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Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

I am using R on unix version 63.0 I am doing an image plot of the following data file: ================================ lag1 lag2 cif2d 0.000 0.000 NaN 0.000 1.000 0.500000 0.000 2.000 0.489831 0.000 3.000 0.492986 0.000 4.000 0.493409 0.000 5.000 0.492727 0.000 6.000 0.494485 1.000 0.000 0.500000 1.000 1.000 NaN 1.000 2.000 0.495098 1.000 3.000 0.489831 1.000 4.000 0.492986 1.000 5.000

Hello everyone, I am using *mlogit* to analyse my choice experiment data. I have *3 alternatives* for each individual and for each individual I have *9 questions*. I have a response from *516 individuals*. So it is a panel of 9*516 observations. I have arranged the data in long format (it contains 100 columns indicating different variables and identifiers). In mlogit I tried the following

I have a binary file(a) with size of (360 720 )for the globe.I wrote the code given below to read and extract an area (south america)from that file. when I use summary for the whole file I got: summary(a, na.rm=FALSE) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's 0.00 1.00 3.00 4.15 7.00 20.00 200083 . But when I used summary for the region(b) which I extracted, I got

Hi, I have more questions about the fft. The application in Excel is very limited. In Excel I can adjust graphs and calibrate the x and y-axis. The input and process, however, is limited compared to R. With a Dataset table where one column is the hour difference and the second are the values with hidden frequencies, how is it possible to plot a frequency spectrum with the x-axis calibrated to

Hi, I have the following problem (library Hmisc loaded, iris data loaded, R Version 1.7.0 (2003-04-16), packages updated, running on a linux Debian i386): > summary(Species~Sepal.Length,method="reverse")->a > a Descriptive Statistics by Species +------------+-----------------+-----------------+-----------------+ | |setosa |versicolor |virginica

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323

My data is called xc and has more than 15 variables. When I used summary(xc) it gave me the detail description of each variable. Summary(xc) Y1 x1 x2 x3 .. Min. :0.0000 Min. : 1.000 Min. : 1.000 Min. : 1.000 1st Qu. :0.0000 1st Qu.: 1.000 1st Qu.: 1.000 1st Qu.: 2.000 Median :1.0000 Median : 1.000

Hola colegas: He estado usando R Studo para obtener unos reportes que espero me ayuden a comprender un ejemplo que estoy tomando de la red. Usando sink("Tabla.txt") , al pedir summay(n), obtengo: sr tl 1 : 1 1: 4 10 : 1 2:13 18 : 1 3:10 30 : 1 4: 4 47 : 1 7: 9 67 : 1 (Other):34 d Min. : 1.000 1st Qu.: 2.000 Median : 5.000

Dear R experts, I have been given data files in the following configuration and have been puzzling about how to read them in. First I will give a snippet of the beginning of file: Data File: W Para File: GABOR_0.gor v 10.6, Date : 29/10/2007 Time : 13:33 3.00 5.000 Noise SD (deg): 15.000 1 -5.321 -5.321 2 -5.321 -3.991 3 -5.321 -2.661 4 -5.321 -1.330 5 -5.321 0.000 6 -5.321 1.330 7

I see um twitter dag's want to start his own RHEL rebuild. Dag how much monies you do need? I make contribution of 5.000 rupee if other join. Ant -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.centos.org/pipermail/centos/attachments/20110411/256d9f0d/attachment-0005.html>

I suspect the weights argument is not having any effect. Package: Matrix Version: 0.995-2 Date: 2006-01-19 Beginning with this: Browse[1]> resp.lmer <- lmer(SensSSC ~ Block + Season + (1 | Plot) + (1 | Ma) + (1 | Pa) + + (1 | MaPa), weights = SensSSC.N, data = xx) I group the output into a table with my ran.eff function and get this:

Dear Sirs, I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: >

Trying to apply the latest updates to an Athalon base fully updated CentOS 4.2. In the past used the update process kicked off by clicking the hand-dandy red icon. This failed, so I tried yum. Somewhere in the process, it complained that there was no GPG key for the lame libraries (IIRC, everything's a little hazy after all the Googling trying to fix this). Anyway, tried "yum

Full_Name: Kevin Wright Version: 1.4 OS: Windows 95 Submission from: (NULL) (170.54.59.160) Note: This was the 1.4 build for Windows that Brian Ripley made available. The first example in the help for reshape doesn't work for me. When I cut and paste, this is what happens: > data(Indometh,package="nls") > summary(Indometh) Subject time conc

Hi all, I have a very easy questions (I hope). I had measure a property of plants, growing in three different substrates (A, B and C). The rest of the conditions remained constant. There was very high variation on the results. I want to do address, whether there is any difference in the response (my measurement) from substrate to substrate?

Dear R list, I've been attempting to interpret the results from a three-way ANOVA. I think I understand contrasts and the R defaults for these (treatment contrasts). My question is: what is the intercept in this test? As far as I can tell, its NOT the expected value of a point that belongs to the first level of all three explanatory factors (because there is only one point that satisfies these

Dear R users, First of all, thanks for the incredibly fast answers and help of Rolf, Marc and Robert. Yes, I noticed that it was a lot of permutacions, but my intention was to make this process automatic and take only 5.000 - 10.000 permutations. Therefore, I wanted only to take that "interesting permutations" with "some information" [inter-block permutations]. The

I think you are not using the best function for what your intentions are. Try: > by(data=mtcars, INDICES=list(as.factor(mtcars$am)), FUN=colMeans) : 0 mpg cyl disp hp drat wt qsec vs 17.1473684 6.9473684 290.3789474 160.2631579 3.2863158 3.7688947 18.1831579 0.3684211 am gear carb 0.0000000

Dear all, I often import data sets from other programs that contain user defined missing values. As an example consider a variable MXPLOCO with a user defined missing value of as -9. > summary(MXPLOCO) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's -9.000 4.750 5.000 5.349 6.000 8.500 2.000 The following command successfully converts the value -9 to NA. However,