similar to: stacked area chart

Hello, everybody! I have a code below that creates a data set and then a stacked bar chart based on that data set. No need to look at it - just notice please that my horizontal axis is a date varible (x=my.data$date). I have a question about the last 2 lines of this code: grid(nx=NULL,ny=NULL,col = "lightgray", lty = "dotted",lwd = par("lwd")) axis(1, las = 2) Could

Hello! My code below creates a data frame and a plot for it. However, I can't figure out how to add a legend that is not ON the plot itself, but outside of it (e.g., to the right of my graph or below it). I tried something: I put a line par(xpd=T, mar=par()$mar+c(0,0,0,4)) right before my plot command), but that screwed up all my gridlines - they covered all graph and do not coincide with

Hello! If you run the whole code below, it'll produce a stacked diagram. And it looks good - because the tick-marks are aligned with the grid. However, if I stretch the graph window, grid becomes misaligned with the tickmarks. Or, rather, it seems aligned for the first and the last tick mark, but not for tickmarks in between. Can it be addressed? Thank you! Dimitri ### Creating a data set

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

I am not sure if this question has been asked before - but is there a procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE regression commands in SPSS? Thanks a lot! -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello, everyone! I have a very simple task - I have a data frame (see MyData below) and I need to stack the data (see result below). I wrote the syntax below - it's very basic and it does what I need. But I am sure what I am trying to do is a very typical task and there must be a much shorter/more elegant way of doing it. Any advice? Thank you very much!

Hello! I am wondering if it's possible to run - in sequence - code that is stored in several R scripts. For example: Script in the file "code1.r" contains the code: a = 3; b = 5; c = a + b Script in the file "code2.r" contains the code: d = 10; e = d - c Script in the file "code3.r" contains the code: result=e/a I understand that I could write those 3 scripts

I know that I can use as.yearmon in the package "zoo" to find the year and the month of a date. I can use as. yearqtr to find the year and the quarter. But how can one find just the year of a date? Thanks a lot! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

I am sorry, I'd like to split my column ("names") such that all the beginning of a string ("X..") is gone and only the rest of the text is left. x<-data.frame(names=c("X..aba","X..abb","X..abc","X..abd")) x$names<-as.character(x$names) (x) str(x) Can't figure out how to apply strsplit in this situation - without using a

Dear R-ers, I have a large data frame (several thousands of rows and about 2.5 thousand columns). One variable ("group") is a grouping variable with over 30 levels. And I have a lot of NAs. For each variable, I need to divide each value by variable mean - by subgroup. I have the code but it's way too slow - takes me about 1.5 hours. Below is a data example and my code that is too

Dear R'rs, is there a function that checks if a given vector contains a certain value. E.g., x<-c(1,2,3,4). How can I get a TRUE or FALSE for whether x contains a 2? -- Dimitri Liakhovitski Ninah.com Dimitri.Liakhovitski at ninah.com

Hello! If I have in my data frame MyFrame a variable saved as a Date and want to translate it into years, I currently do it like this using "zoo": library(zoo) as.year <- function(x) as.numeric(floor(as.yearmon(x))) myFrame$year<-as.year(myFrame$date) Is there a function that would do it directly - like "as.yearmon" - but for years? Thank you! -- Dimitri

Hello. Until today I've been using R2.9 and since today R2.10 (on a PC). In both of them it takes about 20 sec for the prompt to appear IN R console after I start R. And every time it says: "Previous saved work space restored" - even if I have not saved any workspace or, in case of R2.10 - even though I have not used it once. In the older versions - R would start within 2-3 sec. Is

Dear R-ers, I have a data frame data with predictors x1 through x5 and the response variable y. I am running a simple regression: reg<-lm(y~x1, data=data) I would like to loop through all predictors. Something like: predictors<-c("x1","x2",... "x10) for(i in predictors){ reg<-lm(y~i) etc. } But it's not working. I am getting an error: Error in

Dear all, I have a function that predicts DV based on one predictor pred: pred<-c(0,3000000,7800000,15600000,23400000,131200000) DV<-c(0,500,1000,1400,1700,1900) ## I define Function 1 that computes the predicted value based on pred values and parameters a and b: calc_DV_pred <- function(a,b) { DV_pred <- rep(0,(length(pred))) for(i in 1:length(DV_pred)){ DV_pred[i] <- a *

Hello, dear R-ers! I have a data frame: x<-data.frame(a=c(4,2,4,1,3,4),b=c(1,3,4,1,5,0),c=c(NA,2,5,3,4,NA),d=rep(NA,6),e=rep(NA,6)) x When x$a==1, I would like to replace NAs in columns d and e with 8 and 9, respectively When x$a != 1, I would like to replace NAs in columns d and e 101 and 1022, respectively. However, I only want to do it for rows 2:5 - while ignoring what's happening in

Hello! I have 2 data frames like this (well, actually, I have 200 of them): df1<-data.frame(location=c("loc 1","loc 2","loc 3"),date=c("1/1/2010","1/1/2010","1/1/2010"), a=1:3,b=11:13,c=111:113) df2<-data.frame(location=c("loc 1","loc 2","loc

Dear everyone, I am trying to run rgenoud on several chips simultaneusly. I used the instructions provided on Jasjeet Sekhon's Homepage (http://sekhon.berkeley.edu/rgenoud/multiple_cpus.html). However, I have the newer version of R (R 2.12) installed - for a 64-bit machine. So, when I tried to install the special version of "snow" from a zip file provided by Jasjeet on his page, R