similar to: Compiling a contingency table of counts by case

Dear all, how can I "reshape"/"cast" the following matrix 00;01;10;11 John.Mike;123;313;12;31 John.Jim;54;57;39;36 John.Steve;135;47;47;74 Mike.Jim;63;37;27;16 Mike.Steve;15;15;5;61 Jim.Steve;6;10;34;35 into a set of stacked 2x2 contingency tables 0;1 John;123;12 Mike;313;31 John;54;39 Jim;57;36 John;135;47 Steve;47;16 ... so that the "fisher.test" and

Here is an example of the data.frame that I have, df<-data.frame("case"=rep(1:5,each=9),"id"=rep(1:9,times=5),"x"=round(runif(length(rep(1:5,each=9))))) "case" represents the cases, "id" the persons, and "x" is the binary state. I would like to know in how many cases any two persons a. both have "1", b. the first has

Hello, how can I get rid of all dimnames so that: $amat Var3 Var2 Var1 8 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 7 1 1 1 0 1 0 0 0 1 0 0 0 0 0 0 6 1 1 0 1 0 1 0 0 0 1 0 0 0 0 0 5 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 4 1 0 1 1 0 0 1 0 0 0 0 1 0 0 0 3 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0

Given an x-vector with, say, 3 elements, I would like to compute the following vector of permutated products (1-x1)*(1-x2)*(1-x3) (1-x1)*(1-x2)*x3 (1-x1)*x2*(1-x3) x1*(1-x2)*(1-x3) (1-x1)*x2*x3 x1*(1-x2)*x3 x1*x2*(1-x3) x1*x2*x3 Now, I already have the correctly sorted matrix of permutations! So, the input looks something like: #input x<-c(0.3,0.1,0.2) Nx<-length(x) Ncomb<-2^Nx

Hi, I am trying to add duplicates of matrix "mat" by row. Commands subset(mat,duplicated(rownames(mat))) or mat[which(duplicated(rownames(mat))),] return only half of the required indices. How can I find the remaining ones, ie the matches, so that I can add them up? Thanks, Serguei ___________________________________________________________________ Austrian Institute of Economic

Hi all, for the data below I would like to 1. generate a dummy variable for each group "gr" of the same composition by people, then save each portion in a separate file, 2. compute the frequency of "1"'s in "x" for each person by group "gr". So, "mike" will have freq=2/3, as he has two "1" and one "0" in 3 groups.

Hi, In the data below, "case" represents cases, "x" binary states. Each "case" has exactly 9 "x", ie is a binary vector of length 9. There are 2^9=512 possible combinations of binary states in a given "case", ie 512 possible vectors. I generate these in the order of the decimals the vectors represent, as:

Say I have a FOR-loop for computing powers (just a trivial example) for(i in 1:5) { x<-i^2 y<-i^3 } How can I create a data.frame and a 3D plot of (i,x(i),y(i)), i.e. for each iteration Thanks, Serguei Kaniovski -- ___________________________________________________________________ ??sterreichisches Institut f??r Wirtschaftsforschung (WIFO) Name: Serguei Kaniovski

Hallo, I have two vectors of different lengths which contain the same set of values: X < -c(2,6,1,7,4,3,5) Y <- c(1,1,6,4,6,1,4,1,2,3,6,6,1,2,4,4,5,4,1,7,6,6,4,4,7,1,2) How can I replace the values in Y with the index (!) of the corresponding values in X. So 2 appears in X in the first coordinate, so all 2’s in Y should be replaced by 1, etc. Thank you for your help, Serguei

Dear All, Suppose I have the following dataframe: country;weight;group bul;10;1 cze;12;1 grc;12;1 hun;12;1 prt;12;1 rom14;1 fra;29;2 ita;29;2 gbr;29;2 aut;10;3 bel;12;3 The "group" variable denotes the id-number of a group of countries. How can I re-label the groups in the descending order of their cumulative "weight", which wound be: country;weight;group fra;29;1 ita;29;1

Dear All! I want to test a coeffcient restriction beta=1 in a univariate model lm (y~x). Entering lm((y-x)~1) does not help since anova test requires the same dependent variable. What is the right way to proceed? Thank you for your help and marry xmas, Serguei Kaniovski ________________________________________ Austrian Institute of Economic Research (WIFO)

Hello, how is, for example, the Schwarz criterion is defined for kmeans? It should be something like: k <- 2 vars <- 4 nobs <- 100 dat <- rbind(matrix(rnorm(nobs, sd = 0.3), ncol = vars), matrix(rnorm(nobs, mean = 1, sd = 0.3), ncol = vars)) colnames(dat) <- paste("var",1:4) (cl <- kmeans(dat, k)) schwarz <- sum(cl$withinss)+ vars*k*log(nobs) Thanks

I've a vector of pairwise correlations in the order low-index element precedes the high-index element, say: corr(1,2)=0.1, corr(1,3)=0.2, corr(2,3)=0.3, corr(3,4)=0.4 How can I construct the corresponding correlation matrix? I tried using the "combn"-function in "combinat" package: library(combinat) combn(c(0.1,0.2,0.3,0.4),2) , but to no avail... Thank you for your

Hi all, I apply "optim" to the function "obj", which minimizes the goodness of fit statistic and obtains Pearson minimum chi-squared estimate for x[1], x[2] and x[3]. The vector "fr" contains the four observed frequencies. Since "fr[i]" appears in the denominator, I would like to substitute "0" in the sum if fr[i]=0. I tried an

Dear R-Users I would like to estimate a constrained bivariate normal density, the constraint being that the means are of equal magnitude but of opposite signs. So I need to estimate four parameters: mu (meanvector (mu,-mu)) sigma_1 and sigma_2 (two sd deviations) rho (correlation coefficient) I have looked at several packages, including Gaussian mixture models in Mclust, but I am not sure

Dear All! my data is on pairs of countries, i and j, e.g.: y,i,j 1,AUT,BEL 2,AUT,GER 3,BEL,GER I would like to create a dummy (indicator) variable for use in regression (using factor?), such that it takes the value of 1 if the country is in the pair (i.e. EITHER an i-country OR an j-country). Thank you for your help, Serguei ________________________________________ Austrian Institute of

Hi all, does anyone know how to automatically update starting values in R? I' m fitting multiple nonlinear models and would like to know how I can update starting values without having to type them in. thank all --- On Fri, 12/26/08, r-help-request@r-project.org <r-help-request@r-project.org> wrote: From: r-help-request@r-project.org <r-help-request@r-project.org> Subject:

Dear All, I have several socio-economic and geographic variables for the 27 EU countries. I would to use these data to derive a correlation matrix between groups of countries (for a different application). I thought of using kmeans to cluster the groups, and then calibrate between group correlations using distances between the centroids, and within group correlations using distances in a cluster

Hello All, I have a panel date - here a small-scale example: df <- data.frame(cbind(rep(c("AUT","BEL","DEN","GER"),4),cbind(rep(c(1999,2000,2001,2002),4)),sample(10,16,replace=T))) names(df) <- c("country","year","x") SORT <- c("GER","BEL","DEN","AUT") I need to compute the

Hello, I cannot figure out how to use "tapply" to compute the correlation matrix in the variable "x" between the states? The data is in long format, e.g.: state,year,x Alabama,2001,0.45 Alabama,2002,0.47 Alabama,2003,0.48 Alabama,2004,0.44 Arizona,2001,0.34 Arizona,2002,0.32 Arizona,2003,0.38 Arizona,2004,0.36 Thank you in advance for your help, Serguei Kaniovski