dplyr mean group on Long Format Data - r

Question

I have trouble figuring out how I can compute a simple mean with dplyr on Long Format data.
My data look like this :
hldid idno sex diary age
1 1294 1294_1 2 1 39
2 1294 1294_1 2 2 39
3 1294 1294_2 1 1 43
4 1294 1294_2 1 2 43
...
With 4 variables : hldid idno sex diary age
idno is the personal identifier but not the unique key.
Each individual is repeated 2 times, one for each diary filled.
What I would like is to simply compute the age mean by sex.
Could you help me out ?
I tried something like :
dta %>%
group_by(sex) %>%
mutate( ng = n_distinct(idno)) %>%
group_by(age, add=TRUE) %>%
summarise(mean = n()/ng[1] )
But it does not work.
The data :
dta = structure(list(hldid = c(1294, 1294, 1294, 1294, 1352, 1352,
1352, 1352, 3741, 3741, 3741, 3741, 3809, 3809, 3809, 3809, 4037,
4037, 4037, 4037), idno = c("1294_1", "1294_1", "1294_2", "1294_2",
"1352_1", "1352_1", "1352_2", "1352_2", "3741_1", "3741_1", "3741_2",
"3741_2", "3809_1", "3809_1", "3809_2", "3809_2", "4037_1", "4037_1",
"4037_2", "4037_2"), sex = c(2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L,
2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L), diary = c(1L,
2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L,
2L, 1L, 2L), age = c(39L, 39L, 43L, 43L, 31L, 31L, 37L, 37L,
33L, 33L, 37L, 37L, 34L, 34L, 37L, 37L, 41L, 41L, 32L, 32L)), .Names = c("hldid",
"idno", "sex", "diary", "age"), class = c("grouped_df", "tbl_df",
"tbl", "data.frame"), row.names = c(NA, -20L), vars = list(hldid), drop = TRUE, indices = list(
0:3, 4:7, 8:11, 12:15, 16:19), group_sizes = c(4L, 4L, 4L,
4L, 4L), biggest_group_size = 4L, labels = structure(list(hldid = c(1294,
1352, 3741, 3809, 4037)), class = "data.frame", row.names = c(NA,
-5L), .Names = "hldid", vars = list(hldid)))
quick update
Maybe this does not apply for this example,
but this kind of issues I have in mind is the following :
Imagine we have data like this :
3 women and 2 men, and a dummy act variable.
If we do and not taking into account the Long format computing the mean, we will have problems.
aggregate(act ~ sex, FUN = mean, data = dtaTime)
What we should do is this :
aggregate(act ~ sex, FUN = sum, data = dtaTime)
6 / 2 # men
10 / 3 # women
Data
dtaTime = structure(list(id = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L),
sex = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L,
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), act = c(1L,
1L, 0L, 1L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L,
1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L)), .Names = c("id", "sex",
"act"), class = "data.frame", row.names = c(NA, -25L))

Answer 1

You are making it too complicated,
dta %>%
group_by(sex) %>%
summarise(meanage = mean(age))
should give you the mean age by sex.
A base R alternative:
aggregate(age ~ sex, dta, mean)
A data.table alternative:
library(data.table)
setDT(dta)[, .(meanage = mean(age)), by = sex]