similar to: Neuman-Keuls

Sorry to ask such a simple question, but I can't find the answer after extensive searching the docs and the web. How do you remove a component from a list? For example say you have: lst<-c(5,6,7,8,9) How do you remove, for example, the third component in the list? lst[[3]]]<-NULL generates an error: "Error: more elements supplied than there are to replace" Also,

Hi, May be this helps: #Creating some dummy data.? set.seed(24) lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE))) set.seed(49) lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE))) ??Find_Max_CCF() #No vignettes or demos or help files found with alias or concept or #title matching ?Find_Max_CCF? using regular expression matching. Found a function with the same

Hi. I have a list where each object in the list has multiple parts. I'd like to take the mean of just one part of each object. Is it possible to do this with lapply? If not, can you recommend another function? Thanks. eric > x1 <- c(0,1,2,3) > x2 <- c(7,8) > x3 <- c(2,6,6,8) > x4 <- c(4,8) > > Lst1 <- list(label1 = x1,label2 = x2) > Lst2 <-

Hi, ?In the example you provided, it looks like the dates in Date2 happens first.? So, I changed it a bit.? DataA<- read.table(text=" ID,Status,Date1,Date2 ??? ??? ?????? 1,A,3-Feb-01,15-May-01 ??? ??? 1,B,15-May-01,16-May-01 ??? ??? 1,A,16-May-01,3-Sep-01 ??? ??? ??? ??? ??? 1,B,3-Sep-01,13-Sep-01 ??? ??? ??? ??? ??? 1,C,13-Sep-01,26-Feb-04 ??? ??? ??? ??? ???

hello, I need to groupe some means and I wonder how to do to get critical range table for tukey or for Neuman-Keuls I think it's possible to do that with the qtukey() function? thanks. _____________________________________________________________________________ [[alternative HTML version deleted]]

Hi. I have 2 tables, with same dimensions (8000 x 5). Something like: tab1: V1 V2 V3 V4 V5 14.23 1.71 2.43 15.6 127 13.20 1.78 2.14 11.2 100 13.16 2.36 2.67 18.6 101 14.37 1.95 2.50 16.8 113 13.24 2.59 2.87 21.0 118 tab2: V1 V2 V3 V4 V5 1.23 1.1 2.3 1.6 17 1.20 1.8 2.4 1.2 10 1.16 2.6 2.7 1.6 11 1.37 1.5 2.0 1.8 13 1.24 2.9 2.7 2.0 18 I need generate a table of averages, the

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi, Try this: final3New<-read.table(file="real_data_cecilia.txt",sep="\t") dim(final3New) #[1] 5369??? 5 #Inside the split within split, dummy==1 for the first row.? For lists that have many rows, I selected the row with dummy==0 (from the rest) using the #condition that the absolute difference between the dimensions of those rows and the first row dimension was minimum

Hi, May be this helps: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun2<- function(dat){?? ????? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])

Dear list I have a matrix composed of islandID as rows and speciesID as columns. IslandID: Island A, B, C….O (15 islands in total) SpeciesID: D0001, D0002, D0003….D0100 (100 species in total) The cell of the matrix describes presence (1) or absence (0) of the species in an island. Now I would like to search the species with absence (0) in all the islands (Island A to Island O.)

Hey, Is it possible that R can calculate each options under each column and return a summary table? Suppose I have a table like this: Gender Age Rate Female 0-10 Good Male 0-10 Good Female 11-20 Bad Male 11-20 Bad Male >20 N/A I want to have a summary table including the information that how many answers in each category, sth like this: X

Writing R Ext says to treat R objects that are arguments to .Call as read only (i.e. don't modify). I have a long list of lists that and I want to avoid the overhead of a copy in my C code. I would just like to modify some of the elements of list by replacing them with elements of exactly the same size/type. below is an example of the essence of the problem. This seems to work. Is this

Hello, having a data frame like test with pairs of characters I would like to create chains. For instance from the pairs A/B and B/I you get the vector A B I. It is like jumping from one pair to the next related pair. So for my example test you should get: A B F G H I C F I K D L M N O P > test V1 V2 1 A B 2 A F 3 A G 4 A H 5 B F 6 B I 7 C F 8 C I 9 C K 10 D L

Hi guys, I still did not solve my problem properly! I have to compare the values of two lists of 250 numbers as a result of using the ?by function! List1 of 250  $ 0   : num [1:28] 22 11 31...  $ 1   : num [1:15] 12 14 9 ... .. .. ..  - attr(*, "dim")= int 250  - attr(*, "dimnames")=List of 1 List2 of 250  $ 0   : num [1:24] 20 12 22...  $ 1   : num [1:17] 11 12 19 ... .. ..

Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for

HI, Using c11<- 0.01 c12<- 0.01 c1<- 0.10 c2<- 0.10 One possible problem is that: dim(res5) #[1] 513? 20 res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max) #Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) : ?# undefined columns selected A.K. ________________________________ From: Joanna Zhang <zjoanna2013 at gmail.com> To: arun <smartpink111 at

I R-helpers #I have a data panel of thousands of firms, by year and industry and #one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not #and another variable the represents the firm dimension (total assets in thousand of euros) #I need to create two separated samples with the same number os firms where #one firm in the first have a corresponding

I'm a bit of an amateur R programmer. I can do simple R scenarios but my handle on complex grammatical issues isn't steady. I have 12 CSV files that I've read into dataframes. Each has 8 columns and over 2000000 rows. Each dataframe has data associated by time component and a date component in the format of: X.DATE and then X.TIME X.DATE is in the format of MMDDYYYY and X.TIME is

Hi, Try this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun1<- function(dat){??? ? ??? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),]) ???

HI, You can do this: dat1<- read.csv("dat7.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t") dat.bru<- dat1[!is.na(dat1$evnmt_brutal),] fun2<- function(dat){? ????? lst1<- split(dat,dat$patient_id) ??? lst2<- lapply(lst1,function(x) x[cumsum(x$evnmt_brutal==0)>0,]) ??? lst3<- lapply(lst2,function(x) x[!(all(x$evnmt_brutal==1)|all(x$evnmt_brutal==0)),])