calc mean/std/ci from column - r

Question

is there a package to easily calculate for each specific n number, the mean/std/ci.
In example starting with the data:
> n = c(0,0,0,0,0,0,0,2,2,2,2,5,5,5,5,8,8,8,8)
> s = c(43,23,65,43,12,54,43,12,2,43,62,25,55,75,95,28,48,68,18)
> df = data.frame(n, s)
> df
n s
1 0 43
2 0 23
3 0 65
4 0 43
5 0 12
6 0 54
7 0 43
8 2 12
9 2 2
10 2 43
11 2 62
12 5 25
13 5 55
14 5 75
15 5 95
16 8 28
17 8 48
18 8 68
19 8 18
resulting as:
data
n mean std ci
0 40 .. ..
2 30 .. ..
5 63 .. ..
8 41 .. ..

Answer 1

dplyr is nice, but not necessary. In base R:
## df() is built-in in R, avoid ...
dd <- data.frame(n=rep(c(0,2,5,8),c(7,4,4,4)),
s = c(43,23,65,43,12,54,43,12,2,43,
62,25,55,75,95,28,48,68,18))
sumfun <- function(x) {
m <- mean(x)
s <- sd(x)
se <- s/sqrt(length(x))
c(mean=m,sd=s,lwr=m-1.96*se,upr=m+1.96*se)
}
(or see smean.cl.normal(), smean.cl.boot(), etc. from the Hmisc package ...)
res <- do.call(rbind,tapply(dd$s,dd$n,sumfun))
res <- cbind(n=unique(dd$n),as.data.frame(res))
Or as pointed out by #thelatemail:
res <- do.call(data.frame,aggregate(s ~ n, data=df, FUN=sumfun ))
You can easily package this into a function if you're going to use it on a regular basis.
For larger data sets/more complex transformations you can search SO for answers comparing solutions from the dplyr, plyr, data.table, doBy packages as well as base-R solutions using combinations of tapply(), ave(), aggregate(), by() ...

Answer 2

You can use the dplyr package.
Here's a code snippet. Note, I'm assuming you want to build the confidence interval using the standard normal approximation at the 95% level but you can make whatever choice you like.
n = c(0,0,0,0,0,0,0,2,2,2,2,5,5,5,5,8,8,8,8)
s = c(43,23,65,43,12,54,43,12,2,43,62,25,55,75,95,28,48,68,18)
df = data.frame(n, s)
df %>%
group_by(n) %>%
summarise(mean = mean(s),
std = sqrt(var(s)),
lower = mean(s) - qnorm(.975)*std/sqrt(n()),
upper = mean(s) + qnorm(.975)*std/sqrt(n()))
Source: local data frame [4 x 5]
n mean std lower upper
1 0 40.42857 17.88721 27.177782 53.67936
2 2 29.75000 27.69326 2.611104 56.88890
3 5 62.50000 29.86079 33.236965 91.76303
4 8 40.50000 22.17356 18.770313 62.22969

Answer 3

Thanks for the advice everyone, I have taken a look into plyr and solved it:
n = c(0,0,0,0,0,0,0,2,2,2,2,5,5,5,5,8,8,8,8)
s = c(43,23,65,43,12,54,43,12,2,43,62,25,55,75,95,28,48,68,18)
dd = data.frame(n, s)
library(plyr)
data <- ddply(dd,.(n),function(dd) c(mean=mean(dd$s),
std = sd(dd$s),
se = sd(dd$s)/sqrt(length(dd$s)),
lower = mean(dd$s)-qnorm(.975)*sd(dd$s)/sqrt(length(dd$s)),
upper = mean(dd$s)+qnorm(.975)*sd(dd$s)/sqrt(length(dd$s))
))
resulting as:
data
n mean std se lower upper
1 0 40.42857 17.88721 6.760731 27.177782 53.67936
2 2 29.75000 27.69326 13.846630 2.611104 56.88890
3 5 62.50000 29.86079 14.930394 33.236965 91.76303
4 8 40.50000 22.17356 11.086779 18.770313 62.22969
Will avoid df() in the future, thanks

Answer 4

Update tidyr 1.0.0
Even though the solution by #user1357015 is totally fine, if you are a tidyverse fan like me, there is an elegant alternative:
The new tidyr 1.0.0 contained a function that didn't get much attention but is very useful: unnest_wider.
With that, you can simplify the code to the following:
df %>%
group_by(n) %>%
nest(data = -"n") %>%
mutate(ci = map(data, ~ MeanCI(.x$s))) %>%
unnest_wider(ci)
which gives
# A tibble: 4 x 5
# Groups: n [4]
n data mean lwr.ci upr.ci
<dbl> <list> <dbl> <dbl> <dbl>
1 0 <tibble [7 × 1]> 40.4 23.9 57.0
2 2 <tibble [4 × 1]> 29.8 -14.3 73.8
3 5 <tibble [4 × 1]> 62.5 15.0 110.
4 8 <tibble [4 × 1]> 40.5 5.22 75.8