similar to: Editing Strings in R

Hi; I am having problems inverting matrices using the function solve() For example R can not invert the following matrix [,1] [,2] [,3] [,4] [,5] [1,] 25 500 11250 275000 7.106250e+06 [2,] 500 11250 275000 7106250 1.906250e+08

I am new to R and have written a function that clusters on subsets of a big data data set with 60,000 points. I am not sure why, but I keep getting a run-time error. Any suggestions would be greatly appreciated. Here is the code: library(cba) d<-read.csv("data.csv", header=TRUE) v<-c(53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,21,72 73,74,75,76,77,78)

Surprisingly, I played around with some test code and below actually creates equations that look correct. tempmat<-matrix(10,nrow=6,ncol=6) restrictmat<-diag(6) colnames(tempmat)<-c("AUD.l1","CHF.l1","CAD.l1","GBP.l1","EUR.l1","JPY.l 1")

I have a data frame. And I'd like to subset according to rownames. subset(mydataframe, rownames(mydataframe) == myrow, select = mycols) it turned out that "myrow" cannot be a vector. But I have multiple rows to pick. Is there a way to get around this problem?? Thank you for your help!! Lei Jiang Department of Chemsitry University of Washington Box 351700 Seattle, WA 98195

An unwanted side effect of the new restrictions on abrreviated names. The anova.coxph command, in a slavish copy of anova.lm etc, returns a data frame with column labels of loglik Chisq Df Pr(>|Chi|) If one tries to extract the final column of the table errors result since it is not a standard R variable name. > afit <- anova(lm(conc ~ uptake, CO2)) > afit$P [1]

I read somewhere that t.test is equivalent to a hypothesis testing for one way ANOVA. But I'm wondering how they are equivalent. In the following code, the p-value by t.test() is not the same from the value in the last command. Could somebody let me know where I am wrong? > set.seed(0) > N1=10 > N2=10 > x=rnorm(N1) > y=rnorm(N2) > t.test(x,y) Welch Two Sample t-test data:

Hello,=0D =0D I just noticed that the "svd" function does not work properly for some=0D sparse matrices.=0D When I replace the 0 by very small noises (let's say 10^-16), it then=0D works.=0D The test I've performed is to compared the singular values to the eigen=0D values (as I work with squarred matrices).=0D =0D Here is the code (I may be wrong!):=0D =0D

Is there easy way (without copying the existing rows to a temporary location and copying back) to add a new row to a specific index location in an existing data frame? Example df = data.frame( A= c('a','b','c'), B=c(1,2,3), C=(10,20,30)) newrow = c('X', 100, 200) I want to add the newrow as the second row to the data frame df Please suggest a solution that is

Hi, I don't understand what the meaning of the following lines returned by model.matrix(). Can somebody help me understand it? What can they be used for? attr(,"assign") [1] 0 1 2 2 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4

Hi All, I was adding a new row of data to my data frame using rbind(). I was surprised to see that after adding new row, I lost my data frame level attibute as well as col level attribute. Please help me to insert a new row at frist or middle position so that my custom attribute is not lost. Here is what I did. age<-c(15,20,18) weight<-c(40,42,30) ### creating my data frame

I have the R code at the end. The last command gives me "1 observation deleted due to missingness". I don't understand what this error message. Could somebody help me understand it and how to fix the problem? > summary(afit) Df Sum Sq Mean Sq F value Pr(>F) A 2 0.328 0.16382 0.1899 0.82727 B 3 2.882 0.96057 1.1136 0.34644 C

I need to implement a conditional in my RJS template which looks something like: if (page.select(''row1'').first != null) page << "new TableRow.MoveAfter(''row1'', ''newrow'');" else page << "new TableRow.MoveAfter(''row2'', ''newrow'');" end Now, dumb question.. My

Hello fellow R sufferers, Is there a way to perform an appending operation in place? Currently, the way my pseudo-code goes is like this for (i in 1:1000) { if (some condition) { newRow <- myFunction(myArguments) X <- rbind(X, newRow) # <- this is the bottleneck!! } } As you can see, it works but as the matrix X gets the size of a few million rows, the

Grateful for any hints as to why I'm not getting the inner loop to cycle the expected number of times. Code and one run's results below. Thanks, Galen > # source("looptest.r") > sp<-numeric() > iter<-numeric() > rn<-numeric() > ds<-data.frame(sp, iter, rn) > > for (sp in c(1:6)) { + i<-1 +

I have the following code. I'm wondering why summary() doesn't show F value and Pr? Rscript multi_factor.R > a=3 > b=4 > c=5 > d=6 > e=7 > > A=1:a > B=1:b > C=1:c > D=1:d > E=1:e > > X=matrix(nr=a*b*c*d*e,nc=5) > colnames(X)=LETTERS[1:5] > > for(i_a in 1:a-1) { + for(i_b in 1:b-1) { + for(i_c in 1:c-1) { + for(i_d in 1:d-1) { +

Hi, I run the following commands. 'A' has 3 levels and 'B' has 4 levels. Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3, B4)? > a=3 > b=4 > n=1000 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > > fr =

Hi, I want to do ANOVA for nested designs like following. I don't understand why the matrix (t(X) %*% X) is singular. Can somebody help me understand it? Regards, Peng > a=2 > b=3 > n=4 > A = as.vector(sapply(1:a,function(x){rep(x,b*n)})) > B = as.vector(sapply(1:(a*b), function(x){rep(x,n)})) > cbind(A,B) A B [1,] 1 1 [2,] 1 1 [3,] 1 1 [4,] 1 1 [5,] 1 2 [6,] 1 2 [7,]

Hello Gurus, I am still new to R. Here is my issue. I was trying to add column to data frame that was populated by read.spss(). When I used cbind to add a new variable(column). library(foreign) mydf<-read.spss(file="C:/myspss.sav",use.value.labels=FALSE, to.data.frame=TRUE,use.missings=FALSE) attr(mydf,"variable.labels") ## it gives you all the labels

Dear friends, I need to fill in (duplicate the whole record) the missing days with the same record values as long as AE is the same value (i.e. "1"), once AE value changes, the process of duplication should proceed with the new AE value till it changes again. e.g. I need to fill in records: day 18-day 44, all the records are carried with the new AE value of "0". At the

Hi, I am wondering if anyone knows how to perform an F-test on the change in R square between hierarchical models in R? SPSS provides this information and a researcher that I am working with is interested in getting this information. Alternatively, if someone knows how I can calculate the test statistic (SPSS calls it F-change?) and dfs that would be helpful as well. The output and the test I am