Making a table that contains Mean and SD of a Dataset - r

Question

I am using this dataset: http://www.openintro.org/stat/data/cdc.R
to create a table from a subset that only contains the means and standard deviations of male participants. The table should look like this:
Mean Standard Deviation
Age: 44.27 16.715
Height: 70.25 3.009219
Weight: 189.3 36.55036
Desired Weight: 178.6 26.25121
I created a subset for males and females with this code:
mdata <- subset(cdc, cdc$gender == ("m"))
fdata <- subset(cdc, cdc$gender == ("f"))
How should I create a table that only contains means and SDs of age, height, weight, and desired weight using these subsets?

Answer 1

The data frame you provided sucked up all the memory on my laptop, and it's not needed to provide that much data to solve your problem. Here's a dplyr/tidyr solution to create a summary table grouped by categories, using the starwars dataset available with dplyr:
library(dplyr)
library(tidyr)
starwars |>
group_by(sex) |>
summarise(across(
where(is.numeric),
.fns = list(Mean = mean, SD = sd), na.rm = TRUE,
.names = "{col}__{fn}"
)) |>
pivot_longer(-sex, names_to = c("var", ".value"), names_sep = "__")
# A tibble: 15 × 4
sex var Mean SD
<chr> <chr> <dbl> <dbl>
1 female height 169. 15.3
2 female mass 54.7 8.59
3 female birth_year 47.2 15.0
4 hermaphroditic height 175 NA
5 hermaphroditic mass 1358 NA
6 hermaphroditic birth_year 600 NA
7 male height 179. 36.0
8 male mass 81.0 28.2
9 male birth_year 85.5 157.
10 none height 131. 49.1
11 none mass 69.8 51.0
12 none birth_year 53.3 51.6
13 NA height 181. 2.89
14 NA mass 48 NA
15 NA birth_year 62 NA

Answer 2

Just make a data frame of colMeans and column sd. Note, that you may also select columns.
fdata <- subset(cdc, gender == "f", select=c("age", "height", "weight", "wtdesire"))
data.frame(mean=colMeans(fdata), sd=apply(fdata, 2, sd))
# mean sd
# age 45.79772 17.584420
# height 64.36775 2.787304
# weight 151.66619 34.297519
# wtdesire 133.51500 18.963014
You can also use by to do it simultaneously for both groups, it's basically a combination of split and lapply. (To avoid apply when calculating column SDs, you could also use sd=matrixStats::colSds(as.matrix(fdata)) which is considerably faster.)
res <- by(cdc[c("age", "height", "weight", "wtdesire")], cdc$gender, \(x) {
data.frame(mean=colMeans(x), sd=matrixStats::colSds(as.matrix(x)))
})
res
# cdc$gender: m
# mean sd
# age 44.27307 16.719940
# height 70.25165 3.009219
# weight 189.32271 36.550355
# wtdesire 178.61657 26.251215
# ------------------------------------------------------------------------------------------
# cdc$gender: f
# mean sd
# age 45.79772 17.584420
# height 64.36775 2.787304
# weight 151.66619 34.297519
# wtdesire 133.51500 18.963014
To extract only one of the data frames in the list-like object use e.g. res$m.
Usually we use aggregate for this, which you also might consider:
aggregate(cbind(age, height, weight, wtdesire) ~ gender, cdc, \(x) c(mean=mean(x), sd=sd(x))) |>
do.call(what=data.frame)
# gender age.mean age.sd height.mean height.sd weight.mean weight.sd wtdesire.mean wtdesire.sd
# 1 m 44.27307 16.71994 70.251646 3.009219 189.32271 36.55036 178.61657 26.25121
# 2 f 45.79772 17.58442 64.367750 2.787304 151.66619 34.29752 133.51500 18.96301
The pipe |> call(what=data.frame) is just needed to get rid of matrix columns, which is useful in case you aim to further process the data.
Note: R >= 4.1 used.
Data:
source('https://www.openintro.org/stat/data/cdc.R')
or
cdc <- structure(list(genhlth = structure(c(3L, 3L, 1L, 5L, 3L, 3L), levels = c("excellent",
"very good", "good", "fair", "poor"), class = "factor"), exerany = c(0,
1, 0, 0, 1, 1), hlthplan = c(1, 1, 1, 1, 1, 1), smoke100 = c(1,
0, 0, 0, 0, 1), height = c(69, 66, 73, 65, 67, 69), weight = c(224L,
215L, 200L, 216L, 165L, 170L), wtdesire = c(224L, 140L, 185L,
150L, 165L, 165L), age = c(73L, 23L, 35L, 57L, 81L, 83L), gender = structure(c(1L,
2L, 1L, 2L, 2L, 1L), levels = c("m", "f"), class = "factor")), row.names = c("19995",
"19996", "19997", "19998", "19999", "20000"), class = "data.frame")