similar to: cluster in R

Hi, I searched the archives and did not find a good solution to that. assume I have 10 sets and I want to have the common character elements of them. how could i do that? -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

Hi, there: I am wondering if there is a quick way to "reverse" a list like this: t0 <- list(a=1, b=1, c=2, d=1) reverst t0 to t1 > t1 $`1` [1] "a" "b" "d" $`2` [1] "c" thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

Hi, I am wondering how to make a function Fun to make the following work: t0 <- (paste("hgu133a", "ENTREZID", sep="")) xx <- as.list(Fun(t0)) # make it work like xx<-as.list(hgu133aENTREZID) thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

hi, if I have 20 x 3 data.frame, how to split it into 10 x 6 (moving the lower part of 10x3 to column) or 5 x 12 thanks -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

Dear listers, a little off-topic: I am looking for and compare algorithms which can calculate "distance" or "similarity" between two gene lists with different lengths. Any paper, any implementation in R and any suggestion is welcome! Thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..."

Hi, there: Assume I have a dataframe with rownames like A with rownames like a to e, > A [,1] [,2] a 1 6 b 2 7 c 3 8 d 4 9 e 5 10 when I use A[1,], I lost the rowname for it, like below. How could I keep it? Is there an easy way instead that I have to modify by myself after I used A[1,] manually. > A[1,] [1] 1 6 Thanks, W. -- Weiwei Shi, Ph.D Research

Hi, there: Suppose I have a couple of data.frames and each one has five columns (one for x-axis, two for y-axis and two for std of y's.) There is another dimensions (besides x and y) which is continuous. My question is, how to plot such series of data frames in one plot (thus, 3-dimensional plot) AND multiple 2-D plots. I am not familar with R's plotting utilities. Thanks. -- Weiwei

Hi, I am using dlda algorithm from supclust package and I am wondering if the output can be a continuous probability instead of discrete class label (zero or one) since it puts some restriction on convariance matrix, compared with lda, while the latter can. thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..."

hi, i have a data frame like this: 0.3 0.7 0.4 0.8 i am trying to plot this data frame and each cell is filled with different colors based on the value. Is there a function which can do this? thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

Hi, I am trying to print out my confusion matrix after having created my random forest. I have put in this command: fit<-randomForest(MMS_ENABLED_HANDSET~.,data=dat,ntree=500,mtry=14, na.action=na.omit,confusion=TRUE) but I can't get it to give me the confusion matrix, anyone know how this works? Thansk! Ruben [[alternative HTML version deleted]]

Dear Listers: I have a couple of data frames to report and each corresponds to different condtions, e.g. conditions=c(10, 15, 20, 25). In this examples, four data frames need to be exported in a "pretty" report. I knew Perl has some module for exporting data to Excel and after googling, I found R does not. So I am wondering if there is a package in R for generating good reports. I

Dear Listers: For example, if I have a .R source file which has more than one function, and I want to just load only one of the functions, how could I do that? (removing the rest after sourcing is not what I intend b/c in my workspace, I might have some of the rest and I don't want to change my workspace: i.e., I only change my workspace by adding one function from a R source file). Thanks,

Dear all, I have a question on Cohen's kappa: Assume I have two datasets, one has 500 objects, 10 methods and the other, 1000 different objects, 20 different methods. Could I compare between the two datasets to conclude the 10 methods are more "concordant" than the 20 ones by looking at some output, for example, cohen.kappa{concord} ? One more, could anyone explain in brief,

Dear R-user; I recently switched from PC to MAC. Is there a compatible editor as Win-editor with package RWinEdit for MAC? Thanks Yihsu Chen [[alternative HTML version deleted]]

Hi, I am wondering how I can write a help page for my own function so when I type ?myOwnfunc, then the help page will show up? thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

hi, I have a conversion table for colnames like this: Probe_ID HUMAN_LLID 1 AF106325_PROBE1 7052 2 NM_019386_PROBE1 7052 3 NM_012907_PROBE1 339 4 AW917796_PROBE1 84196 5 L27651_PROBE1 10864 The Probe_ID contains a list of colnames for another data.frame, say x1. I need to convert such colnames to another ID's system, HUMAN_LLID by using the table.

Dear Listers: I have a question on loading objects from workspace: suppose I have two workspaces for two approaches. My old workspace has some objects I need for the new workspace but I don't want to load the whole old workspace and remove most of the old objects and get what I want. Is there an easier way to do like this: load "some needed obj" from old workspace, which has been

Hi, anyone interested in this: I tried to simply this loop with lapply or something but haven't figured it out: mapt = c("203929_s_at", "203930_s_at", "203928_x_at", "206401_s_at") mapt.combn <- lapply(1:4, function(i) combn(mapt, i)) out = list() k = 1 for (i in 1:length(mapt.combn)){ for (j in 1:ncol(mapt.combn[[i]])){ out[[k]] =

Dear Listers: I have this task and suppose a0 is a list of 10 data.frames, I want to calculate like this > (a0[[1]]+a0[[2]]+..+a[[10]])/10 Thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..." ---Matrix III

hi, i have another question on tapply: i have a dataset z like this: 5540 389100307391 2600 5541 389100307391 2600 5542 389100307391 2600 5543 389100307391 2600 5544 389100307391 2600 5546 381300302513 NA 5547 387000307470 NA 5548 387000307470 NA 5549 387000307470 NA 5550 387000307470 NA 5551 387000307470 NA 5552 387000307470