similar to: print()

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

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:

Hi, I am using Rattle 2.6.4 with R 2.12.2 on win64, is this a bug ? Following is the content after execute the associate analysis process: Summary of the Apriori Association Rules: Number of Rules: 23351 Summary of the Measures of Interestingness: support confidence lift Min. :0.1250 Min. :1 Min. :2.667 1st Qu.:0.1250 1st Qu.:1 1st Qu.:2.667 Median :0.1250

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

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: >

Passing in two dates to a sql statement (sqldf). Is returning a factor. Tried setting back to a Date via as.Date, but get an error the error: character string is not in a standard unambiguous format. Any thoughts appreciated. Code/Results listed below: > summary(df.possible.combos) Date Hour Min. :2011-03-01 Min. : 0.00 1st Qu.:2011-03-23 1st Qu.: 5.75

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, Thanks for your help. However, I would like to draw your attention to the following: Actually, I was replicating the Example 2.3, using the dataset "brainsize.txt" given in Section 2.3.3 ("Summarize by group") at page 55, of a famous book "R by Example" written by "Jim Albert and Maria Rizzo" published in Springers (2012) in a Use R! Series. The

Dear All, I have got your core message, that it is my responsibility to determine whether any particular function in my version of R satisfies the language requirements at the time of your use. Jim Albert and Maria Rizzo must have used their code, which was permitted in the R-code of their time (2012). Therefore, I have now modified my R-code, as per R-3..2.4 version, according to my requirement

> On Apr 15, 2016, at 1:16 AM, Akhilesh Singh <akhileshsingh.igkv at gmail.com> wrote: > > Dear All, > > Thanks for your help. However, I would like to draw your attention to the > following: > > Actually, I was replicating the Example 2.3, using the dataset > "brainsize.txt" given in Section 2.3.3 ("Summarize by group") at page 55, > of a

Buenas tardes, Estoy interesado en generar observaciones de una distribucion binomial bivariada en la que hay _cierto_ grado de correlacion (denotemoslo rho). Podria por favor alguien indicarme como hacerlo en R? Este es el contexto. Supongamos que se tienen dos experimentos en los que la variable respuesta _sigue_ una distribucion binomial, i.e., X_i ~Binomial(n_i, p_i), i=1,2 y que, por ahora,

> On Apr 16, 2016, at 2:03 AM, Akhilesh Singh <akhileshsingh.igkv at gmail.com> wrote: > > Dear All, > > I have got your core message, that it is my responsibility to determine whether any particular function in my version of R satisfies the language requirements at the time of your use. Jim Albert and Maria Rizzo must have used their code, which was permitted in the R-code

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

Dear R Help: I am trying to sum data from one column in a dataframe based on a value in another. I do not know how to do this easily in R. For example: Col A Col B 1 0 3 0 2 1 2 0 1 0 4 0 1 1 9 1 3 0 5 0 2 1 I would like to cumsum the values in Col A for all rows where Col B is 0, and a value of 1 in Col B will reset

Dear R users; Is there a function in R that I can put "text" with proper alignments in a fixed format. For instance, if I have three fields: A, B and C, where both A and C are text with 3 characters and left alignment; B is a numeric one with 2 decimals and 3 integer space digits. How can I put the following row in a file? (note there is a space between a and 2, and after b.)

This is my first post to R-help and I look forward receiving some advice for a novice like me... I?ve got a simple repeated (4 periods so far) 10-question survey data that is very easy to work on Excel. However, I?d like to move the compilation to R but I?m having some trouble operating on count list data in a neat way. The data C > str(C) 'data.frame': 551 obs. of 13

Hi, I am trying to reproduce a tukey test in R ========================== x=c(145,40,40,120,180, 140,155,90,160,95, 195,150,205,110,160, 45,40,195,65,145, 195,230,115,235,225, 120,55,50,80,45 ) y2=c( rep(as.character(1),5), rep(as.character(2),5), rep(as.character(3),5), rep(as.character(4),5), rep(as.character(5),5), rep(as.character(6),5) ) crd2=data.frame(x,y2)

Hello there, I am doing analysis on survival data. How do I pick out a probability of survival at a chosen landmark time(for example, 3 years, 4 years) from the result of "survfit"? or any other functions? As I know, the result of 'survfit' only have the probabilities for all event or cencor time. Thank you very much Lisa Wang Princess Margaret Hospital Toronto, Ca tel:

Hello, I am trying to print a crosttabulation of mean,sd,n for a continuous variable crossclassified by anoother/s grouping variables. I came up with: xtab2 <- function(x,g1,g2) { funy <- function(z) list(mean(z,na.rm=T),sd(z,na.rm=T),length(z)) aa <- by(x,list(g1,g2),funy) bb <- matrix(unlist(aa),nrow=3 ,dimnames=list(c("mean","sd","n"),

Hi guys My data is Tasmania txt There are *N *= 16 samples, consisting of 8 replicate cores (taken from different areas across the sandflat) from each of 2 natural 'treatments' (either disturbed "D" or undisturbed "U" by soldier crab burrowing activity. The abundances of each of *p *= 56 species were recorded from each core (variables 1 to 39 in the file are