search for: lst2

...t;-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, how do you remove a row from a data frame? For example, say you have: lst1<-c(1,2,3,4,5) lst2<-c(6,7,8,9,10) frame<-data.frame(lst1,lst2) How do you remove, for example, the second row of frame? Thanks, - Jason

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

...-lapply(seq_len(nrow(DataB)),function(i) {x1<-unlist(mapply(function(x,y) which(x==y),DataA[,3:4],DataB[i,2]));x2<-if(length(x1)==2) DataA[x1[which.min(x1)],!names(DataA)%in%names(x1[which.max(x1)])] else if(length(x1)==1) DataA[x1,c("ID","Status",names(x1))] else NULL}) ?lst2<-lapply(lst1,data.frame) lst2<-lst2[lapply(lst2,nrow)!=0] ?lst2 #[[1]] #? ID Status??? Date2 #3? 1????? A 3-Sep-01 #[[2]] #? ID Status???? Date2 #5? 1????? C 26-Feb-04 #[[3]] #? ID Status??? Date1 #9? 2????? A 6-Mar-02 library(plyr) ?dataNew<-do.call(rbind,lapply(lst2,function(x) {colnam...

...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 <- list(label1 = x3, label2 = x4) > > BigList <- list(Lst1, Lst2) > > lapply(BigList$label1, mean) list() > [[alternative HTML version deleted]]

hello, I have programmed this function to calculate the Neuman-Keuls test but I have a problem the function return an empty list and I don't know why. summary(fm1) E <- sqrt((summary(fm1)[[1]]["Residuals","Mean Sq"])/length(LR)) lst <- list() lst1 <- list() lst2 <- list() NK <- function (x) { if (length(x) == 2) { Tstudent <- t.test(subset(exple, groupe == names(x)[1])$vd,subset(exple, groupe == names(x)[2])$vd) t <- as.numeric(Tstudent$statistic) if (t >= Tstudent$conf.int[1:2][1] & t <= Tstudent$conf.int[1:2][2]) { lst1[[...

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

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.)

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 name from http://r.789695.n4.nabble.com/ccf-functi...

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

...lied the first #constraint).? I guess you wanted to select only a pair of rows (dummy=0 and dummy=1) for each lists. fun1<- function(dat,percent,number){??? ??? lst1<- split(dat,list(dat$year,dat$industry)) ??? lst1New<- lapply(lst1,function(x) x[!(all(x$dummy==0)|all(x$dummy==1)),]) ??? lst2<- lst1New[lapply(lst1New,nrow)>0] ??? lst3<- lapply(lst2,function(x){ ??? ??? ??? ??? lapply(x$dimension,function(y){ ??? ??? ??? ??? ? x1<- x[(y < (x$dimension+(x$dimension*percent))) & (y > (x$dimension-(x$dimension*percent))),]??? ??? ??? ??? ??? ??? })??? ??? ??? ??? ???...

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

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)),]) ??? lst4<- lst3[lapply(lst3,nrow)!=0] ??? lst5<- lapply(seq_along(lst4),function(i){ ??? ??? ??? ??? ???? do.call(rbind,lapply(which(lst...

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

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

...t;>>>data.frame(Qm2,Qn2)})) >>>> >>>>res5<-cbind(res3,res4) >>>>head(res5,5) >>>> >>>>################################################# calculate cumulative sum ################################# >>>> >>>>lst2<- split(res5,list(res5$m1,res5$n1,res5$m,res5$n)) >>>> >>>>res5<-do.call(rbind,lapply(lst2[lapply(lst2,nrow)!=0], >>>>function(x){? >>>>x[,21:26]<-NA;? >>>>x[,21:22][x$Qm2<=c1 & x$Qm>c11,]<-cumsum(x[,17:18][x$Qm2<=...

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

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 ... .. ..

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$evnmt_brutal==0)),]) ??? lst4<-lapply(lst3,function(x) {vect.brutal=c() ??? ??? ??? ??? for(line in which(x$evnmt_brutal==1)){ ??? ??? ??? ??? ?? if(x$evnmt_bru...

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)),]) ??? lst4<- lst3[lapply(lst3,nrow)!=0] ??? lst5<- lapply(seq_along(lst4),function(i){ ???????????????????? do.call(rbind,lapply(which(lst...

I am using expression() to incorporate text into graphics. To create a superscript, I use the '^' character. Can someone please tell me the character to use to create a subscript? Thank you ----------------------------------------------------- Christine Donnelly Statistical Consulting Unit The Graduate School John Dedman Mathematical Sciences Building (Bldg 27) Australian National