[R] (no subject)

Bert Gunter bgunter@4567 @end|ng |rom gm@||@com
Mon Sep 16 20:08:03 CEST 2024 

Incidentally, if you intend to use these means for further analytical
purposes, you may produce irreproducible conclusions: after all, a
mean of 10 things is more *meaningful* than a mean of 2. However, that
is a discussion too far for this list. Consult your local statistical
resources if you need to go there and need help.

Cheers,
Bert

On Mon, Sep 16, 2024 at 11:02 AM Bert Gunter <bgunter.4567 using gmail.com> wrote:
>
> It's NA *not* Na. Details matter.
>
> Ah, but note:
> > mean(c(NA,NA), na.rm = TRUE)
> [1] NaN
>
> So if that might happen, you'll have to write your own mean function,
> say mymean(), to do what you want. I leave that (simple) pleasure to
> you.
>
> -- Bert
>
> On Mon, Sep 16, 2024 at 8:05 AM Francesca <francesca.pancotto using gmail.com> wrote:
> >
> > All' Na Is Na.
> >
> >
> > Il lun 16 set 2024, 16:29 Bert Gunter <bgunter.4567 using gmail.com> ha scritto:
> >>
> >> See the na.rm argument of ?mean
> >>
> >> But what happens if all values are NA?
> >>
> >> -- Bert
> >>
> >>
> >> On Mon, Sep 16, 2024 at 7:24 AM Francesca <francesca.pancotto using gmail.com> wrote:
> >> >
> >> > 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 <ruipbarradas using 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
> >> > >
> >> >
> >> >
> >> > --
> >> >
> >> > Francesca
> >> >
> >> >
> >> > ----------------------------------
> >> >
> >> >         [[alternative HTML version deleted]]
> >> >
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