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.panco...@gmail.com> wrote:
>
> All' Na Is Na.
>
>
> Il lun 16 set 2024, 16:29 Bert Gunter <bgunter.4...@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.panco...@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 <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
>> > >
>> >
>> >
>> > --
>> >
>> > Francesca
>> >
>> >
>> > ----------------------------------
>> >
>> > [[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.
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