similar to: Using apply to get group means

Dear all Please, can you advice me how to compute an error, standard deviation or another measure of variability of computed value. I would like to do something like: var(y) = some.function(var(x1),var(x2),var(x3)) for level F1 (2,3,...) Let say I have some variables - x1, x2, x3 (two computed for levels of factor F and one which is same for all levels) and I want to compute y =

Dear all, I have a question about how to set arguments in my own function. For example, I have a function that looks like this: my.f <- function(a = x1, b = x2) { x1 = equation 1 x2 = equation 2 x3 = equation 3 y = a + b } x1, x2, and x3 are temporary variables (intermediate results) calculated from other variables within the funciton. I want to use two of these three

This might be a trivial question, but I would appreciate if anybody could suggest an elegant way of plotting a function such as the following (a simple distribution function): F(x) = 0 if x<=0 =(x^2)/2 if 0<x<=1 =2x-((x^2)/2)-1 if 1<x<=2 =1 if x>2 This is just an example. In this case it is a continuous function. But how to do it in general in an elegant way.

I am trying to write a function that will run a linear model and plot the regression coeficients with their corresponding means. I am having two problems. I can get the plot with the function below, but I am having trouble labeling the points. function(y,x1,x2,x3,x4){ outlm<-lm(y~x1+x2+x3+x4) imp<-as.data.frame(outlm$coef[-1]) meanvec<-c(mean(x1),mean(x2),mean(x3),mean(x4))

Dear list, Below I've written a clunky for loop that counts NA's in a row, replacing all with NA if there are more than 3 missing values, or keeping the values if 4 or more are present. This is sample code from a very large dataframe I'm trying to clean up. I know there are many simpler more elegant solutions to this little problem. Would someone be willing to show me how to

The following dataframe will illustrate the problem DF<-data.frame(name=rep(1:5,each=2),x1=rep("A",10),x2=seq(10,19,by=1),x3=rep(NA,10),x4=seq(20,29,by=1)) DF$x3[5]<-50 # we have a data frame. we are interested in the columns x2,x3,x4 which contain sparse # values and many NA. DF name x1 x2 x3 x4 1 1 A 10 NA 20 2 1 A 11 NA 21 3 2 A 12 NA 22 4 2 A 13 NA

I seem to be doing something really stupid or missing something really obvious but what? I have a simple three column data.frame that I would like to reshape to wide preferably using reshape2. An example from http://stackoverflow.com/questions/9617348/reshape-three-column-data-frame-to-matrix looked perfect except I wanted a data frame but it seemed okay. I just changed acast to dcast and it

Hi all, I?m fitting GLM?s and I can?t interprete the coefficients when I run a model with interaction terms. When I run the simpliest model there is no problem: Model1<-glm (Fishes ~ Year + I(Year^2) + Kind.Geographic + Kind.Fishers + Zone.2 + Hours + Fishers + Month, family = poisson(log)) # Fishes, Year, Hours, and Fishers are numeric, Kind.Geographic, Kind.Fishers, Zone.2 and

HI All, I am unable to solve a optimization Problem Please Help Me out of this to solve. The Optimization problem is as follows :- My objective function is linear and one of the constraint is quadratic. Min z = 5 * X1 + 9* X2 + 7.15 *X3 + 2 * X4 subject to X1 + X2 + X3 +X4 = 9 X1 + X4 < = 6.55 X3(X3 - 3.5) >=0 X1,X2,X3,X4 >=0 Now the problem is how to solve this kind of

Hello, I have a csv file with many variables, both characters and integers. I would like to load it on R and do some operations on integer variables, the problem is that R loads the entire dataset considering all variables as characters, instead I would like that R makes the distinction between the two types, because there are too many variables to do: x1<-as.integer(x1)

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1]

R 2.2 on windows XP I have a dataset with multiple columns. Some of the columns represent independent variables, some represent dependent variables. I would like to run the same analyses on a fixed set of independent variables, changing only the dependent variable, e.g. y1-y2=x1+x2+x3 y3-y4=x1+x2+x3 y5-y6=x1+x2+x3, etc. I know I can write a function to perform the analyses, however in order to

Hi R-helpers! I have the following dataframe: firm<-c(rep(1:3,4)) year<-c(rep(2001:2003,4)) X1<-rep(c(10,NA),6) X2<-rep(c(5,NA,2),4) data<-data.frame(firm, year,X1,X2) data So I want to obtain the same dataframe with a variable X3 that is: X1, if X2=NA X2, if X1=NA X1+X2 if X1 and X2 are not NA So my final data is X3<-c(15,NA,12,5,10,2,15,NA,12,5,10,2)

Hi All, I'm trying to do the following constrained optimization example. Maximize x1*(1-x1) + x2*(1-x2) + x3*(1-x3) s.t. x1 + x2 + x3 = 1 x1 >= 0 and x1 <= 1 x2 >= 0 and x2 <= 1 x3 >= 0 and x3 <= 1 which are the constraints. I'm expecting the answer x1=x2=x3 = 1/3. I tried the "constrOptim" function in R and I'm running into some issues. I first start off

Hi, I recently ran into an inconsistency in the way model.matrix.default handles factor encoding for higher level interactions with categorical variables when the full hierarchy of effects is not present. Depending on which lower level interactions are specified, the factor encoding changes for a higher level interaction. Consider the following minimal reproducible example: -------------- >

Hi R-helpers, I have a character object called dd that has 32 elements each of which is a model formula contained within quotation marks. Here's what it looks like: > dd [1] "lm(y ~ 1,data=Cement)" "lm(y ~ X,data=Cement)" "lm(y ~ X1,data=Cement)" [4] "lm(y ~ X2,data=Cement)" "lm(y ~

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1"

I ran into strange behavior when removing names. Two ways of removing names: i <- rep(1:4, length.out=20000) k <- c(a=1, b=2, c=3, d=4) x1 <- unname(k[i]) x2 <- k[i] x2 <- unname(x2) Are they identical? identical(x1,x2) # TRUE but no identical(serialize(x1,NULL),serialize(x2,NULL)) # FALSE But problem is with serialization type 3, cause:

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Dear All, I have a data frame with n columns: X1, X2, ., Xn. Now I want to create a new column: if X1 = X2 = . = Xn, the value is 1; Otherwise, the value is 0. How to do that in a quick way instead of doing (n choose 2) comparisons? Thank you, Frank [[alternative HTML version deleted]]