Hi,
May be this helps:
dat1<- read.table(text="
ID county date company 
1?????? x????? 1?????? comp1
2?????? y????? 1?????? comp3
3?????? y????? 2?????? comp1
4?????? y????? 3?????? comp1
5??????? x????? 2????? comp2
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- dat1
dat1$answer<-unsplit(lapply(split(dat1,dat1$county),function(x)
do.call(rbind,lapply(seq_len(nrow(x)),function(i) {x1<-x[1:i,];
x2<-table(x1$company)/sum(table(x1$company));sum(x2^2)}))),dat1$county)
?dat1
#? ID county date company??? answer
#1? 1????? x??? 1?? comp1 1.0000000
#2? 2????? y??? 1?? comp3 1.0000000
#3? 3????? y??? 2?? comp1 0.5000000
#4? 4????? y??? 3?? comp1 0.5555556
#5? 5????? x??? 2?? comp2 0.5000000
#or
dat2$answer<-with(dat2,unlist(ave(company,county,FUN=function(x)
lapply(seq_along(x),function(i) {x1<-table(x[1:i]);sum((x1/sum(x1))^2)}))))
?dat2
#? ID county date company??? answer
#1? 1????? x??? 1?? comp1 1.0000000
#2? 2????? y??? 1?? comp3 1.0000000
#3? 3????? y??? 2?? comp1 0.5000000
#4? 4????? y??? 3?? comp1 0.5555556
#5? 5????? x??? 2?? comp2 0.5000000
A.K.
Hi - 
I have a seemingly complex data summarizing problem that I am having a hard time
wrapping my mind around.
What I'm trying to do is sum the square of all company market 
shares ?in a given county, UP TO that corresponding time. Sum of market 
share is defined as: Number of company observations/ Total observations. 
Here is example data and desired answer: 
ID	county	date	company answer
1	 ? ? ? x	 ? ? ?1	 ? ? ? comp1	 ? ? ? ? ? 1
2	 ? ? ? y	 ? ? ?1	 ? ? ? comp3	 ? ? ? ? ? 1
3	 ? ? ? y	 ? ? ?2	 ? ? ? comp1	 ? ? ? ? ? 0.5
4	 ? ? ? y	 ? ? ?3	 ? ? ? comp1	 ? ? ? ? ? 0.55556
5	 ? ? ? ?x	 ? ? ?2	 ? ? ?comp2	 ? ? ? ? ?0.5
For example, to get the answer for ID 4, we look at county y, dates 1, 2, 3 and
sum: ?[(2/3)comp1]^2 +[(1/3)comp3]^2 = 0.55556
I've tried cumsum, but am simply stuck given all of the 
different conditions. ?I have a large matrix of data for this with 
several hundred companies, tens of counties and unique dates. 
Any help would be extremely appreciated. 
Thank you,