similar to: taking rows from data.frames in list to form new data.frame?

Hallo, I have this dataset with repeated measures. There are two within-subject factors, "formant" (2 levels: 1 and 2) and "f2 Ref" (25 levels: 670, 729, 788, 846, 905, 1080, 1100, 1120, 1140, 1170, 1480, 1470, 1450, 1440, 1430, 1890, 1840, 1790, 1740, 1690, 2290, 2210, 2120, 2040, 1950), and one between-subject factor, lang (2 levels:1 and 2). The response variable

Hi all, I try to represent a multiple curve graphic where the x-axis is the temperature and the different y-axes are the different X (X22,X43,X44...) some X corresponds to the same molecule (22 and 44 are for CO2 for instance) so I use the same colour for them. I wanna mix the linetype with the colour to be able to visually see the difference between X43 and X45 The best I have done up to now

Hi everyone, I'm sorry if this turns out to be more a statistical question than one specifically about R - but would greatly appreciate your advice anyway. I've been using a logistic regression model to look at the relationship between a binary outcome (say, the odds of picking n white balls from a bag containing m balls in total) and a variety of other binary parameters:

I have a huge frame holding holding model results for a number of locations and time series: > str(tonedata) `data.frame': 434 obs. of 339 variables: $ VALUE : int 101 104 105 106 111 118 119 121 122 123 ... $ COUNT : int 2443 184 1539 1016 132 1208 1580 654 864 560 ... $ AREA : num 6.11e+08 4.60e+07 3.85e+08 2.54e+08 3.30e+07 ... $ D1_1958 : num 470 446 452 457 407 ...

Hi all, I am trying to run Weibull PH model in R. Assume in the data set I have x1 a continuous variable and x2 a categorical variable with two classes (0= sick and 1= healthy). I fit the model in the following way. Test=survreg(Surv(time,cens)~ x1+x2,dist="weibull") My questions are 1. Is it Weibull PH model or Weibull AFT model? Call: survreg(formula = Surv(time, delta) ~ x1

Hello, I bumped into the following funny use-case. I have too much data for a given plot. I have the following data frame df: > str(df) 'data.frame': 5015 obs. of 5 variables: $ n : Factor w/ 5 levels "1000","2000",..: 1 1 1 1 1 1 1 1 1 1 ... $ iter : int 10 20 30 40 50 60 70 80 90 100 ... $ Error : num 1.05e-02 1.24e-03 3.67e-04 1.08e-04

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Hi, I've data from an incomplete fatorial design. One level of a factor doesn't has the levels of the other. When I use lm(), the summary() return NA for that non estimable parameters. Ok, I understant it. But I use contrast::contrast(), gmodels::estimable(), multcomp::glht() and all these fail when model has NA estimates. This is becouse vcov() and coef() has different dimensions. Is

Hello!   I estimate vector error correction model (vecm) model. I have only one cointegratio relationship. I write :   joh.vecm.rls <- cajorls(joh.vecm, r=1) The output estimation is : Call: lm(formula = substitute(form1), data = data.mat) Coefficients:                up.d            expl.d        upd.d           r.d      ect1      -1.34e-01   4.55e+02   6.91e+00   2.43e+03 constant 

Upgrading an ancient customer installation... was running 1.4.23.1 (Trixbox) with Zaptel 1.4.12.9 and a Sangoma A102D, which has been running fine for 5+ years. Customer getting anxious about hardware failure, so we built a new box and installed 1.8.24.0, Dahdi 2.7.0.1, and a new Sangoma A104D. The single active span is an RBS T1 B8ZS/ESF/E&M Wink. I tried to move one span over one

The PDF file on the web, which is an appendix on nonlinear regression associated with the CAR book, is very nice. When I ran through the code presented there, I found something odd. The code does a certain model in 3 ways: Vanilla NLS (using numerical differentation), Analytical derivatives (where the user supplies the derivatives) and analytical derivatives (using automatic differentiation). The

Hello, i have the following model, poi1<-glm(F~S+T+L+C,family=poisson,x=T) where F,S,T,L are metric and C is a factor variable with the levels "0", "1", "2", "3", "4", "5" and "6" if i do summary(poi1), i get the following Call: glm(formula = F ~ S + T + L + C, family = poisson, x = T) Deviance Residuals: Min

Good Afternoon, I am relatively new to R and have been trying to figure out how to estimate regression coefficients using the MLE method. Some background: I am trying to examine scenarios in which certain estimators might be preferred to others, starting with MLE. I understand that MLE will (should) produce the same results as Ordinary Least Squares if the assumption of normality holds. That

Hello, I am trying to shrink the coefficients of a logistic regression for a sparse dataset, I am using the lasso (lasso2) and I am trying to determine the shrinkinage factor by cross-validation. I would like please some of the experts here to tell me whether i'm doing it correctly or not. Below is my dataset and the functions I use w= a b c d e P A 0 0 0 0 0 1 879 1 0 0 0 0 1 3 0 1 0 0 0 7 7

Dear Help, I'm trying to calculate a weighted correlation matrix from a data frame with 6 columns (variables) and 297 observations extracted from the regression. The last column is a weight column which I want to apply. $ model :'data.frame': 297 obs. of 6 variables: ..$ VAR1 : num [1:297] 5.21 9.82 8.08 0.33 8.7 6.82 3.94 4 0 5 ... ..$ VAR2 : num [1:297]

Hi, I'm using geeglm function to account for the repeated measure. fit1<- geeglm( binary.outcome ~ age + race + gender + fever.yes.no, data=mydata, id=ID, family=binomial, corstr="exchangeable") summary(fit1)$coef gives too large estimates and standard deviation: Estimate Std.err Wald Pr(>|W|) (Intercept) 3.07e+16

I have a large number of troff help files which are automatically converted to .Rd files. Many of these have examples which contain the printed output of the examples. This makes example(functionname) not work. Is there a way to avoid removing all this output by delimiting it in some way such that the output will not appear in R-ex? Secondly, it would be nice if there was a place where output

Hi! I have some questions about MARS model's coefficient of determination. I use the MARS method in my master's thesis and I have noticed some problems with the MARS model's R^2. You can see the following example that the MARS model's R^2 is too big when i have used mars() -function for MARS model building, and when I have made MARS-model using a linear regression, it

Howdy I apologize for duplicated posting. But I decided to correct my previous posting. I had the regression results using r <- lm(Y ~ nemp + as.factor(devt), data=d). First, there is the result of anova(r). Here I could not find regression coefficients. Response: Y Df Sum Sq Mean Sq F value Pr(>F) nemp 1 58.2 58.2 1233.23 < 2e-16 ***

Hi, this is my script > anova <- aov(data ~ Ts*Te*t + Error(R/Ts*Te*t)) > results <- summary(anova$Within) this is results Df Sum Sq Mean Sq F value Pr(>F) Ts 2 1232.2 616.11 53.606 3.965e-10 *** Ts:Te 4 4889.5 1222.37 106.356 4.075e-16 *** Ts:t 4 6472.1 1618.01 140.780 < 2.2e-16 *** Ts:Te:t 8 4181.0 522.63 45.473 1.088e-13 ***