similar to: please show me simple example how to plot "Distance-Weighted Least Squares" fitting

many languages have shorthands for that operation like: variable += 1 or ++variable is there something like that in R ? -- View this message in context: http://r.789695.n4.nabble.com/increase-or-decrease-variable-by-1-tp3076390p3076390.html Sent from the R help mailing list archive at Nabble.com.

x= c(1,5,7,-3,4) y= c(2,4,-5,2,5) plot(x,y,ylim=c(-20,20),xlim=c(min(x),max(x)),pch='X',col = rgb(0, 0, 0, 0.5),yaxt="n", ann=FALSE) and this code produces: http://i53.tinypic.com/ffd7d3.png Where I marked in red areas that I want to get rid of and use as much real screen estate as I can. -- View this message in context:

I know that by setting alpha to for example col = rgb(0, 0, 0, 0.1) it is possible to see how many overlapping is in the plot. But disadvantage of it is that single points are barely visible on the background. So I wonder if there is possible to make setting that single points would be almost black, but with more and more data on the same spot it would get more and more whiteish. Or maybe it is

I have tried quit(), and return() but first exits from whole graphical interface and second is only for functions. What I need only to exit from current script and return to the console. -- View this message in context: http://r.789695.n4.nabble.com/beginner-simple-keyword-to-exit-script-tp3052417p3052417.html Sent from the R help mailing list archive at Nabble.com.

In datamatrix[, "y"] == datamatrix[, "y"][-1] : longer object length is not a multiple of shorter object length out = c(FALSE,datamatrix[,'y'] == datamatrix[,'y'][-1]) and I do not know why I get that error, the resulting out matrix is somehow one row larger than datamatrix... all I try to do is filter matrix by dropping rows where [,'y'][-1] ==

consider this matrix: [,1] [,2] [1,] 3 7 [2,] 6 5 [3,] 7 5 [4,] 3 5 [5,] 7 5 [6,] 5 5 [7,] 8 4 [8,] 2 4 [9,] 7 4 [10,] 0 6 I need to delete all rows where column 2 above and below has the same value, so the effect would be: [,1] [,2] [1,] 3 7 [2,] 6 5 [6,] 5 5 [7,] 8 4 [9,] 7 4 [10,] 0 6

I have got a list with 3 colums x,y,z, now I want do delete whole rows where column z has NA values I am not intereted in x and y columns if there are also NA values or not, moreover I do not want to touch them because if there would be NA that would mean a more serious error for me, and I have cath for that somewhere else I am asking because I only saw an example where function loops over all

for example I have matrix with two columns x,y 1,0.56 2,9.55 2,7.56 5,2.55 5,0.56 3,0.55 2,0.56 2,1.56 so I need to take average from y values placed where x==2 -- View this message in context: http://r.789695.n4.nabble.com/how-to-make-partial-mean-of-a-matrix-only-when-second-value-matching-some-logic-tp3078449p3078449.html Sent from the R help mailing list archive at Nabble.com.

I know little of statistics and have created this function out of intuition. But since this algorithm is so basic I wonder what is the proper name of this function and is it build in R. here is some code in PHP to illustrate what the function is doing, it uses some function I created but the meaning is obvious: #get csv file and interchange rows with columns to get two arrays $csv =

Dear All, Which package/function could i use to solve following linear least square problem? A over determined system of linear equations is given. The nnls-function may would be a possibility BUT: The solving is constrained with a inequality that all unknowns are >= 0 and a equality that the sum of all unknowns is 1 The influence of the equations according to the solving process is

Hello, is it possible in R to calculate a Weighted averaging partial least squares regression? I'm not firm in statistics and didn't found anything about weighted averaging in combination with PLS in the help archives. Or is it possible to develop a workaround with the pls-package? thanks for help in advance Andreas Plank -- _____________________________________________ Dipl. Biol.

As John noted, there are different kinds of weights, and different terminology: * inverse-variance weights (accuracy weights) * case weights (frequencies, counts) * sampling weights (selection probability weights) I'll add: * inverse-variance weights, where var(y for observation) = 1/weight (as opposed to just being inversely proportional to the weight) * weights used as part of an

Howdy, In SPSS, there are 2 ways to weight a least squares regression: 1. You can do it from the regression menu. 2. You can set a global weight switch from the data menu. These two options have no, in my experience, been equivalent. Now, when I run lm in R with the weights= switch set accordingly, I get the same set of results you would see with option #1 in SPSS. Does anybody know how to

Hello, I have a data set of data which is best fit by an exponential decay function. I would like to use a nonlinear weighted least squares regression. What function should I be using? Thank you! [[alternative HTML version deleted]]

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector

I need to find out the difference between the way R calculates weighted regression and standard regression. I want to plot a 95% confidence interval around an estimte i got from least squares regression. I cant find he documentation for this ive looked in ?stats ?lm ?predict.lm ?weights ?residuals.lm Can anyone shed light? thanks Chris. -- View this message in context:

Hi everyone, I admit I am a bit of an R novice, and I was hoping someone could help me with this error message: Warning message: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : extra arguments weigths are just disregarded. My equation is: lm( Y ~ X1 + X2 + X3, weigths = seq(0.1, 1, by = 0.1)) -- View this message in context:

What is the best way to produce a barplot from my data? I would like the barplot to show each person with the values stacked val1+val2+val3, so there is one bar for each person When I use barplot(data.matrix(realdata)), it shows one bar for each value instead. To post here, I created an artificical data set, but it works fine. fakedata <- as.data.frame(list(LETTERS[1:3])) colnames(fakedata)

Hi there, I'm trying to fit a growth curve to some data and need to use a weighted least squares estimator to account for heteroscedasticity in the data. A weights argument is available in nls that would appear to be appropriate for this purpose, but it is listed as 'not yet implemented'. Is there another package which could implement this procedure? Regards, Robert Brown

Hi all, I met with a problem about the weighted least square regression. 1. I simulated a Normal vector (sim1) with mean 425906 and standard deviation 40000. 2. I simulated a second Normal vector with conditional mean b1*sim1, where b1 is just a number I specified, and variance proportional to sim1. Precisely, the standard deviation is sqrt(sim1)*50. 3. Then I run a WLS regression without the