similar to: missing handling

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

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

Stock returns and other financial data have often found to be heavy-tailed. Even Cauchy distributions (without even a first absolute moment) have been entertained as models. Your qq function subtracts numbers on the scale of a normal (0,1) distribution from the input data. When the input data are scaled so that they are insignificant compared to 1, say, then you get essentially the

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

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 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

Hi, I have a dataset and want to build a generalized linear model on it. Unfortunately, complete.cases(df) returns null, which means I have to find a way to "fill" those missings. One way is following my previous post to use median to replace(or use most freq. of level to replace for catergorical case), but I am wondering if there are other ways, when glm or something like it is

Hi there, I need a network builder and it can change the node size and color; I am not sure if network package in R can do this or not. The other functions I wanted have been found in that package. BTW, if there is another package in R relating to this, please suggest too. Thanks, Weiwei -- Weiwei Shi, Ph.D Research Scientist "Did you always know?" "No, I did not. But I

hi, is there some good summary on clustering methods in R? It seems there are many packages involving it. And I have two questions on clustering here: 1. Is there a way of evaluate the effecitives (or seperation) of clustering (rather than by visualization)? 2. Is there a search method (like genetic search) which can help find the best subset of attributes which gives best seperation? Thanks,

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

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, 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

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: 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 have a question on read.table. I have a dataset with 273,000 lines and 195 columns. I used the read.table to load the data into R: trn<-read.table('train1.dat', header=F, sep='|', na.strings='.') I found it takes forever. then I run 1/10 of the data (test) using read.table again. And this time it finished quickly. So, there might be something wrong in my data

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]]

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 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..."

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,

Hi, there: I spent some time on this but I think I really cannot figure it out, maybe I missed something here: my data looks like this: > dim(trn3) [1] 7361 209 > dim(val3) [1] 7427 209 > mg.rf2<-randomForest(x=trn3[,1:208], y=trn3[,209], data=trn3, xtest=val3[, 1:208], ytest=val3[,209], importance=T) my test data has 7427 observations but after prediction, > dim(mg.rf2$votes)