If you want a row for each unique item:
> example <- data.frame(id=rep(
+ ( abs(round(rnorm(50,mean=500,sd=250),digits=0)))
+ ,3), group=rep(1:15,10))
> example <-example[with(example,order(id,group)),]
>
> uniqueIDs <- do.call(rbind
+ , lapply(split(example, example$group), function(.grp){
+ data.frame(group = .grp$group[1L]
+ , unique = unique(.grp$id)
+ )
+ })
+ )
>
> head(uniqueIDs, 25)
group unique
1.1 1 15
1.2 1 282
1.3 1 361
1.4 1 497
1.5 1 553
1.6 1 583
1.7 1 618
1.8 1 729
1.9 1 817
1.10 1 842
2.1 2 135
2.2 2 253
2.3 2 290
2.4 2 351
2.5 2 418
2.6 2 457
2.7 2 518
2.8 2 594
2.9 2 686
2.10 2 804
3.1 3 208
3.2 3 327
3.3 3 442
3.4 3 516
3.5 3 536
On Thu, Feb 16, 2012 at 11:15 AM, Matt Spitzer
<matthewjspit...@gmail.com> wrote:
> Dear R experts,
> I am having difficulty using loops productively and would like to please
> ask for advice. I have a dataframe of ids and groups. I would like to
> break down the dataframe into groups, find the unique sets of ids, then
> reassemble. My thought was to use a loop, but I have been unable to finish
> this loop in a logical way. I would like to find the unique ids for group
> 1, group 2, etc., and rbind these back together. However, I am unclear how
> to do this because in other attempts my final product is always a part of
> the last run loop. My way of working around this has been to write.csv and
> then re read at the end, which is so clumsy. Previously, I have used a
> store matrix for individual cells. 1. Is there a better way to approach
> this? 2. How can I combine parts of matrices to other parts created in
> prior loops?
> I have created a primitive example below. Each of the groups varies in
> number, so my repetitive example below is not accurate. In my real data,
> ids repeat often within groups.
> Thank you so much, Matt
>
> example <- data.frame(id=rep(
> ( abs(round(rnorm(50,mean=500,sd=250),digits=0)))
> ,3), group=rep(1:15,10))
> example <-example[with(example,order(id,group)),]
>
> for (i in 1:15) {
> ai <- example[example[,2]==i,][!duplicated ( example[example[,2]==i,][,1]
> ),]
> write.csv(ai, paste('a',i,'.csv',sep=""))
> }
> b1<-read.csv('a1.csv')
> b2<-read.csv('a2.csv')
> b3<-read.csv('a3.csv')
> b4<-read.csv('a4.csv')
> b5<-read.csv('a5.csv')
> b6<-read.csv('a6.csv')
> b7<-read.csv('a7.csv')
> b8<-read.csv('a8.csv')
> b9<-read.csv('a9.csv')
> b10<-read.csv('a10.csv')
> b11<-read.csv('a11.csv')
> b12<-read.csv('a12.csv')
> b13<-read.csv('a13.csv')
>
> unis2 <-
> rbind(rbind(rbind(rbind(rbind(rbind(rbind(rbind(rbind(rbind(rbind(rbind
> (b1,b2),b3),b4),b5),b6),b7),b8),b9),b10),b11),b12),b13)
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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