similar to: tabulating 2 factors weighting by a third var

Being italians when writing comments/instructions we use accented letters - like ?, ?, ?, etc.... when running R scripts using such characters I get and error saying: invalid multibyte character in parser I have been looking at the help and searched the r-help archives but I haven't find anything that I could intelligibly apply to my case. Can anyone suggest a fix for this error? Thanks,

Hi, I have been trying to get the label under the total column - i.e. a mean value of columns 2 to 6 - in a barplot I generate with this script: t1 <- tapply(A, B, sum) t1[8] <- mean(t1[2:6]) t1 <- as.table(t1) barplot(t1, ylim=c(0,3000)) mtext("Var1", side = 1, line = 3) mtext("Var2", side = 2, line = 3) I have been trying to use axis(1, at=1:8,

Hello, I have a multi-year dataset (see below) with date, a data value and a flag for the data value. I want to find the monthly median for each month in this dataset and then plot it. If anyone has suggestions they would be greatly apperciated. It should be noted that there are some dates with no values and they should be removed. Thanks Emily > print ( str(data$flow$daily) )

I am trying to pass a couple of variable names to a xtabs formula: > tab <- function(x,y){ xtabs(time~x+y, data=D) } But when I run: > tab(A,B) I get: Error in eval(expr, envir, enclos) : object "A" not found I am quite sure that there is some easy way out, but I have tried with different combinations of deparse(), substitute(), eval(), etc without success, can someone

Hi, I first should say I am new to R. I have searched without success the R-archives to see if I could find an answer to what I am about to ask you. My dataset is like: x fine 1 A 2010-12-09 07:57:33 2 B 2010-12-09 08:05:00 3 C 2010-12-08 20:42:00 ... that is: 'data.frame': 3 obs. of 2 variables: $ x : Factor w/ 3 levels "A","B","C":

Dear R-users, I have a factor variable within my data frame which I derive week after week from a POSIXct variable using the cut(var,"weeks") command I have found in the chron package. The levels() command gives me: [1] "2009-03-30 00:00:00" "2009-04-06 00:00:00" "2009-04-13 00:00:00" "2009-04-20 00:00:00" "2009-04-27 00:00:00"

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Hello R-experts, I would like to know if there is a solution to read files with extension .gsheet directly into R - see http://www.fileinfo.com/extension/gsheet for more info on this file format. Thank you, Luca Mr. Luca Meyer www.lucameyer.com R 2.15.1 Mac OS X 10.8.2 [[alternative HTML version deleted]]

Hello R experts, I have go this data frame: 'data.frame': 1 obs. of 20 variables: $ Anno : chr "PREVISIONI VS TARGET" $ OreTot: num 41 $ GioTot: logi NA $ OrGTot: logi NA $ OreCli: num 99 $ GioCli: logi NA $ OrGCli: logi NA $ OreFor: num -27 $ GioFor: logi NA $ OrGFor: logi NA $ OreOrt: num -18 $ GioOrt: logi NA $ OrGOrt: logi NA $ OreSpo: num -6 $ GioSpo: logi

*** APOLOGIZES FOR THOSE READING THE LIST THROUGH NABBLE THIS WAS ALREADY POSTED THERE BUT NOT FORWARDED TO THE LIST FOR SOME UNKNOWN REASON *** I have a dataset that looks like: $ V1: factor with 4 levels $ V2: factor with 4 levels $ V3: factor with 2 levels $ V4: num (summing up to 100 within V3 levels) $ V5: num (nr of cases for each unique combination of V1*V2*V3 levels) Quite new to

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Dear R-help, First of all, thank you VERY much for any help you have time to offer. I greatly appreciate it. I would like to write a function that, given an arbitrary number of factors from a data frame, tabulates the number of occurrences of each unique combination of the factors. Cleary, this works: > table(horse,date,surface) <SNIP> , , surface = TURF

I asked the question in Rmetrics subforum, but for some reason, almost two weeks later, it keeps saying I haven't been approved to post yet there. I'll try again here on the open forum. I wanted to have a script that tabulates results by trade. I.e. instead of tabulating each day as a trade event, you enter on one date, exit another, and simply tabulate metrics based on one trade period.

I have a data set with many missing observations. When I run a regression, R of course discards the observations (the whole row) that have "NA". I want to tabulate some baseline characteristics (column means) but only for the observations that R used for the regression. I tried to recreate this data frame by using na.omit on the original data frame, but this will not work as this will

Hello, I wonder if someone could send me suggestions on how to solve the following problem: I have a vector of an arbitrary size (ex. data<-c(10,10,11,10,12,11,10,12,11,11,10,11)) and use the table function, which gives the following result 10 11 12 5 5 2 that''s fine, but what I would like to do now is: construct new classes based on the number of classes from table, 10

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

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Dear R-helpers, I intend to upgrade R in Mac OS with updated R version and updated Mac OS version. I think my Mac notebook is produced with Mac x86_64, darwin9.8.0. I have updated my Mac OS to Mac OS X version 10.6.5. But, when I installed R 2.12.1, the "version" function still gave me information that R is based on old Mac OS. I need to know how can I update R to let it to fit for

I am having hard time tabulating data in a dataframe, and getting a single "table" for an answer. I am trying to tabulate all "counts" for given "status" on a given date. I have a data frame such as: delta_ts status count 1 2008-02-27 CLOSED 3 2 2008-02-27 NEW 56 3 2008-02-27 RESOLVED 5 4 2008-02-21 ASSIGNED 1 5

R-helpers: I have a data frame containing 4 factor variables (let's say A,B,C, and D) and 1 numerical variable (N). I would like to produce a cross-tabulated data frame in which A,B,C are individual columns, each factor of D is its own column, and the field is calculated as a given function of N (I would like to have two output data frames, one with the mean(N) and one with the