search for: 0.462

Hello, I did a cross-validation using cvlm from DAAG package but wasn't sure how to assess the result. Does this result means my model is a good model? I understand that the overall ms is the mean of sum of squares. But is 0.0987 a good number? The response (i.e. gailRel5yr) has min,1st Quantile, median, mean and 3rd Quantile, and max as follows: (0.462, 0.628, 0.806, 0.896, 1.000, 2.400) ?

Hi, I've got the following edf: *** x = c(1.6,1.8,2.4,2.7,2.9,3.3,3.4,3.4,4,5.2) F2.5 <- ecdf(x) plot(F2.5, verticals= TRUE, do.p = TRUE, lwd=3, ylab = "", xlab = "", xlim = c(1,5.5)) abline(h= (0:5)*0.2) #mean abline(v=mean(x), lwd=2) mtext(text=expression(bar(x) == 3.07), side=1, adj=0.462, padj=3, cex=1) *** Now I would like to

With rw2010dev I get a strange protect(): protection stack overflow error with a small data frame which otherwise is usable: If anybody wants to have a look I can provide an RData file with the problematic data frame. Doesn't seem to be necessary, the following simulated example generates the error: > testmat <- matrix(1:80, 20,4) > dim(testmat) [1] 20 4 > str(testmat) int

Well pointed out, Jim! It is infortunate that the documentation for options(digits=...) does not mention that these are *significant digits* and not *decimal places* (which is what Joshua seems to want): "?digits?: controls the number of digits to print when printing numeric values." On the face of it, printing the value "0,517" of 'ccc' looks like printing 4

I did an update of both rstudio and my packages. I had some trouble but was able to move a lot of the packages so most troubles seem to be behind me. But having a problem with code that previously ran fine. See below: require(quantmod) Loading required package: quantmod Loading required package: Defaults Loading required package: xts Loading required package: zoo Attaching package: ?zoo? The

>>>>> Ted Harding >>>>> on Thu, 31 May 2018 07:10:32 +0100 writes: > Well pointed out, Jim! > It is infortunate that the documentation for options(digits=...) > does not mention that these are *significant digits* > and not *decimal places* (which is what Joshua seems to want): Since R 3.4.0 the help on ?options *does* say

Hi All, Please pardon me if I am missing something obvious here. How do I get the Kaplan-Meier estimate function that is created by survfit and plotted by the code. fit <- survfit(Surv(time, status) , data=aml) plot(fit) That is, I need a function that will give me the survival estimate at a given time: \hat{S}(t). Thanks in advance. Ritwik Sinha ritwik.sinha at gmail.com | +12033042111 |

Hi Joshua, Because there are no values in column ddd less than 1. itemInfo[3,"ddd"]<-0.3645372 itemInfo aaa bbb ccc ddd eee skill 1.396 6.225 0.517 5.775 2.497 predict 1.326 5.230 0.462 5.116 -2.673 waiting 1.117 4.948 NA 0.365 NA complex 1.237 4.170 0.220 4.713 5.642 novelty 1.054 4.005 0.442 4.260 2.076 creative 1.031 3.561 0.362 3.689

R version 3.5.0 (2018-04-23) -- "Joy in Playing" Platform: x86_64-pc-linux-gnu (64-bit) options(digits=3) itemInfo <- structure(list("aaa" = c(1.39633732316667, 1.32598263816667, 1.11658324066667, 1.23651072616667, 1.05368679983333, 1.03100737383333, 0.9630728395, 0.7483865045, 0.620086646166667, 0.5411017985, 0.496397607833333, 0.459528044666667, 0.427877047833333,

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]

Hi I have a follow up question, relating to subsetting to list items. After using the list and min(sapply()) method to adjust the length of the variables, I specify a dynamic regression equation using the variables in the list. My list looks like this: Dcr<- list(Dcre1=DCred1,Dcre2=DCred2,Dcre3=DCred3,Dbobc1=DBoBC1,Dbobc2=DBoBC2,Dbobc3=DBoBC3,...) By specifying the list items with names, I

Dear all I need to compute percentage changes of my data, but unfortunately they contain both negative and zero values, and I am quite confused on how to proceed. Searching the internet I found that many people ran into similar issues, with no obvious solution available. The last couple of weeks I've been playing with all the data transformations that I could think of. Below I will expose on

Dear R-Users My problem is quite simple: I need to use a fitted model to predict the next point (that is, just one single point in a curve). The data was divided in two parts: identification (x and y - class matrix) and validation (xt and yt - class matrix). I don't use all values in x and y but only the 10 nearest points (x[b,] and y[b,]) for each regression (b is a vector with the

Hello, I've fitted a parametric survival model by > survreg(Surv(Week, Cens) ~ C(Treatment, srmod.contr), > data = poll.surv.wo3) where srmod.contr is the following matrix of contrasts: prep auto poll self home [1,] 1 1 1.0000000 0.0 0 [2,] -1 0 0.0000000 0.0 0 [3,] 0 -1 0.0000000 0.0 0 [4,] 0 0 -0.3333333 1.0 0 [5,] 0 0

Hi, I have the following type of data: 86 subjects in three independent groups (high power vs low power vs control). Each subject solves 8 reasoning problems of two kinds: conflict problems and noconflict problems. I measure accuracy in solving the reasoning problems. To summarize: binary response, 1 within subject var (TYPE), 1 between subject var (POWER). I wanted to fit the following model:

Hello, Apologies if this is the wrong list, I am a first-time poster here. I have an experiment in which an output is measured in response to 42 different categories. I am only interested which of the categories is significantly different from a reference category. Here is the summary of the results: summary(simple.fit) Call: lm(formula = as.numeric(as.vector(TNFa)) ~ Mutant.ID, data =

?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]

Hi Everyone, I'm continuing to run into trouble with polyfit. I'm using the fitting function of the form; fit <- lm(y ~ poly(x,degree,raw=TRUE)) and I have found that in some cases a polynomial of certain degree can't be fit, the coefficient won't be calculated, because of a singularity. If I use orthogonal polynomials I can fit a polynomial of any degree, but I don't get

Hi All. I'd like to send some control arguments to the nls function when performing a nlsList analysis. I'm fitting a power model to some grouped data and would like to impose lower bounds on the estimates using the "port" algorithm. Obtaining the lower bound constraint works fine with a direct call to nls for a single level of the grouping variable. ?However, the bounds

To the purpose of fitting a 2nd order polynomial (a + b*x + c*x^2) to the chunk of signal falling in a 17 consecutive samples window I wrote the following very crude script. Since I have no previous experience of using Least Square Fit with R I would appreciate your supervision and suggestion. I guess the returned coefficients of the oolynomial are: a = -1.3191398 b = 0.1233055 c = 0.9297401