similar to: bar chart issue

He everybody, I want to add 1 to some elements of a vector: x is a vector u is a vector of idices, that is, integers assumed to be within the range 1..length(x) and I want to add 1 to the elements of x each time their index appears in u x[u]<-x[u]+1 works only when there are no duplicated values in u I found this solution: tu <- table(u) indices <- as.numeric(names(tu)) x[indices]

Hi, a colleague ran a stepwise discriminant analysis twice in a row and got different results, suggesting some "sochasticity" in the algorithms involved. I looked at her data and found that there was a lot of collinearity, so that I reckoned that maybe "stepclass" (klaR) cannot find a clear winner when trying to include a new variable and makes a random choice. Is that true?

Hi, I would like some extra information on the 'cgh' package in R. I noticed that there isn't much activity regarding this package on the R and BioC mailing list (I googled it). I started using this package and I have few questions: 1/ As I have a custom tiling like array @8um features resolution (affy), I have a lot of probes to work with. I'm assuming it is correct to

Hello R-help I don’t know how to interpret significance from the contr.poly() function . From the example below : how can I tell if data has a significant Linear/quadratic/cubic trend? > contr.poly(4, c(1,2,4,8))               .L         .Q          .C [1,] -0.51287764  0.5296271 -0.45436947 [2,] -0.32637668 -0.1059254  0.79514657 [3,]  0.04662524 -0.7679594 -0.39757328 [4,]  0.79262909 

--- Miroslav Lichvar <lichvarm@phoenix.inf.upol.cz> wrote: > On Thu, Oct 17, 2002 at 09:51:02AM -0500, Brady Patterson wrote: > > ... But it seems to me that you could make a decent guess > > about when something "new" happens based on the second derivative > of the signal > > (where the first derivative is the difference between a given > sample and the

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

Hello, not understanding the output of prcomp, I reduce the number of components and the output continues to show cumulative 100% of the variance explained, which can't be the case dropping from 8 components to 3. How do i get the output in terms of the cumulative % of the total variance, so when i go from total solution of 8 (8 variables in the data set), to a reduced number of

I have a data set like this: if I want to less than 200000 obs from this data set. How can I do these? > str(p1982) 'data.frame': 465979 obs. of 6 variables: $ p : Factor w/ 1982 levels "154l_aa","1A0P_aa",..: 1 1 1 1 1 1 1 1 1 1 ... $ aa : Factor w/ 19 levels "ALA","ARG","ASN",..: 2 16 4 5 18 3 19 3 2 9 ... $ as : num 152.0

I have a data set p1982, its structure is the following Then I take 20 observations from this dataset, and assign to pr. in p1982, p has 1982 levels, in dataset pr, p should have 1 levels. But I do str(pr), it shows that p still has 1982 levels. also for these > pr$aa [1] ARG THR ASP CYS TYR ASN VAL ASN ARG ILE ASP THR THR ALA SER CYS LYS THR ALA LYS Levels: ALA ARG ASN ASP CYS GLN

I have a data set like this. I want to do glm, but I get this error: Error in model.matrix.default(mt, mf, contrasts) : cannot allocate vector of length 932889958 I am wondering if my data set is too large or I did something wrong. Is there some limitation for data size for R? thanks, Aimin > p1982<- read.csv("p_1982_aa.csv") > names(p1982) [1] "p"

Dear all,   I did ANOVA for many response variables (Var1, Var2, ....Var75000), and i got the result of p-value like below. Now, I want to select those predictors, which have pvalue less than or equal to 0.05 for each response variable. For example, X1, X2, X3, X4, X5 and X6 in case of Var1, and similarly, X1, X2.......X5 in case of Var2, only X1 in case of Var3 and none of the predictors in case

Hello, I'm very intersted in developing a site using your platform. I was wondering if you've tested the theoretical maximums for this software and can show me some numbers. How many simultaneous users could I expect to handle at 128k (hardware not being the limiting factor)? If you'd like to speak please send a number and a good time to call. Sincerely, -- Eric Haddad

I am trying to use the survfit() function with the newdata argument to produce predicted survivor curves for a particular covariate profile. The main purpose of the plot will be to visualise the effect of snp1, coded 0 and 1. In my Cox model I have stratified by one variable, edu, and so I know I will automatically get a separate curve for each strata. My problem is how to deal with the

Hey, R-listers I am a new user of R and just found the package of PCURVE which can estimate principal curve for arbitrary dimensional data set. Now I have some 2-Dimensional data set X, which is stored as an Nx2 matrix in data.txt file and looks as following: -1.5551 2.4183 1.0051 1.0102 0.90644 0.82163 1.3248 1.7551 -1.3626 1.8568 -1.3554 1.8371 0.039396 0.0015521 -0.99622 0.99245 0.2314

hi there!! i am trying to implement the code of the e1071 package for naive bayes, but it doens't really work, any ideas?? i am very glad about any help!! i need a naive bayes with 10-fold cross validation: code: library(e1071) model <- naiveBayes(code ~ ., mydata) tune.control <- tune.control(random = FALSE, nrepeat = 1, repeat.aggregate = min, sampling = c("cross"),

Hi List, I have spent more than 30 minutes, but failed to read in this file using the read.table() function. I could not figure out how to fix the following error. Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements Any help would be be appreciated. Thanks, Pradip Muhuri ####### below is the reproducible example xd1 <-

Hello R community, I am computing weighted average statistic by using ddply function: My data set is: N1 T1 S1 I1 C1 FY-4 ROE11 EPS11 MKT11 N1 T1 S1 I1 C1 FY-3 ROE12 EPS12 MKT12 N1 T1 S1 I1 C1 FY-2 ROE13 EPS13 MKT13 N1 T1 S1 I1 C1 FY-1 ROE14 EPS14 MKT14 N1 T1 S1 I1 C1 FY0 ROE15 EPS15 MKT15 N1 T1 S1 I1 C1 FY1 ROE16 EPS16 MKT16 N1 T1 S1 I1 C1 FY2 ROE17

I would be incredibly grateful to anyone who'll help me translate some SAS code into R code. Say for example that I have a dataset named "dat1" that includes five variables: wshed, site, species, bda, and sla. I can calculate with the following SAS code the mean, CV, se, and number of observations of "bda" and "sla" for each combination of

My operating system is Solaris 8 SPARC and it runs Samba 3.0.1 Any suggestions? I have included configuration/logs and traces. This is the smb.conf file: # Samba config file created using SWAT # from 151.184.34.182 (151.184.34.182) # Date: 2003/09/17 20:34:48 # Global parameters [global] netbios name = dali workgroup = DALIUNIX passdb backend = tdbsam os level

I was trying to fit a curve to the number of people who identify as liberal by age. I got some puzzling results which suggested to me that I don't really understand how local polynomial fitting works. Why, I am wondering, is lowess producing a local fit of zero for every age? > liberal.bin [1] 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0 0