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The editor in R version 1.2.0 returns an error after editing a data frame that has row names. This is on a Windows 2000 system, but I've encountered the same problem under Windows 95. Here's a simple example: ------------------------------------------------------------------------------------ R : Copyright 2000, The R Development Core Team Version 1.2.0 (2000-12-15) R is free

On Sun, 31 Dec 2000 jfox@mcmaster.ca wrote: > The editor in R version 1.2.0 returns an error after editing a data frame > that has row names. This is on a Windows 2000 system, but I've encountered > the same problem under Windows 95. Here's a simple example: Actually, that's not where the error is. If you use traceback() you will see it is in edit.data.frame. That tries

Hello, I am trying to filter a data set like below so that the peaks in the Phase value are more obvious and can be identified by a peak finding function following the useful advise of Carl Witthoft. I have written the following for(i in length(data$Phase)){ newphase=if(abs(data$Phase[i+1]-data$Phase[i])>6){ data$Phase[i+1] }else{data$Phase[i] } } I get the following error which I have not

At present the example data sets in R libraries are to be given as expressions that can be read directly into R. For example, the acid.R file in the main library looks like acid <- data.frame( carb = c(0.1, 0.3, 0.5, 0.6, 0.7, 0.9), optden = c(0.086, 0.269, 0.446, 0.538, 0.626, 0.782), row.names = paste(1:6)) This is great when you have only a few observations. I have one example data

Hello, I'm very new to using R, but I was told it could do what I want. I'm not sure how best to enter the information but here goes... I'm trying to transfer the following integral into R to solve for ln(gamma_1), on the left, for multiple instances of gamma_i and variable N_i. gamma_i is, for example, (0, 0.03012048, 0.05000000, 0.19200000, 0.44000000, 0.62566845) N_i (N_1 or

I am a total newbie at R but experienced with computers. If this is not the right forum for this question, please let me know one that is. I searched in the R manuals in the Help section, the Nabble help and Rseek but no luck. I am trying to install the package gcl on my recently installed R 2.8.1 I searched on cran for the windows version and found a zip file for the package In RGui, I ran

days=Sys.Date()-1:70 price=abs(rnorm(70)) regular=rep(c(0,0,0,0,1,0,1,0,0,1),c(7,7,7,7,7,7,7,7,7,7)) y=data.frame(cbind(days,price,regular)) y is like days price regular 1 14990 0.16149463 0 2 14989 1.69519358 0 3 14988 1.57821998 0 4 14987 0.47614311 0 5 14986 0.87016180 0 6 14985 2.55679229 0 7 14984 0.89753533 0 the output I want:

Dear Thomas, The head of my dataset > head(wsuv) parcel sp time censo treatment species 1 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 2 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 3 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1 1 4 S8 Poecilanthe effusa ( Hub. ) Ducke. 1 1 1

Hi I have a table that goes data cluster_ac clockrate age class 7337 0.9 0.001 alpha_proteins 7888 0.1 0.78 beta proteins etc The class column can have 7-8 different unique values While the clockrate and age columns are floats varying from 0 to 1. I wish to get the average clockrate across each of the classes for this data. I would appreciate your help

I've been trying to run some analysis using the nls function in R but keep coming up with an error message which I don't understand how to fix. The message follows here: Error in nls(formula = f.p, data = n.data, start = list(S = 1, a = -0.1, : singular gradient In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing

To get the number of TRUE in logical vector 'x', tabulate(x, 1) can be used. Actually, it gives count of 1, but TRUE as integer is 1. Since R 3.4.0, it gives a correct answer, too, when the count is 2^31 or more.

Ah, yes, of course. tabulate(x, 1) doesn't work, too, in R 3.4.0. Sorry, I didn't actually try. I thought of an alternative when TRUE count is 2^31 or more. sum(x) returns NA with a warning. sum(as.numeric(x)) works, but requires a quite large memory. -------------------------------------------- On Thu, 8/6/17, Bert Gunter <bgunter.4567 at gmail.com> wrote: Subject: Re: [R]

I am encountering difficulty with NextMethod in 0.50-a4. We created a class of groupedData objects which are data.frames with additional attributes. The most important attribute is a formula describing roles of some of the variables in the experimental design. The class of such objects ends in "groupedData", "data.frame". The print method for the groupedData class simply

Hi, I make a weibull survival regression using suvreg function. Bu when I try to get the P values from anova, it give me NAs: I'm using R 2.4.0 and survival 2.29 Look: m <- survreg(Surv(tempo,censor)~grupo*peso) anova(m) Df Deviance Resid. Df -2*LL P(>|Chi|) NULL NA NA 148 966.6416 NA grupo -2 25.6334407 146 941.0081 NA

Hi, I make a weibull survival regression using suvreg function. Bu when I try to get the P values from anova, it give me NAs: I'm using R version 2.4.0 Patched (2006-11-25 r39997) and survival 2.30 library Look: m <- survreg(Surv(tempo,censor)~grupo*peso) anova(m) Df Deviance Resid. Df -2*LL P(>|Chi|) NULL NA NA 148 966.6416 NA grupo

Dear useRs, I am trying to model a time series with a transfer function. I think it can be put into the ARMA framework, and estimated with the 'arima' function (and others have made similar comments on this list). I have tried to do that, but the results have so far been disappointing. Maybe I am trying to make 'arima' do something it can't... The data are time series of

Hello, Thank you to all those great folks that have helped me in the past (especially Dennis Murphy). I have a new challenge. I often generate time-series data sets that look like the one below, with a variable ("Phase") which has a series of flat-topped peaks (sample data below with 5 "peaks"). I would like to calculate the phase value for each peak. It would be great to

Good afternoon, I?m in need of an advice regarding a proper use of glm.nb, zeroinfl or hurdle with my dataframe. I can not provide a self-contained example, since I need an advice on this current dataset and its ?contradictory? results. So.... i have a dataset which contains 1309 cases and 11 variables, highly right-skewed and heavily zeroinflated (with over 1100 cases that have 0 value

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

Hi all, Why subscripting a one column matrix drops one dimension? > x<- matrix(rnorm(100), ncol=1) > str(x) num [1:100, 1] -0.413 -0.845 -1.625 -1.393 0.507 ... > str(x[20:30,]) num [1:11] -0.315 -0.693 -0.771 0.448 0.204 ... > str(x[20:30]) num [1:11] -0.315 -0.693 -0.771 0.448 0.204 ... This breaks: > cov(x) [,1] [1,] 0.9600812 >