Dunno. But if I understand correctly, here's a base R way to do it:
(## using your m) > m$total <- with(m,ave(score,id,FUN = cumsum)) > m id score total 1 1 6 6 2 1 6 12 3 2 6 6 4 2 3 9 5 3 3 3 6 3 3 6 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Jul 9, 2018 at 2:35 PM, Sumitrajit Dhar <s-d...@northwestern.edu> wrote: > Hi Folks, > > I am trying to get a group_by cumsum using: > > R version 3.5.0 (2018-04-23) -- "Joy in Playing" > Copyright (C) 2018 The R Foundation for Statistical Computing > Platform: x86_64-apple-darwin15.6.0 (64-bit) > > Here is an example of a simple construct that is not working. > > m <- data.frame( id = rep(1:3, each=2), score = rep(c(6,3), each=3) ) > > m %>% group_by(id) %>% mutate(total = cumsum(score)) > > My output: > > # A tibble: 6 x 3 > # Groups: id [3] > id score total > <int> <dbl> <dbl> > 1 1 6 6 > 2 1 6 12 > 3 2 6 18 > 4 2 3 21 > 5 3 3 24 > 6 3 3 27 > > What am I missing? Thanks in advance. > > Regards, > Sumit > > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/ > posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
