Rui's solution is good. Bert's suggestion is also good!
For Berts suggestion you'd make the list bit list(mean = mean_narm) But prior to that define a function: mean_narm<- function(x) { m <- mean(x, na.rm = T) if (!is.Nan (m)) { m <- NA } return (m) } Would do what you suggested in your reply to Bert. On Mon, 16 Sep 2024, 19:48 Rui Barradas, <ruipbarra...@sapo.pt> wrote: > Às 15:23 de 16/09/2024, Francesca escreveu: > > Sorry for posting a non understandable code. In my screen the dataset > > looked correctly. > > > > > > I recreated my dataset, folllowing your example: > > > > test<-data.frame(matrix(c( 8, 8, 5 , 5 ,NA ,NA , 1, 15, 20, 5, NA, 17, > > 2 , 5 , 5, 2 , 5 ,NA, 5 ,10, 10, 5 ,12, NA), > > c( 18, 5, 5, 5, NA, 9, 2, 2, 10, 7 , 5, > 19, > > NA, 10, NA, 4, NA, 8, NA, 5, 10, 3, 17, NA), > > c( 4, 3, 3, 2, 2, 4, 3, 3, 2, 4, 4 ,3, 4, 4, 4, > 2, > > 2, 3, 2, 3, 3, 2, 2 ,4), > > c(3, 8, 1, 2, 4, 2, 7, 6, 3, 5, 1, 3, 8, 4, 7, > 5, > > 8, 5, 1, 2, 4, 7, 6, 6))) > > colnames(test) <-c("cp1","cp2","role","groupid") > > > > What I have done so far is the following, that works: > > test %>% > > group_by(groupid) %>% > > mutate(across(starts_with("cp"), list(mean = mean))) > > > > But the problem is with NA: everytime the mean encounters a NA, it > creates > > NA for all group members. > > I need the software to calculate the mean ignoring NA. So when the group > is > > made of three people, mean of the three. > > If the group is two values and an NA, calculate the mean of two. > > > > My code works , creates a mean at each position for three subjects, > > replacing instead of the value of the single, the group mean. > > But when NA appears, all the group gets NA. > > > > Perhaps there is a different way to obtain the same result. > > > > > > > > On Mon, 16 Sept 2024 at 11:35, Rui Barradas <ruipbarra...@sapo.pt> > wrote: > > > >> Às 08:28 de 16/09/2024, Francesca escreveu: > >>> Dear Contributors, > >>> I hope someone has found a similar issue. > >>> > >>> I have this data set, > >>> > >>> > >>> > >>> cp1 > >>> cp2 > >>> role > >>> groupid > >>> 1 > >>> 10 > >>> 13 > >>> 4 > >>> 5 > >>> 2 > >>> 5 > >>> 10 > >>> 3 > >>> 1 > >>> 3 > >>> 7 > >>> 7 > >>> 4 > >>> 6 > >>> 4 > >>> 10 > >>> 4 > >>> 2 > >>> 7 > >>> 5 > >>> 5 > >>> 8 > >>> 3 > >>> 2 > >>> 6 > >>> 8 > >>> 7 > >>> 4 > >>> 4 > >>> 7 > >>> 8 > >>> 8 > >>> 4 > >>> 7 > >>> 8 > >>> 10 > >>> 15 > >>> 3 > >>> 3 > >>> 9 > >>> 15 > >>> 10 > >>> 2 > >>> 2 > >>> 10 > >>> 5 > >>> 5 > >>> 2 > >>> 4 > >>> 11 > >>> 20 > >>> 20 > >>> 2 > >>> 5 > >>> 12 > >>> 9 > >>> 11 > >>> 3 > >>> 6 > >>> 13 > >>> 10 > >>> 13 > >>> 4 > >>> 3 > >>> 14 > >>> 12 > >>> 6 > >>> 4 > >>> 2 > >>> 15 > >>> 7 > >>> 4 > >>> 4 > >>> 1 > >>> 16 > >>> 10 > >>> 0 > >>> 3 > >>> 7 > >>> 17 > >>> 20 > >>> 15 > >>> 3 > >>> 8 > >>> 18 > >>> 10 > >>> 7 > >>> 3 > >>> 4 > >>> 19 > >>> 8 > >>> 13 > >>> 3 > >>> 5 > >>> 20 > >>> 10 > >>> 9 > >>> 2 > >>> 6 > >>> > >>> > >>> > >>> I need to to average of groups, using the values of column groupid, and > >>> create a twin dataset in which the mean of the group is replaced > instead > >> of > >>> individual values. > >>> So for example, groupid 3, I calculate the mean (12+18)/2 and then I > >>> replace in the new dataframe, but in the same positions, instead of 12 > >> and > >>> 18, the values of the corresponding mean. > >>> I found this solution, where db10_means is the output dataset, db10 is > my > >>> initial data. > >>> > >>> db10_means<-db10 %>% > >>> group_by(groupid) %>% > >>> mutate(across(starts_with("cp"), list(mean = mean))) > >>> > >>> It works perfectly, except that for NA values, where it replaces to all > >>> group members the NA, while in some cases, the group is made of some NA > >> and > >>> some values. > >>> So, when I have a group of two values and one NA, I