see that the result is really far from the behavior of my time

dear all i have this dataset: x,y, datavalue > dati[,c(1,2,5)] [,1] [, 2] [,3] [1,] 2.386 3.077 1.740 [2,] 2.544 1.972 1.335 [3,] 2.807 3.347 1.610 [4,] 4.308 1.933 2.150 [5,] 4.383 1.081 1.565 [6,] 3.244 4.519 1.145 [7,] 3.925 3.785 0.894 [8,] 2.116 3.498 0.525 [9,] 1.842 0.989 0.240 [10,] 1.709 1.843 0.625 [11,] 3.800 4.578 3.873 [12,] 2.699 1.199 1.425

Hello R-help I don’t know how to interpret significance from the contr.poly() function . From the example below : how can I tell if data has a significant Linear/quadratic/cubic trend? > contr.poly(4, c(1,2,4,8))               .L         .Q          .C [1,] -0.51287764  0.5296271 -0.45436947 [2,] -0.32637668 -0.1059254  0.79514657 [3,]  0.04662524 -0.7679594 -0.39757328 [4,]  0.79262909 

?s 02:10 de 02/08/2024, Steven Yen escreveu: > Good Morning. Below I like statement like > > j<-grep(".r\\b",colnames(mydata),value=TRUE); j > > with the \\b option which I read long time ago which Ive found useful. > > Are there more or these options, other than ? grep? Thanks. > > dstat is just my own descriptive routine. > > > x > ?[1]

I thought the point of adjusting the R^2 for degrees of freedom is to allow comparisons about goodness of fit between similar models with different numbers of data points. Someone has suggested to me off-list that this might not be the case. Is an ADJUSTED R^2 for a four-parameter, five-point model reliably comparable to the adjusted R^2 of a four-parameter, 100-point model? If such values

Hello R experts I have two linear regressions for sexes (Male, Female, Unknown). All have a good correlation between body length (response variable) and head length (explanatory variable). I know it is not recommended, but for a good practical reason (the purpose of study is to find a single conversion factor from head length to body length), the regressions need to go through the origin (0

Starting with the head of a 499 element matrix whose column names are now the labels trom a cut() operation, I needed to get to a vector of midpoints to serve as the basis for plotting a calibration curve ( exp(linear predictor) vs. : > dput(head(dimnames(mtcal)[2][[1]])) # was starting point testvec <- c("(-8.616,-3.084]", "(-3.084,-2.876]",

f <- (structure(list(X = structure(96:97, .Label = c("119DAmm", "119DN", "119DNN", "119DO", "119DOC", "119Flow", "119Nit", "119ON", "119OPhos", "119OrgP", "119Phos", "119TKN", "119TOC", "148DAmm", "148DN", "148DNN", "148DO",

dear all, I am a new comer to R and statistic. Now I have a little confuse about the package pls. I have to use 5 components to form a model. There are strong relationship between some of the components, which leads to the changes of the sign of each coeficeince, of course this is unwanted when using the normal regression way. So I choose the way of PLS, which is good at solve this kind of

I've defined the following for objects of a class called jml summary.jml <- function(object, ...){ tab <- cbind(Estimate = coef(object), StdError = object$se, Infit = object$Infit, Outfit = object$Outfit) res <- list(call = object$call, coefficients = tab, N = nrow(object$Data), iter = object$Iterations) class(res) <- "summary.jml" res }

Hello,

Dear all, I'm pleased to announce the release on CRAN of the "mmand" package (for Mathematical Morphology in Any Number of Dimensions). It provides functions for performing mathematical morphology (erode, dilate, etc.), smoothing, and other kernel-based operations on array-like objects of any dimensionality. The package is centred around a flexible function called morph(), which can

Dear all, I'm pleased to announce the release on CRAN of the "mmand" package (for Mathematical Morphology in Any Number of Dimensions). It provides functions for performing mathematical morphology (erode, dilate, etc.), smoothing, and other kernel-based operations on array-like objects of any dimensionality. The package is centred around a flexible function called morph(), which can

I was noticing mainly sign differences amoung the solutions to QR decomposition. For example R: > x <- matrix(c(12,-51,4,6,167,-68,-4,24,-41),nrow=3,byrow=T) > x [,1] [,2] [,3] [1,] 12 -51 4 [2,] 6 167 -68 [3,] -4 24 -41 > r <- qr(x) > r$qr [,1] [,2] [,3] [1,] -14.0000000 -21.0000000 14 [2,] 0.4285714 -175.0000000 70 [3,]

I am getting the following erro rmessage in ordistep. I have a number of similarly structured datasets using ordistep in a loop, and the message only occurs for some of the datasets. I cannot include a reproducible sample - the specific datasets where this is occur ing are fairly large and there are several pcnm's in the rhs of the formula. thanks for any pointers that may allow me to

Hello, and thank you in advance. I would like to capture the expected survival from a coxph model for subjects in an observational study with recurrent events, but the survexp statement is failing due to memory. I am using R version 2.13.1 (2011-07-08) on Windows XP. My objective is to plot the fitted survival with the Kaplan-Meier plot. Below is the code with output and [unfortunately]