Re: [R] Sum Question

2011-06-30 Thread Dennis Murphy
Hi: Here's a data.table solution. After I read in your data as a data frame named dd, I used str() to check its contents: > str(dd) 'data.frame': 19 obs. of 5 variables: $ RID : int 43 95 230 230 235 235 247 247 321 321 ... $ SCRNO : Factor w/ 6 levels "HBA0020036","HBA0020087",..: 1 2

Re: [R] Sum Question

2011-06-30 Thread Marc Schwartz
On Jun 30, 2011, at 12:30 PM, Marc Schwartz wrote: > On Jun 30, 2011, at 11:20 AM, Edgar Alminar wrote: > I did this: library(data.table) dd <- data.table(bl) dd[,sum(as.integer(CONTTIME)), by = SCRNO] (I used as.integer because I got an error message:

Re: [R] Sum Question

2011-06-30 Thread Marc Schwartz
On Jun 30, 2011, at 11:20 AM, Edgar Alminar wrote: >>> I did this: >>> >>> library(data.table) >>> >>> dd <- data.table(bl) >>> dd[,sum(as.integer(CONTTIME)), by = SCRNO] >>> >>> (I used as.integer because I got an error message: sum not meaningful for >>> factors) >>> >>> And got this: >>>

[R] Sum Question

2011-06-30 Thread Edgar Alminar
>> I did this: >> >> library(data.table) >> >> dd <- data.table(bl) >> dd[,sum(as.integer(CONTTIME)), by = SCRNO] >> >> (I used as.integer because I got an error message: sum not meaningful for >> factors) >> >> And got this: >> >>SCRNO V1 >> [1,] HBA0020036 111 >> [2,] HBA0020

Re: [R] Sum Question

2011-06-29 Thread Rolf Turner
The simplest way is: xxx <- with(clyde,tapply(CONTTIME,SCRNO,sum)) You could also do: xxx <- by(clyde,clyde[["SCRNO"]],function(x){sum(x[["CONTTIME"]])}) but this gives somewhat messy output; the aforesaid output may be convenient for some purposes, not for others. cheers, Rolf

[R] Sum Question

2011-06-29 Thread Edgar Alminar
Hello, I have the following dataset (this is a piece of a much larger set): RID SCRNO VISCODE RECNO CONTTIME 2318 HBA0190012 bl 15 2418 HBA0190012 bl 35 2818 HBA0190012 bl 55 2918 HBA0190012 bl 25 321