would like that for > >>> those with a value, the mean is replaced, for those with NA, the NA is > >>> replaced. > >>> Here the mean function has not the na.rm=T option associated, but it > >>> appears that this solution cannot be implemented in this case. I am not > >>> even sure that this would be enough to solve my problem. > >>> Thanks for any help provided. > >>> > >> Hello, > >> > >> Your data is a mess, please don't post html, this is plain text only > >> list. Anyway, I managed to create a data frame by copying the data to a > >> file named "rhelp.txt" and then running > >> > >> > >> > >> db10 <- scan(file = "rhelp.txt", what = character()) > >> header <- db10[1:4] > >> db10 <- db10[-(1:4)] |> as.numeric() > >> db10 <- matrix(db10, ncol = 4L, byrow = TRUE) |> > >> as.data.frame() |> > >> setNames(header) > >> > >> str(db10) > >> #> 'data.frame': 25 obs. of 4 variables: > >> #> $ cp1 : num 1 5 3 7 10 5 2 4 8 10 ... > >> #> $ cp2 : num 10 2 1 4 4 5 6 4 4 15 ... > >> #> $ role : num 13 5 3 6 2 8 8 7 7 3 ... > >> #> $ groupid: num 4 10 7 4 7 3 7 8 8 3 ... > >> > >> > >> And here is the data in dput format. > >> > >> > >> > >> db10 <- > >> structure(list( > >> cp1 = c(1, 5, 3, 7, 10, 5, 2, 4, 8, 10, 9, 2, > >> 2, 20, 9, 13, 3, 4, 4, 10, 17, 8, 3, 13, 10), > >> cp2 = c(10, 2, 1, 4, 4, 5, 6, 4, 4, 15, 15, 10, > >> 4, 2, 11, 10, 14, 2, 4, 0, 20, 18, 4, 3, 9), > >> role = c(13, 5, 3, 6, 2, 8, 8, 7, 7, 3, 10, 5, > >> 11, 5, 3, 13, 12, 15, 1, 3, 15, 10, 19, 5, 2), > >> groupid = c(4, 10, 7, 4, 7, 3, 7, 8, 8, 3, 2, 5, > >> 20, 12, 6, 4, 6, 7, 16, 7, 3, 7, 8, 20, 6)), > >> class = "data.frame", row.names = c(NA, -25L)) > >> > >> > >> > >> As for the problem, I am not sure if you want summarise instead of > >> mutate but here is a summarise solution. > >> > >> > >> > >> library(dplyr) > >> > >> db10 %>% > >> group_by(groupid) %>% > >> summarise(across(starts_with("cp"), ~ mean(.x, na.rm = TRUE))) > >> > >> # same result, summarise's new argument .by avoids the need to group_by > >> db10 %>% > >> summarise(across(starts_with("cp"), ~ mean(.x, na.rm = TRUE)), .by = > >> groupid) > >> > >> > >> > >> Can you post the expected output too? > >> > >> Hope this helps, > >> > >> Rui Barradas > >> > >> > >> -- > >> Este e-mail foi analisado pelo software antivírus AVG para verificar a > >> presença de vírus. > >> www.avg.com > >> > > > > > Hello, > > Something like this? > > > test <- > structure(list( > cp1 = c(1, 5, 3, 7, 10, 5, 2, 4, 8, 10, 9, 2, > 2, 20, 9, 13, 3, 4, 4, 10, 17, 8, 3, 13, 10), > cp2 = c(10, 2, 1, 4, 4, 5, 6, 4, 4, 15, 15, 10, > 4, 2, 11, 10, 14, 2, 4, 0, 20, 18, 4, 3, 9), > role = c(13, 5, 3, 6, 2, 8, 8, 7, 7, 3, 10, 5, > 11, 5, 3, 13, 12, 15, 1, 3, 15, 10, 19, 5, 2), > groupid = c(4, 10, 7, 4, 7, 3, 7, 8, 8, 3, 2, 5, > 20, 12, 6, 4, 6, 7, 16, 7, 3, 7, 8, 20, 6)), > class = "data.frame", row.names = c(NA, -25L)) > > library(dplyr) > > test %>% > group_by(groupid) %>% > mutate(across(starts_with("cp"), list(mean = ~ mean(.x, na.rm = TRUE)))) > #> # A tibble: 25 × 6 > #> # Groups: groupid [11] > #> cp1 cp2 role groupid cp1_mean cp2_mean > #> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> > #> 1 1 10 13 4 7 8 > #> 2 5 2 5 10 5 2 > #> 3 3 1 3 7 6.17 5.17 > #> 4 7 4 6 4 7 8 > #> 5 10 4 2 7 6.17 5.17 > #> 6 5 5 8 3 10.7 13.3 > #> 7 2 6 8 7 6.17 5.17 > #> 8 4 4 7 8 5 4 > #> 9 8 4 7 8 5 4 > #> 10 10 15 3 3 10.7 13.3 > #> # ℹ 15 more rows > > > Hope this helps, > > Rui Barradas > > > -- > Este e-mail foi analisado pelo software antivírus AVG para verificar a > presença de vírus. > www.avg.com > > ______________________________________________ > 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 > https://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 https://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.