similar to: duration analysis

Hi there, I am trying to export a regression output to Latex. I am using the xtable function in the xtable library. Doing myfit <- lm(myformula, mydata) print.xtable(xtable(myfit), file="myfile") only returns the estimated coefficients and the correspondent standard erros, t-statiscs and p-values. But I wish to get a bit more, say, the number of observations used in the

Hi there, From the vector X of integers, X = c(11999, 122000, 81997) I would like to make these two vectors: Z= c(1999, 2000, 1997) Y =c(1 , 12 , 8) That is, each entry of vector Z receives the four last digits of each entry of X, and Y receives "the rest". Any suggestions? Thanks in advance, Dimitri [[alternative HTML version deleted]]

Hi, x and y are (numeric) vectors. I wonder if one of the following is more efficient than the other: x%*%y or sum(x*y) ? Thanks, Dimitri Szerman

Hi all. If I run the simple regression when x is a categorical variable ( x <- factor(x) ): > MyFit <-coxph( Surv(start, stop, event) ~ x ) How can I get the overall p-value on x other than for each dummy variable? > anova(MyFit) does NOT provide that information as previously suggested on the list. All the best, Kare [[alternative HTML version deleted]]

Hi, Suppose I I wish to run lm( y ~ x + z + log(w) ) where w assumes non-negative values. A problem arises when w=0, as log(0) = -Inf, and R doesn't accept that (as it "accepts" NA). Is there a way to tell R to do with -Inf the same it does with NA, i.e, to ignore it? ( Otherwise I have to do something like w[w==0] <- NA which doesn't hurt, but might be a bit

Hi there, I am trying to compute Gini coefficients for vectors containing income classes. The data I possess look loke this: yit <- c(135, 164, 234, 369) piit <- c(367, 884, 341, 74 ) where yit is the vector of income classes, and fit is the vector of associated frequencies.(This data is from Rustichini, Ichino and Checci (Journal of Public Economics, 1999) ). In ineq pacakge, Gini( )

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

I'm drawing a fitted normal distribution over a histogram. The use case is trivial (fitting normal distributions on densities) but I want to extend it to other fitting scenarios. What has stumped me so far is how to take the list that is returned by nls() and use it for curve(). I realize that I can easily do all of this with a few intermediate steps for any specific case. But I had expected

Hi, I searched for this in the mailing list, but found no results. I have a large dataframe ( dim(mydata)= 1297059 16, object.size(mydata= 145280576) ) , and I want to perform some calculations which can be done by a factor's levels, say, mydata$myfactor. So what I want is to split this dataframe into nlevels(mydata$myfactor) = 80 levels. But I must do this efficiently, that is, I

Dear all, I have a list with three (named) numeric vectors: > lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) ) > lst $a A B 1 8 $b A B C 2 3 0 $c B D 2 0 Now, I'd love to use this list to create the following data frame: > dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA), + b=c(A=2,B=3,C=0,D=NA), + c=c(A=NA,B=2,C=NA,D=0) ) > dtf a b

Hi, I have two data frames. One is like > dtf = data.frame(y=c(rep(2002,4), rep(2003,5)), + m=c(9:12, 1:5), + def=c(.74,.75,.76,.78,.80,.82,.85,.85,.87)) and the other dtf2 = data.frame(y=rep( c(2002,2003),20), m=c(trunc(runif(20,1,5)),trunc(runif(20,9,12))), inc=rnorm(40,mean=300,sd=150) ) What I want is to divide

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I haven't followed your example closely, but can't you use the predict() method for this? To draw a curve, the function that will be used in curve() sets up a newdata dataframe and passes it to predict(fit, newdata= ...) to get predictions at those locations. Duncan Murdoch On 17/10/2020 5:27 a.m., Boris Steipe wrote: > I'm drawing a fitted normal distribution over a

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

Hi, I have problem getting the standard deviation from the manova output. I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1 + x2 + x3, data=mydata) . I tried to get the predicted values and their standard deviation by using: predict(myfit, type="response", se.fit=TRUE) But the problem is that I don't get the standard deviation values, I only

Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit <- tree (y ~ x1 + x2 + x3 + x4 ) # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as:

Dear List, I think I'm going crazy here...can anyone explain why do I get the same predictions in train and test data sets below when the second has a missing input? y <- rnorm(1000) x1 <- rnorm(1000) x2 <- rnorm(1000) train <- data.frame(y,x1,x2) test <- data.frame(x1) myfit <- glm(y ~ x1 + x2, data=train) summary(myfit) all(predict(myfit, test) == predict(myfit, train))

Hi R-Helpers: I?m trying to fit the same linear model to a bunch of variables in a data frame, so I was trying to adapt the codes John Fox, Spencer Graves and Peter Dalgaard proposed and discused yesterday on this e-mail list: for (y in df[, 3:5]) { mod = lm(y ~ Trt*Dose, data = x, contrasts = list(Trt = contr.sum, Dose = contr.sum)) Anova(mod, type = "III") } ## by John Fox or for