similar to: Aggregate Values for All Levels of a Factor

Hello Everyone, I have recently been introduced to the ggplot package by Hadley Wickham and must say I am quite impressed so far at how easy it is to make attractive plots, but one thing I am struggling over is how to consolidate legends. I have 3 plots that I would like to put on a single page and all 3 map the same dimension of the data to the colour aesthetic. Right now, when I plot all

I have downloaded an application at "http://www.philhealth.gov.ph/employers/download/pprs.htm" this is a custom windows application, it loads its own mouse pointer Hand.avi which cannot be opened in wine. Can anyone help me on what to do? I am using Wine version 1.1.16 because I am also running MS Office 2007 with wine. I am also using Ubuntu Jaunty.

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hello R-Help: I would really appreciate some help on this. I have spent a good deal of time searching for an answer for what I would think to be a common practice: Subclasses an S4 class from a data.frame. But have found very little. Here is some sample code that I've been trying: setClass("MyDataFrame",

hi, say i have a function f and i'd like to to call it like this: 1) f(list(a=...,b=...)) but i can't do it, because f is defined as: 2) f<-function(a=NULL,b=NULL){...} is there a way that i can approximate 1), such as mapping list(a=,...b=...) to "list(a=,...b=...)" and then replacing "list" by "f", and then evaluating the expression? thanks.

I'd like to avoid looping through an array in order to change values in the array as it takes too long. I red from an earlier post it can be done by "do.call" but never got it to work. The Idea is to change the value of "y" according to values in "x". Wherever "x" holds the value 3, the corresponding value in "y" should be set to 1. So I

Hi all, I have roughly fifty dataframes and a dataframe with the names of the fifty dataframes. I want to perform the same set of manipulations on all fifty dataframes, but can't find a way to batch process from a list with the dataframe names using a loop. Is there a way to read the file names from the dataframe with the names and then call the referenced dataframe? This would save me a

Dear R-helpers, I've got two functions; callTimes() calls times(), passing it an optional argument (bar) by name (bar=harry). times() then believes it has been passed a name, rather than a value ? but I want the value, not the name. Worse, if I evaluate the name, it is evaluated in the environment times() was defined, not where it is called. How can I call times(), defining its optional

I installed samba 3.0.6 today. I had to upgrade samba.schema in openldap as well to include sambaPasswordHistory and sambaLoginTime, otherwise certain things would not work (e.g. smbpasswd or users changing passwords) openldap failed to start because both attributes have the same OID, ...2.1.50. In order to get started anyway, I changed the sambaPasswordHistory OID to ...2.1.52 (which, in my

Hi, I am using aggregate() to add up groups of data according to year and month. It seems that the function aggregate() automatically sorts the levels of factors of the grouping elements, even if the order of the levels of factors is supplied. I am wondering if this is a bug, or if I missed something important. Below is an example that shows what I mean. Does anyone know if this is just the way

Hello! I have larger and a smaller data frame with 1 factor in each - it's the same factor: large.frame<-data.frame(myfactor=LETTERS[1:10]) small.frame<-data.frame(myfactor=LETTERS[c(9,7,5,3,1)]) levels(large.frame$myfactor) levels(small.frame$myfactor) table(large.frame$myfactor) table(small.frame$myfactor) myfactor has 10 levels in large.frame and 5 levels in small.frame. All 5

Hello, I have a vector x containing letters ("a", "b" etc.). Now I want to convert it to factor and group some letters into one common level. If I do it by factor function, giving the same label names for all values I want to group, it doesn't work: > x<-letters[1:5] > x [1] "a" "b" "c" "d" "e" >

Dear List, I need to print out each of 'k' levels of a factor 'n' times each, where 'n' is the number of elements belonging to each factor. I know that this can normally be done using the gl() command, but in my case, each level 'k' has an unequal number of elements. Example with code is as below: vc<-read.table("P:\\Transit\\CORRECT

Dear R helpers: Sorry to bother for a basic question about model.matrix. Basically, I want to apply the dummy coding rule in a dataframe with complete factor levels to another dataframe with incomplete factor levels. I used model.matrix, but could not get what I want. The following is an example. #Suppose I have two dataframe A and B

Hi, Database model of my web app is as follows, users(id,name) Taskcategories(id,user_id,name) Tasks(id,taskcategory_id,name) TasksData(id,tasks_id,durationInHrs,date) I want to collect and show following data from the above tables, in the view, 1. Category wise duration total for a user for a week. 2. Category wise duration total for a user for each of the 7 days of the week. 3. Total

Hi, I''m developing a dashboard like page and basically want to combine the result of multiple actions on a page. I''ve done this so far: def index r1 = render_to_string :action => "action1" r2 = render_to_string :action => "action2" render :text => r1+r2 end This works, but now I want to render from other controllers and this is where

Hi, I have 3 internet connexions : Router 1 : 192.168.212.2 Router 2 : 192.168.212.3 Router 3 : 196.200.8.73 I can''t change IP adress of router 3 because I haven''t admin password and my provider don''t want to give it. I want only 1 PC to aggregate the 3 connexions. How I can make that Router 3 work on my 192.168.212.x network? Thanks Jeremy

Hi, I have a data.frame which contains timeseries from several different locations, which I want to compare against each other for example calculating RMSE, or normalized mean bias of each location against the others. An example of this is the cor function where I can put in a data.frame and get the correlation of each location compared to the others in the data.frame eg.

an example available on the net goes like > df identifier quantity 1 1 10 2 1 20 3 2 30 4 1 15 5 2 10 6 3 20 > aggregate(df$quantity, by=list(df$identifier), sum) Group.1 x 1 1 45 2 2 40 3 3 20 I'd like Group.1 to retain the name "identifier" and would like to

try this: scores.melt = data.frame(grade = floor(runif(100, 1,10)), variable = 'score', value = rnorm(100)); cast(scores.melt, grade ~ variable, fun.aggregate = c(mean, length)) it has the nice column names of: grade score_mean score_length 1 1 0.08788535 8 2 2 0.16720313 15 3 3 0.41046299 7 4 4 0.13928356 13 ... but

I have a data frame and I would like to reshape it to wide format while at the same time applying different aggregate functions to each column AND at times multiple aggregate functions: test1 = data.frame( id = c(rep('101',8),rep('102',8)), phase = rep(c('D','D','L','L'),4), day =