similar to: chisq.test

Dear list! I work with multiple Kruskal-Wallis test (kruskalmc, package pgirmess), which evaluates differences in medians among groups (5 groups). A result of a test is significant differences among some groups, while median values are the same for 4 groups (using tapply). Why? p.s.: number of samples in groups vary from 50 to 4900. Thanks to all, OV .

Dear list!   I would like to write a function to transform matrix, which is input argument of a written function. It is easy with new matrix (see below), but my idea is to transform input argument (matrix) of function without any additional matrixes. Here is an example: fun1 <- function(xy) { xy <- cbind(xy[,1], xy[,2], xy[,1] + xy[,2]) return(xy) }   df1 <- matrix(c(1,2,3,1,2,3), ncol =

Dear list members! I am looking for ''nice solution'' for (maybe) simple problem. I need a code (small program) to calculate row index for max value (example below: df1$values) by groups (example below: df1$groups). df1 <- data.frame( groups = c(1,1,1,1,1,2,2,2,3,3,3,3), values = c(1,1,1,2,1,1,2,3,2,1,4,3) ) df1 expected results > 4 8 11 # row index of max values by group

Dear list! I have problem with buffer size (width) in package rgeos. I would like to expand given geometry (points) to specified width based on the z value from attribute table. Here is example: point <- data.frame(x=c(10,20), y=c(10, 10), z = c(2,7)) point_shp <- SpatialPointsDataFrame(point[,1:2],point) plot(point_shp, xlim = c(0,30), ylim = c(0,20)) plot(gBuffer(point_shp, width = 5,

Dear list! ? I have question of?'correct function formation'. Which function (fun1 or fun2; see below) is written more correctly? Using ''structure'' as output or creating empty ''data.frame'' and then transform it as output? (fun1 and fun1 is just for illustration). ? Thanks a lot, OV ? code: input <- data.frame(x1 = rnorm(20), x2 = rnorm(20), x3 =

Hi R-helper! I have problem with setMethods for "[". Here is example : setClass("myClass", representation(ID.r = "numeric", ID.c = "character", DAT = "matrix")) to.myClass <- function(ID.r, ID.c, DAT) {     out <- new("myClass", ID.r = ID.r, ID.c = ID.c, DAT = DAT)     return(out)       }       setMethod("[",

Hi everybody! I have a matrix of class "myClass", for example: myMat <- matrix(rnorm(30), nrow = 6) attr(myMat, "class") <- "myClass" class(myMat) When I extract part of ''myMat'', the corresponding class ''myClass'' unfortunately disappear: myMat.p <- myMat[,1:2] class(myMat.p) Please for any advice / suggestions, how

Dear R helpers! I have a vector 'x1' and data.frame 'df1'. Do you have any suggestion how to get vector x2, which will be a result of matching values from vector 'x1' and values from 'df1'? Please, see the example: x1 <- c(rep(1,3), rep(NA,2), rep(2,4)) df1 <- data.frame(c1 = c(1,2), c2 = c(5,6)) I would like to get vector x2: > x2 [1]  5  5  5 NA NA  6 

Hi all!   My question in about using mapply instead for loop. Below is a example with for loop: Is it posible to give same results with mapply function?   Thanks for help!   OV   x <- 1:10 y <- 1:10 xyz <- data.frame(expand.grid(x,y)[1], expand.grid(x,y)[2], z = rnorm(100)) names(xyz) <- c("x", "y", "z") head(xyz) size <- 2 output <- NULL   ### for

Hi all!   My question is how to colour pixels by z value in persp plot in raster package. Here is an example:     x <- seq(-1.95, 1.95, length = 30) y <- seq(-1.95, 1.95, length = 35) z <- outer(x, y, function(a,b) a*b^2) r1 <- raster(nrows=35, ncols=30, xmn=0, xmx=30, ymn = 0, ymx = 35) r1[] <- c(z) persp(r1)   There already exist some function to produce persp plot for anothe

Dear List, If any of observed and/or expected data has less than 5 frequencies, then chisq.test (Pearson's Chi-squared Test for Count Data from package:stats) gives warning messages. For example, x<-c(10, 14, 10, 11, 11, 7, 8, 4, 1, 4, 4, 2, 1, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1) y<-c(9.13112391745095, 13.1626482033341, 12.6623267638188, 11.0130706413029, 9.16415925139016,

Hello, I received the following warning when running chi-square; n Is there a way to catch the 'error' code of 'warning' after run chisq.test(x)? n What does this error mean? Thank you for your help. [[alternative HTML version deleted]]

Full_Name: Reginaldo Constantino Version: 2.8.0 OS: Ubuntu Hardy (32 bit, kernel 2.6.24) Submission from: (NULL) (189.61.88.2) For many tables, chisq.test with simulate.p.value=TRUE gives a p value that is obviously incorrect and inversely proportional to the number of replicates: > data(HairEyeColor) > x <- margin.table(HairEyeColor, c(1, 2)) >

Full_Name: foo ba baz Version: R2.2.0 OS: Mac OS X (10.4) Submission from: (NULL) (219.66.32.183) chisq.test(matrix(c(9,10,9,11),2,2)) Chi-square value must be 0, and, P value must be 0 R does over correction when | a d - b c | < n / 2 &#65292;chi-sq must be 0

I am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

Hi, In the chisq.test(), if the expected frequency for some categories is <5, there will be a warning message which says Warning message: Chi-squared approximation may be incorrect in: chisq.test(x, p = probs) I am wondering whether there are some methods to get rid of this mistake... Seems the ?chisq.test() doesn''t provide more options to solve this problem. Or, the only choice is

Hi, I have a text mining project and currently I am working on feature generation/selection part. My plan is selecting a set of words or word combinations which have better discriminant capability than other words in telling the group id's (2 classes in this case) for a dataset which has 2,000,000 documents. One approach is using "contrast-set association rule mining" while the

Hello, I am trying to see whether there has been a significant difference in whether people experienced damages from wildlife in two different years. I therefore have two columns: year 1: yes no no no yes yes no year 2: no yes no yes I wanted to do a chisq.test, but if I enter it this way: chisq.test(year1, year2) I get the error saying the columns are two different lengths. So then I tried

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,