similar to: How can I fixe convergence=1 in optim

Dear all R-users, I wanted to fit a Garch(1,1) model to a dataset by: >garch1 = garch(na.omit(dat)) But I got a warning message while executing, which is: >Warning message: >NaNs produced in: sqrt(pred$e) The garch parameters that I got are: > garch1 Call: garch(x = na.omit(dat)) Coefficient(s): a0 a1 b1 1.212e-04 1.001e+00 1.111e-14 Can any one

I've searched the Web with Google and do not find what might cause this particular error from an invocation of cenboxplot: cenboxplot(cu.t$quant, cu.t$ceneq1, cu.t$era, range=1.5, main='Total Recoverable Copper', ylab='Concentration (mg/L)', xlab='Time Period') Error in if (n > 0) (1L:n - a)/(n + 1 - 2 * a) else numeric() : argument is of length zero I do

I have a data set for different time intervals. The data has three comment lines before data for each time interval. For each time interval there are 500 data points. I want to change the dataset such that I have the following format: t1 t2 t3 ................ 0.00208 0.00417 0.00625 ................. a1 a2 a3 ...................

#is there a way to get NA in the table of descriptive statistics instead of the function stopping Thank you in advance #data x.f <- structure(list(Site = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L, 9L), .Label = c("BC", "HC", "RM119", "RM148", "RM179", "RM185",

I have a data frame (morgan) of hourly river flow, river levels and wind direction and speed thus: Time hour lev.morgan lev.lock2 lev.lock1 flow direction velocity 1 2009-07-06 15:00:00 15 3.266 3.274 3.240 1710.6 180.282 4.352 2 2009-07-06 16:00:00 16 3.268 3.272 3.240 1441.8 192.338 5.496 3 2009-07-06 17:00:00 17 3.268

I am wondering how to interpret the parameter estimates that lm() reports in this sort of situation: y = round(rnorm(n=24,mean=5,sd=2),2) A = gl(3,2,24,labels=c("one","two","three")) B = gl(4,6,24,labels=c("i","ii","iii","iv")) # Make both observations for A=1, B=4 missing y[19] = NA y[20] = NA data.frame(y,A,B) nonadd = lm(y ~

Friends- I wanted to let people know that there is a new experimental release of this plugin. I would love feedback on syntax and features. There is now a full test suite with fixtures that covers all the available syntax. Look at the test suite for more syntax possibilities. There have been many additions since my last release. Fabien Atelier has been working on this with me and has

When I do ARMA(2,2) using one lag of LCPIH data This is eview result > > *Dependent Variable: DLCPIH > **Method: Least Squares > **Date: 08/12/11 Time: 12:44 > **Sample (adjusted): 1970Q2 2010Q2 > **Included observations: 161 after adjustments > **Convergence achieved after 14 iterations > **MA Backcast: 1969Q4 1970Q1 > ** > **Variable Coefficient Std.

Hi, Can anybody tell me what the message below means and how to overcome it. Thanks, Neil Warning message: X matrix deemed to be singular; variable 2 in: coxph(Surv(age_at_death, death) ~ project$pluralgp + project$yrborn + ......... >

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966

Ok I guess it's time to ask. So I want to plot my data. It's my data from a frequency table, "temp". My formula is just a Gaussian eq. I have done the nls function to get my parameters and now I want to do the whole plot (...) and then lines(..) This is what I have done. > temp bin x 1 -4.0 0 2 -3.9 0 3 -3.8 0 4 -3.7 0 5 -3.6 0 6 -3.5 0 .... and so

Full_Name: Karel Zvara Version: 1.8.1 OS: MS Winows 2000 Submission from: (NULL) (195.113.30.163) The test statistics of the fligner.test (ctest package) depends on the order of cases: > fligner.test(count~spray,data=InsectSprays) Fligner-Killeen test for homogeneity of variances data: count by spray Fligner-Killeen:med chi-squared = 14.4828, df = 5, p-value = 0.01282 >

Hi, I am analyzing a data set with greater than 1000 independent cases (collected in an unrestricted manner), where each case has 3 variables associated with it: one, a factor variable with 0/1 levels (called XX), another factor variable with 8 levels (X) and a third response variable with two levels (Y: 0/1). I am trying to see if X1 has an effect on the relationship between X2 and the

Greetings: I'm running R-1.9.1 on Fedora Core 2 Linux. I tested a proportional odds logistic regression with MASS's polr and Design's lrm. Parameter estimates between the 2 are consistent, but the standard errors are quite different, and the conclusions from the t and Wald tests are dramatically different. I cranked the "abstol" argument up quite a bit in the polr

I'm fitting nonlinear functions to some growth data but I'm getting radically different results in R to another program (Prism). Furthermore the values from the other program give a better fit and seem more realistic. I think there is a problem with the results from the r nls function. The differences only occur with weighted data so I think I'm making a mistake in the weighting.

Hello all, I'm using R2.7.0 (on Windows 2000) and I'm trying do run a robust regression on following model structure: model = "Y ~ x1*x2 / (x3 + x4 + x5 +x6)" where x1 and x2 are both factors (either 1 or 0) and x3.....x6 are numeric. The error code I get when running rlm(as.formula(model), data=daymean) is: error in rlm.default(x, y, weights, method = method, wt.method =

Hi, I am having trouble with the survival library, particualrily the coxph function. the following works coxph(jtree9$cph.call,z,rep(1,dim(z)[1])) Call: coxph(formula = jtree9$cph.call, data = z, weights = rep(1, dim(z)[1])) coef exp(coef) se(coef) z p SM 0.2574 1.294 0.0786 3.274 1.1e-03 Sex -0.1283 0.880 0.1809 -0.709

On 18/09/17 17:23, Ben Turner wrote: > Do you want tuned or untuned? If tuned I'd like to try one of my tunings for metadata, but I will use yours if you want. (Re-CC'd list) I would be interested in both, if possible: To confirm that it's not only my machines that exhibit this behaviour given my settings, and to see what can be achieved with your tuned settings. Thank you!

by considering this example from barley dataset #code start dotplot(variety ~ yield | site, data = barley, scales=list(x=list(log=TRUE)), layout = c(1,6), panel = function(...) { panel.dotplot(...) #median.values <- tapply(x, y, median) # medians for each variety #panel.abline(v=median.values, col.line="red") # but this

Hi, I'm trying to work through a Nested ANOVA for the following scenario: 20 males were used to fertilize eggs of 4 females per male, so that female is nested within male (80 females used total). Spine length was measured on 11 offspring per family, resulting in 880 measurements on 80 families. I used the following two commands: summary(aov(Spinelength ~ Male*Female)) and