similar to: RKWARD

Dear list, Im trying to do the following operation but im not able to do it This is my table: 1 2 3 1 0 7 4 2 0 2 0 3 0 1 3 4 0 3 4 what i would like to do is divide each row values with the corresponding column' sum,namely: 1 2 3 1 0 0.54 0.36 2 0 0.15 0 3 0 0.08

Dear list, I'm looking for an inverse function of melt(which is in package reshape).Namely, I had a data frame like this (Table1) YEAR VAR1 VAR2 VAR3 1995 7 3 45 1996 5 6 32 1997 6 10 15 I transformed my data by using the melt function and my data was reshaped in the following format: (Table2) YEAR variable

Dear list, I'm trying to implement the following function, but what I get is an error message and I don't understand where is the error: #outliers'identification: iqr=lapply(bb,function(){ inner_fencesl=quantile(x,0.25)-1.5*IQR(x) inner_fencesh=quantile(x,0.75)+1.5*IQR(x) outer_fencesl=quantile(x,0.25)-3*IQR(x) outer_fencesh=quantile(x,0.75)+3*IQR(x)}) where bb is a dataframe

Dear list, I have a problem with the cumsum function. I have a data frame like the following one variable Year value EC01 2005 5 EC01 2006 10 AAO1 2005 2 AAO1 2006 4 what I would like to obtain is variable Year value cumsum EC01 2005 5 5 EC01

Dear List I would like to ask you something concenting a better print of the R output: I have a bit data frame which has the following structure: CFISCALE RAGSOCB ANNO VAR1 VAR2......... 9853312 astra 2005 6 45 9853312 astra 2006 78 45

Dear list, I'll try to be more clear in explaining my problem. I have a data frame like this called X: CLUSTER YEAR variable value1 value2 M1 2005 EC01 NA NA M1 2006 EC01 2 5 M1 2007

Hi, I would like to have an index for a column in a matrix encoded in a cell of the same matrix. For example: x = matrix(c(11,12,13,1, 21,22,23,3, 31,32,33,2),byrow=T,ncol=4) In this case, column 4 is the index. I then access the column specified in the index by: > for (i in 1:3) print(x[i,x[i,4]]) [1] 11 [1] 23 [1] 32 > > for (i in 1:3) {x[i,x[i,4]] <- x[i,x[i,4]] + 5} > x

I've noticed that I get a warning message every time a package masks some functions from Hmisc. The warning message says : Warning message: In identical(get(., i), get(., lib.pos)) : ignoring non-pairlist attributes This happens with eg: library(plyr) library(xtable) I think I've seen this passing by before, but I'm not sure any more. Just thought I'd mention it. Cheers Joris

?typeof? is your friend here: > typeof(`[`) [1] "special" > typeof(mc[[1]]) [1] "symbol" > typeof(mc2[[1]]) [1] "special" so mc[[1]] is a symbol, and thus not a primitive. - Lukas > On 28 Mar 2017, at 14:46, Michael Lawrence <lawrence.michael at gene.com> wrote: > > There is a difference between the symbol and the function (primitive >

I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this

Dear gurus, I was utterly surprised to learn that one of my examples illustrating the need of match.fun() doesn't give me the expected result. center <- function(x,FUN) FUN(x) center(1:10, mean) mean <- 4 center(1:10, mean) Used to give me the error message "could not find function FUN". Now it just works, even though I didn't expect it to. I believe this is at least

Im developing an application with R and Gtk+. It's just a simple GUI which helps new users to interactuate with R. Thing is, when you do a statistical analysis, I also want to provide a HTML report, but HTMLStart doesnt work propperly when executing from TinnR. It does create the file but not empty, I've tried some examples from different websites, and it's always the same.. it works

can anyone know where i can find information on compile hmisc on windows, especially 64 windows? thanks, _________________________________________________________________ The New Busy is not the too busy. Combine all your e-mail accounts with Hotmail. ID28326::T:WLMTAGL:ON:WL:en-US:WM_HMP:042010_4 [[alternative HTML version deleted]]

Hi, I want to give a summary or anova for "arima" model in R, as "summary", and "anova" for "lm". As including various intervention factors in arima(xreg = ) part, I want to assess the significancy of thse factors. I can do it using interrupted analysis of time series by linear regression, but want to see whether arima model works for the data first.

HI, I want to import 1.5G CSV file in R. But the following error comes: 'Victor allocation 12.4 size' How to read the large CSV file in R . Any one can help me? -- View this message in context: http://r.789695.n4.nabble.com/Large-Data-tp2254130p2254130.html Sent from the R help mailing list archive at Nabble.com.

All, How can we find a distance matrix for categorical data ie. given a csv below var1 var2 var3 var4 element1-1 yes x a k element1-2 no y b l element1-3 maybe y c m how can i compute the distance matrix between all the elements Actually i need it to create clusters on top

Dear Mr. or Ms.,   I used the R-software to run the zero-inflatoin negative binomial model (zeroinfl()) .   Firstly, I introduced one dummy variable to the model as an independent variable, and I got the estimators of parameters. But the results are not satisfied to me. So I introduced three dummy variables to the model. but I could not get the results. And the error message is

Apparently, as.POSIXlt takes one o'clock as the start of the day : > as.POSIXlt(0,origin="1970-01-01") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 00:00:00") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 23:59:59") [1] "1970-01-02 00:59:59 CET" Cheers -- Joris Meys Statistical

Dear all, Before I file this as a bug, I wanted to check if I didn't miss something. The help page of lag() says that the function returns a time series object. It actually does return something that looks like a ts object (the attribute tsp is set). But when using a vector, the class "ts" is not added to the result: > avec <- 1:10 > lag(avec) [1] 1 2 3 4 5 6 7 8

Hi all, I have a distribution, and take a sample of it. Then I compare that sample with the mean of the population like here in "Wilcoxon signed rank test with continuity correction": > wilcox.test(Sample,mu=mean(All), alt="two.sided") Wilcoxon signed rank test with continuity correction data: AlphaNoteOnsetDists V = 63855, p-value = 0.0002093 alternative hypothesis: