search for: rowtot

Recently, I noted a post and replies on R-Help from Professor Marc Feldesman regarding a cross tabulation function that generates row and column proportions, marginal values, expected cell values and tests for independence presumably similar in a fashion to the output of the S-Plus crosstabs() function or the SAS Proc Freq. Martin Maechler had posted some code in reply for folks to update and

Hi, In Splus if I call the function crosstabs() the output is a contigency table; in each cell of the table is printed: N, N/RowTotal, N/ColTotal, N/Total. N is the number of observations in each cell. The same call to xtabs() in R will produce the contigency table but the only entry in each cell is N. How can I get the same relative frequencies that crosstabs() gives? Thanks, mike -- ************************************...

Hi, I need to compare two matrices with each other. If you can get one of them out of the other one by resorting the rows and/or the columns, then both of them are equal, otherwise they're not. A matrix could look like this: [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 0 1 1 1 0 1 1 0 [2,] 1 1 0 0 0 1 0 1 [3,] 1 0 1 0 0

i have a data frame(dat) which has many variables.and i use the following script to get the crosstable. >danx2<-c("x1.1","x1.2","x1.3","x1.4","x1.5","x2","x4","x5","x6","x7","x8.1","x8.2","x8.3","x8.4","x11",

Hi all, I have a 3x4 contingency table with row totals all being 100. I want to generate 3 x 4 tables from the null distribution. Which R function can do this? -- Thanks, Jim. [[alternative HTML version deleted]]

hi, is there a way of calculating of measuring dependence between two categorical variables. i tried using the chi square test to test for independence but i got error saying that the lengths of the two vectors don't match. Suppose X and Y are two factors. X has 5 levels and Y has 7 levels. This is what i tried doing >temp<-chisq.test(x,y) but got error "the lengths of the two