similar to: Wrap a loop inside a function

Hello R-Users, its again me with a question. I´m using R 2.5.0 on Mac Power Book running Mac OS X 10.4.10 I try to calculate distances betweeen two data tables looking like this V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35 V36 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0

Hello, I am trying to get a linear model of y ~ log(x). *> lm (y~log(x))* However, I always get an error report: /Error in model.frame.default(formula = y ~ log(x), drop.unused.levels = TRUE) : variable lengths differ (found for 'log(x)')/ *Here was my y:* > y [1] 0.4500000 0.0500000 0.5000000 0.4000000 0.0000000 0.5000000 0.4000000 [8] 0.0500000

Dear list, I just made a very simple mistake, but it was hard to spot. And I think that I should warn other people, because it is probably so simple to make... === R code === # Let us create a matrix: (a <- cbind(c(0,1,1), rep(1,3))) # [,1] [,2] # [1,] 0 1 # [2,] 1 1 # [3,] 1 1 # That is a MISTAKE: a/colSums(a) # [,1] [,2] # [1,] 0.0000000 0.3333333

Question 2a) I am also working with arules package and I have the following problem let suppose the matrix b like: b<-matrix(c(1,1,1,1,1,1,0,0,1,1,1,1,0,0,1,1,0,1,1,1,1,1,1,1),nrow=6) rownames(b)=c("T1", "T2", "T3", "T4", "T5", "T6") colnames(b)=c("It1", "It2", "It3", "It4") bt<-as(b,

Dear R list users, the singluar value decomposition of a symmetric matrix M is UDV^(T), where U = V. La.svd(M) gives as output three elements: the diagonal of D and the two orthogonal matrices u and vt (which is already the transpose of v). I noticed that the transpose of vt is not exactly u. Why is that? thank you for your attention and your help Stefano AVVISO IMPORTANTE: Questo messaggio di

Dear R helpers: I have tried many times to find some way to read a list into R. But I faid. Here is an example: I have a file 'List.txt' which includes data as follows: [[1]] [1] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.5 0.0 0.0 0.0 [19] 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 [[2]] [1] 0.0000000 0.5000000 0.0000000 0.0000000

Dear all, mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3)) cor(mydata) x1 x2 x3 x1 1.0000000 -0.5960396 0.3973597 x2 -0.5960396 1.0000000 0.5000000 x3 0.3973597 0.5000000 1.0000000 I wonder if it is possible to fill only lower triangle of this correlation matrix? Using 'dist' doesn't seem to be useful as it doesnt allow to convert this table

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

Hi, xtabs() creates a table of counts. I want a table of proportions -- that is, I want to divide every vector (along a particular dimension) by its sum. The tiny example below does that. The call to xtabs() creates a matrix "A" with dimensions ("x1","x2","y"). I transform "A" using aperm() and aaply() to get the matrix "B". The

Dearing author: I am a student. I am a beginner in learning R-project. I have a queation!! If we want to solve following equation, how can i do first?? We want to solve X's and Y's value!! When we have two equation: For example: 2X+3Y=5 X+Y=2 I just take a simple example!! MY form is more difficult than above example!! Please help me!! And give me a reference to solve it!! THANK YOU!!

Fellow R-helpers, Suppose we create a histogram as follows (although it could be any vector with zeroes in it): R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> lenh$counts [1] 0 0 0 0 0 1 0 3 0 1 0 4 0 2 0 5 0 6 0 10 0 9 0 4 0 [26] 1 0 6 0 7 0 6 0 8 0 7 0 3 0 6 0 6 0 4 0 9 0 7 0 5 [51] 0 2 0 8 0 3 0 4 0 1 0 1 0 3

Dear R-users >prob <- c(0.5,0.4,0.3,0.1,0.0) >cal <- prob * log(prob,base=2) >cal [1] -0.5000000 -0.5287712 -0.5210897 -0.3321928 NaN Is there any way to change NaN to zero ? I did come up with this by applying Ripley's relpy to my previous question cal <-prob*log(pmax(prob,0.00000001),base=2) Any suggestion ? Thank you Taka

Dear all, I have two data frames- x1 and y1 with same row names and column names(actually the names of the patients). x1 a b c d e a 1.0000000 0.4730679 0.6226994 0.6036036 0.6433333 b 0.4730679 1.0000000 0.6227273 0.6303855 0.5730858 c 0.6226994 0.6227273 1.0000000 0.7290503 0.6900585 d 0.6036036 0.6303855 0.7290503 1.0000000

Dear sir, I have a matrix like a<-matrix(c(0,2,0,4,0,6,5,8,0),nrow=3) colnames(a)<-c("F1","F2","F3") rownames(a)<-c("A1","A2","A3") a F1 F2 F3 A1 0 4 5 A2 2 0 8 A3 0 6 0 I want to extract all pairs (rownames, columnames) from which the value in the matrix is 0 The result should be something like this A1, F1 A2,

Dear all, let say I have following data: dat <- structure(list(V1 = structure(c(1L, 4L, 5L, 3L, 3L, 5L, 6L, 6L, 4L, 3L, 5L, 6L, 5L, 5L, 4L, 4L, 6L, 2L, 3L, 4L, 3L, 3L, 2L, 5L, 3L, 6L, 3L, 3L, 6L, 3L, 6L, 1L, 6L, 5L, 2L, 2L), .Label = c("C", "G", "I", "O", "R", "T"), class = "factor"), V2 = c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L,

Dear R-help subscribers Would you tell me how to apply SSfpl with binary data as below? Unfortunately, there is not the EXAMPLE in help(SSfpl) for binary data but for quantitative data(Chick). V1: dose V2: log-transformed dose V3: response (rate) V1 V2 V3 1 0.775 -0.2548922 0.1666667 2 5.000 1.6094379 0.8148148 3 10.000 2.3025851 0.5000000 4 20.000 2.9957323

Hallo R Users, Please help I have some distance matrix data like > M[1:10,] [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.8750000 0.5000000 0.5000000 0.3750000 0.6250000 0.00000000 [2,] 0.8928571 1.0000000 0.0000000 0.8928571 0.1071429 0.00000000 [3,] 0.8928571 1.0000000 0.0000000 0.8928571 0.1071429 0.00000000 [4,] 0.8928571 1.0000000 0.0000000 0.8928571

Hi, I'm trying to find an implementation of monotonic regression in R and I haven't been able to find anything that's really related to this. isoMDS in the MASS package uses monotonic regression, however, I was wondering if there is any standalone function for monotonic regression? Basically what I'm trying to do is implement monotonic regression where I can see not just the

Hi, I would like to construct a transition matrix from a data frame with annual transitions of marked plants. plant<-c(1:6) class<-c("seed","seed", "seed", "veg", "rep", "rep") fate<-c("dead", "veg","veg","rep", "rep", "veg") trans<-data.frame(plant, class, fate)

Hi, a problem involving postscript bounding boxes: I'm composing three scatterplots into a single figure, postsript for publication. The individual scatterplots should be square, so the overall figure should have a roughly 1:3 sort of aspect ratio. By default however, the overall figure comes out nearly square, and the scatterplots are stretched vertically. I fixed this by adding