similar to: Matrix of dummies from a vector

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

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

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

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

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

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.

I have asked a similar question before but this time the problem is somewhat more involved. I have the following data: case;name;x 1;Joe;1 1;Mike;1 1;Zoe;1 2;Joe;1 2;Mike;0 2;Zoe;1 2;John;1 3;Mike;1 3;Zoe;0 3;Karl;0 I would like to count the number of "case" in which any two "name" a. both have "x=1", b. the first has "x=0" - the second has

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

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)

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

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

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:

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

Hallo All, I would like to apply a function to all permutations of variables in a dataframe (except the first). What is the best way to achieve this? I produce the permutations using: nvar <- ncol(dat) - 1 perms <- as.matrix( expand.grid(rep( list(1:0) , nvar ))[ , nvar:1] ) Thanks in advance Serguei Test-dataframe, comma-delimited: code,wav,w,area,gdp,def,pop,coast,milspend,agr

Hi, I have a more complex example, but the problem boils down to this FOR-loop not filling in the res-matrix run_rows<-seq(0,1,0.05) run_cols<-seq(0.3,0.6,0.05) res<-matrix(NA,length(run_rows),length(run_cols)) for(i in run_rows) { for(j in run_cols) { res[i,j]=i+j #niether the above, nor res[[i,j]]=i+j work, why? } } Thank you, Serguei

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

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

Dear list, I would like to rename an object as follows: SimLUall <- matrix(c(0,1,0,0, 1,0,0,0, 0,0,1,0, 0,0,0,1),nrow=4, ncol=4) j <- 2 SimLUall2 <- SimLUall and j The value of j should be assigned automatically to SimLUall. How can I achieve this? Regards, Ulrich

Hallo, suppose I have a vector: x <- c(1,1,1,2,2,3,3,3,3,3,4) How can I generate a vector/sequence in which a fixed number of zeroes (say 3) is inserted between the consecutive values, so I get 1,1,1,0,0,0,2,2,0,0,0,3,3,3,3,3,0,0,0,4 thanks a lot, Serguei [[alternative HTML version deleted]]

Dear R-users, i try to solve to following linear programm in R 0 * x_1 + 2/3 * x_2 + 1/3 * x_3 + 1/3 * x_4 = 0.3 x_1 + x_2 + x_3 + x_4 = 1 x_1, x_2, x_3, x_4 > 0, x_1, x_2, x_3, x_4 < 1 as you can see i have no objective function here besides that i use the following code. library(lpSolve) f.obj<-c(1,1,1,1) f.con<-matrix(c(0,2/3,1/3,1/3, 1,1,1,1,