similar to: adding a legend to the plot (but outside of it)

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

Dear R-ers! Asking for your help with building the stacked area chart for the following simple data (several variables - with date on the X axis): ### Creating a data set my.data<-data.frame(date=c(20080301,20080402,20080503,20090301,20090402,20090503,20100301,20100402,20100503), x=c(1.1,1.0,1.6,1,2,1.5,2.1,1.3,1.9),y=c(-4,-3,-6,-5,-7,-5.2,-6,-4,-4.9),

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! The code below works - if you run it you'll see a stacked area chart generated based on the data example. I only have one understanding question about the legend location (the very last snippet of code): legend(par()$usr[2], mean(par()$usr[3:4]), rev(order.of.vars), xpd=T, bty="n", pch=15, col=all.colors[rev(order.of.colors)]) I see that par()$usr[2] = 14763.72

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),

Hi Suppose I have a matrix (cohort are rows and years are columns) [2000] [2001] [2002] [2003] [C1] 0.01 0.03 0.02 0.09 [C2] 0.06 0.05 0.07 0.11 [C3] 0.1 0.5 0.4 0.98 [C4] 0.7 0.6 0.2 0.77 I want to extracts the diagonals to get a matrix which looks like this (C1 becomes C2 in 2002, C2 becomes C3 in 2003

Hello! I've tried for a while - but can't figure it out. I have data frame x: y=c("a","b","c","d","e") z=c("m","n","o","p","r") a=c(0,0,1,0,0) b=c(2,0,0,0,0) c=c(0,0,0,4,0) x<-data.frame(y,z,a,b,c,stringsAsFactors=F) str(x) Some of the values in columns a,b, and c are >0: I need to

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

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

Dear R-ers, In my question there are no statistics involved - it's all about data manipulation in R. I am trying to write a code that should replace what's currently being done in SAS and SPSS. Or, at least, I am trying to show to my colleagues R is not much worse than SAS/SPSS for the task at hand. I've written a code that works but it's too slow. Probably because it's

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

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

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

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, I am using the following code to create ppp files from csv data and map shape files, but I am getting some errors which I have been unable to fix by searching them online: library(spatstat) library(maps) library(maptools) NYC2<-readShapePoly("nybb.shp") # this is a map of the NYC boroughs without waterways and no census tract divisions (but it does include lines separating

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

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-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