similar to: Select part of character row name in a data frame

Quoting Francesca PANCOTTO <f.pancotto at unimore.it>: > Dear R contributors, > > I have a problem in selecting in an efficient way, rows of a data > frame according to a condition, > which is a part of a row name of the table. > > The data frame is made of 64 rows and 2 columns, but the row names > are very long but I need to select them according to a small

(Re-)read the discussion of indexing (both `[` and `[[`) and be sure to get clear on the difference between matrices and data frames in the Introduction to R document that comes with R. There are many ways to create numeric vectors, character vectors, and logical vectors that can then be used as indexes, including the straightforward way: df[ c( "Unique to strat ", "Unique

Thanks a lot, so simple so efficient! I will study more the grep command I did not know. Thanks! Francesca Pancotto > Il giorno 19 ott 2017, alle ore 12:12, Enrico Schumann <es at enricoschumann.net> ha scritto: > > df[grep("strat", row.names(df)), ] [[alternative HTML version deleted]]

Dear contributors I have tried this experiment: x<-c() for (i in 1:12){ x[i]<-list(cbind(x1[i],x2[i])) #this is a list of 12 couples of time series I am using to perform a test } # that compares them 2 by 2 # ################# #trace statistic test<-data.frame() cval<-array( , dim=c(2,3,12)) for (i in 2:12){ for (k in 1:2){ for (j in 1:3){ result[k,j,i]<-

Dear Contributors, I have an easy question for you which is puzzling me instead. I am running loops similar to the following: for (i in c(100,1000,10000)){ print((mean(i))) #var<-var(rnorm(i,0,1)) } This is what I obtain: [1] 100 [1] 1000 [1] 10000 In this case I ask the software to print out the result, but I would like to store it in an object. I have tried a second loop, because if I

Dear Contributors, thanks for collaboration. I am trying to reorganize data frame, that looks like this: n1.Index Date PX_LAST n2.Index Date.1 PX_LAST.1 n3.Index Date.2 PX_LAST.2 1 NA 04/02/07 1.34 NA 04/02/07 1.36 NA 04/02/07 1.33 2 NA 04/09/07 1.34 NA 04/09/07

Dear Contributors, I am asking help on the way how to solve a problem related to loops for that I always get confused with. I would like to perform the following procedure in a compact way. Consider that p is a matrix composed of 100 rows and three columns. I need to calculate the sum over some rows of each column separately, as follows: fa1<-(colSums(p[1:25,])) fa2<-(colSums(p[26:50,]))

Dear Contributors, I am trying to perform a simulation over sample data, but I need to reproduce the same simulation over 4 groups of data. My ability with for loop is null, in particular related to dimensions as I always get, no matter what I try, "number of items to replace is not a multiple of replacement length" This is what I intend to do: replicate this operation for four

Terry - your example didn't demonstrate the problem because the variable that interacted with strata (zed) was not a factor variable. But I had stated the problem incorrectly. It's not that there are too many strata terms; there are too many non-strata terms when the variable interacting with the stratification factor is a factor variable. Here is a simple example, where I have

Terry - your example didn't demonstrate the problem because the variable that interacted with strata (zed) was not a factor variable. But I had stated the problem incorrectly. It's not that there are too many strata terms; there are too many non-strata terms when the variable interacting with the stratification factor is a factor variable. Here is a simple example, where I have

Dear Contributors I have a problem with the collection of data from the results of a test. I need to perform a comparative test over groups of data , recall the value of the pvalue and create a table. My problem is in the way to replicate the analysis over and over again over subsets of data according to a condition. I have this database, called y: gg t1 t2 d 40 1 1

Terry Therneau has been very helpful on r-help but we can't figure out what change in R in the past months made extra columns appear in model.matrix when the terms object is subsetted to remove stratification factors in a Cox model. Terry has changed his logic in the survival package to avoid this issue but he requires generating a larger design matrix then dropping columns. A simple

Dear Contributors I would like to perform this operation using a loop, instead of repeating the same operation many times. The numbers from 1 to 4 related to different groups that are in the database and for which I have the same data. x<-c(1,3,7) datiP1 <- datiP[datiP$city ==1,x]; datiP2 <- datiP[datiP$city ==2,x]; datiP3 <- datiP[datiP$city ==3,x] datiP4 <-

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Hi All, I am looking for help in rotating species titles produced using the strat.plot( ) function in the rioja package. This function produces a stratigraphic plot of paleoenvironmental data. Currently the titles of each species are plotted vertically while they are typically plotted at a 45 degree angle in other programs. Does anyone have any idea of how to rotates these titles? Below is an

I have a function with the follow signare: apply.strategy(instr, strat) where instr and strat are both objects of classes instrument and strategy respectively. I want to apply this function to a list that holds objects of the class instrument. Currently I am doing this by explicit looping: for(i in length(instr.list) ) { apply.strategy(instr.list[[i]], my.strat) } Is it possible to

Dear Helpers   I have a sample frame and i have sampled from it using three methods and now i want to calculate the statistics but i only get the population parameters.   H <- matrix(rnorm(100, mean=50000, sd=5000)) sampleframe=data.frame(type=c(rep("H",100)),value=c(H)) sampleframe   str=strata(sampleframe,c("type"),size=c(20,), method="srswor")

Samba shares work for windows 7 and Server 2008, but XP and Server 2000 recieve the following error when trying to map samba shares: "The trust relationship between this workstation and the primary domain failed." tail -f /var/log/messages Apr 10 07:38:03 samba01 smbd[23581]:?? connect_to_domain_password_server: unable to open the domain client session to machine ad1.strat.com. Error

Yo no conozco ningún paquete para hacer esto aunque probablemente exista. De todas maneras las cuentas de este tipo de calculos no suelen ser complicadas. Personalmente le recomendaría pasarse por una biblioteca y adquirir el siguiente libro donde explica con todo lujo de detalles, en castellano y simpliciddad como hacer estos calculos

Dear R users, I tried to fit a cox proportional hazard model to get estimation of stratified survival probability. my R code is as follows: cph(Surv(time.sur, status.sur)~ strat(colon[,13])+colon[,18] +colon[,20]+colon[,9], surv=TRUE) Error in if (!length(fname) || !any(fname == zname)) { : missing value where TRUE/FALSE needed Here colon[,13] is the one that I want to stratify and the