
Dan,
 
Thank you very much. That is exactly what I need. What a clever solution.
 
David, thanks for trying.
 
Thank you all. I would never be able to figure it out on my own. 
 
Lin

 

> From: dwinsemius@comcast.net
> To: NordlDJ@dshs.wa.gov
> Date: Tue, 14 Feb 2012 14:28:21 -0500
> CC: r-help@r-project.org
> Subject: Re: [R] sequential sum
> 
> 
> On Feb 14, 2012, at 1:38 PM, Nordlund, Dan (DSHS/RDA) wrote:
> 
> >> -----Original Message-----
> >> From: r-help-bounces@r-project.org [mailto:r-help-bounces@r-
> >> project.org] On Behalf Of baccts
> >> Sent: Tuesday, February 14, 2012 9:04 AM
> >> To: r-help@r-project.org
> >> Subject: [R] sequential sum
> >>
> >> Dear R users,
> >>
> >> I am trying to sum number that exist in another vector up to i, then
> >> increment i and repeat.
> >> Sorry. It's hard to explain but basically I am trying to do the
> >> following:
> >> test <- c(1,2,3,4);
> >> test2 <- c(3,5,6,7,2,8,8,4,4);
> >> test3 <- c(10,20,30,40);
> >> tmp <- 0;
> >> for (i in 1:length(test)){
> >> tmp[i] <- sum(test3[which(test[1:i] %in% test2)]);
> >> }
> >>
> >> so when i = 1, tmp[i] = 0 because test[1]=1 is not in test2 and
> >> when i = 3, tmp[i] = 50 = test3[2] + test3[3] because test[2:3] is in
> >> test2
> >>
> >> Problem is test has 5000+ entries. How do I do the same thing without
> >> the
> >> loop?
> >>
> >> Thanks in advance.
> >>
> >
> > Your example data is not very extensive so I don't know if this 
> > solution is general enough. But, it does work with your data.
> 
> I had exactly that concern so tested my solution, which was a bit 
> different than yours:
> 
> test <- c(1,2,3,4,10, 5);
> test2 <- c(3,5,6,7,2,8,8,4,4);
> test3 <- c(10,20,30,40, 50, 60);
> tmp <- 0;
> for (i in 1:length(test)){
> tmp[i] <- sum(test3[which(test[1:i] %in% test2)]);
> }
> 
> > tmp
> [1] 0 20 50 90 90 150
> > tmp2 <- ifelse(test %in% test2, cumsum(test3[test %in% test2]), 0)
> > tmp2
> [1] 0 50 90 150 0 50 # was going to throw away.
> 
> 
> > tmp3 <- cumsum(ifelse(test %in% test2, test3, 0))
> > tmp3
> [1] 0 20 50 90 90 150
> 
> So yours is better at spanning the non-match entries in the same 
> manner as the for-loop.
> 
> 
> >
> >
> > Hope this is helpful,
> >
> > Dan
> >
> > Daniel J. Nordlund
> 
> 
> David Winsemius, MD
> West Hartford, CT
> 
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