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
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
>
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