similar to: Efficiency: speeding up unlist that is currently running by row

Hello! I have a matrix a with 2 variables (see below) that contain character strings. I need to create a 3rd variable that contains True if the value in column x is equal to the value in column y. The code below does it. a<-data.frame(x=c("john", "mary", "mary", "john"),y=c("mary","mary","john","john"))

Hello everyone! I have a list X with 3 elements, each of which is a data frame, for example: a<-data.frame(a=1,b=2,c=3) b<-data.frame(a=c(4,7),b=c(5,8),c=c(6,9)) c<-data.frame(a=c(10,13,16),b=c(11,14,17),c=c(12,15,18)) X<-list() X[[1]]<-a X[[2]]<-b X[[3]]<-c (X) How can I most effectively transform X into a data frame with columns a, b, and c? I would love to find a generic

Hello, I have a matrix that is a product of tapply on a larger data set. Let's assume it looks like this: X<-matrix(c(10,20,30,40,50,60),2,3) dimnames(X)<-list(c("1","2"),c("1","2","3")) (X) 1 2 3 1 10 30 50 2 20 40 60 Is there an efficient way of transforming this matrix into the following matrix: rows columns entries 1

Thank you for providing advice on this graphics question. I am building an interaction.plot. d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.2,4.4,3.5,3.3,-1.1,-1.3) d[[1]]<-as.factor(d[[1]]) d[[2]]<-as.factor(d[[2]]) print(d) interaction.plot(d$xx, d$yy, d$zz, type="b", col=c("red","blue"), legend=F, lty=c(1,2), lwd=2, pch=c(18,24),

Hello! I apologize - I never used lattice before, so my question is probably very basic - but I just can't find the answer in the archive nor in the documentation: I have a named numeric vector p of 6 numbers (of the type 6 numbers with people's names to whom those numbers belong). I want a simple bar chart. I am doing: library(lattice) trellis.par.set(fontsize=list(text=12)) #

Dear R-list, maybe some of you could point me in the right direction: Are you aware of any FREE Fortran or Java libraries/actual pieces of code that are VERY efficient (time-wise) in running the regular linear least-squares multiple regression? More specifically, I have to run small regression models (between 1 and 15 predictors) on samples of up to N=700 but thousands and thousands of them. I

Redhat's Security Enhanced Linux (in some modes) generates messages like that when code in a shared library is not compiled to have position independent code. I.e., the -fpic or -fPIC flag was not supplied when compiling. Your build log showed that -fpic was used when compiling your code, but the 3rd party library, libnetcdf.a may not have been compiled this way. The command

Hello! I have encountered a really weird problem. Maybe you've encountered it before? I have a large data frame "importances". It has one factor ($A) with 3 levels: 3, 9, and 15. $B is a regular numeric variable. Below I am picking a really small sub-frame (just 3 rows) based on "indices". "indices" were chosen so that all 3 levels of A are present:

Hello - and sorry for what might look like a simple graphics question. I am building an interaction plot for d: d=data.frame(xx=c(3,3,2,2,1,1),yy=c(4,3,4,3,4,3),zz=c(5.1,4.4,3.5,3.3,-1.1,-1.3)) d[[1]]<-as.factor(d[[1]]) d[[2]]<-as.factor(d[[2]]) print(d) interaction.plot(d$xx, d$yy, d$zz, type="b", col=c("red","blue"), legend=F, lty=c(1,2), lwd=2,

Hello, everyone! The code below allows me to produce the graph I want (I know - the colors are strange, but it's just for the sake of an example). After you run the plot<- part and then do print(plot) - that's what I want. However, when I run the bits of code below (with graphics devices) - what they print is different from the original plot. In .png, .emf, and .tiff - my dots change

Hello! I am using function "curve" to create a line graph. I was wondering, if it's possible to "turn off" the default tick marks and introduce those tick marks in specific locations. For example, currently in my X axis tick marks are (automatically) at 10, 11, 12, 13 but I want them to be in 5 specific locations, like 9.89, 10.34, etc. Any hint would be greatly

Hello! I have a question about my lattice barchart that I am trying to build in Section 3 below. I can't figure out a couple of things: 1. When I look at the dataframe "test" that I am trying to plot, it looks right to me (the group "Total" is always the first out of 5). However, in the chart it is the last. Why? 2. How can I make sure the value labels (on y) are not

Dear Random Forests gurus, I have a question about R^2 provided by randomForest (for regression). I don't succeed in finding this information. In the help file for randomForest under "Value" it says: rsq: (regression only) - "pseudo R-squared'': 1 - mse / Var(y). Could someone please explain in somewhat more detail how exactly R^2 is calculated? Is "mse"

Dear everyone, I don't know much about Integer Programming but am afraid I am facing a problem that can only be solved via Integer Programming. I was wondering if those of you who have experience with it could recommend an R package. I found the following R packages: Rglpk glpk lpSolve lpSolveAPI Are there any others? Are some of them easier to use than others for a beginner? Any advice

Hello! I've run a simple linear model: result<-lm(DV~A+B+C,data=Data) My Data$A,Data$B, and Data$C are factors. So, lm automatically recoded them into dummy variables. I have all the results I need but one. Question: Where could I see the variance explained by all A dummy variables together, then all B dummy variables together, and all C dummy variables together - when other predictors

Dear everybody! Need help with graphics. I am runnig a simple lm and then using allEffects from 'effects' package: require(effects) model<-lm(Y~A+B, data=mydataframe) I am trying to build (for each predictor - A and then B) a plot of means on Y. I was successful doing it like this - in one swoop: ml.eff<-allEffects(ml1, se=F) plot(ml.eff,ylab="Title of Y") Is it

Hello! I think I am relatively clear on how predictor importance (the first one) is calculated by Random Forests for a Classification tree: Importance of predictor P1 when the response variable is categorical: 1. For out-of-bag (oob) cases, randomly permute their values on predictor P1 and then put them down the tree 2. For a given tree, subtract the number of votes for the correct class in the

Hello! I am experiencing a problem with section 3 of the code below. I want to generate barplots (based on data generated in Sections 1 and 2) in a loop - for each variable in "data" - and save them as .emf files in my current directory. But it's not working - it's printing values to be plotted to the screen but does not print the plots themselves anywhere (and not in the

Dear R'ers, I have a very large data frame (over 4000 rows and 2,500 columns). My task is very simple - I have to replace all NAs with a zero. My code works fine on smaller data frames - but I have to deal with a huge one and there are many NAs in each column. R runs out of memory on me ("Reached total allocation of 1535Mb: see help(memory.size)"). Is there any other, more efficient

Hello! I promise I looked into help files before asking. Still cannot figure it out. I think it's because I am totally confused what packages use lettice, which use trellis, etc. Sections 1 and 2 below produce the data and the data to plot. My question is about barplot in Section 3. I am trying to: 1. add only horizontal gridlines and manipulate the type and color of that line. tck = 1 is not