[R] What don't I understand about sample()?

[Rui Barradas]

Hello,

I have been following this thread and though answers have been given, 
some of them address R coding in general, not necessarily the sample() 
function. Here are some random notes I think the OP could use, prompted 
by the text linked to, chap3.pdf.

1.
Throughout the text, assignments use the equal sign instead of the left 
arrow. The left arrow is generally considered more idiomatic and there 
is an important diference beteewn he wo, see help("assignOps").


time.mean = with(CommuteAtlanta, mean(Time))
B = 1000
n = nrow(CommuteAtlanta)

# This should be used, not the above.
time.mean <- with(CommuteAtlanta, mean(Time))
B <- 1000
n <- nrow(CommuteAtlanta)


2.
Systematic use of apply(., 1, mean).
rowMeans (and colMeans) are much faster.


boot.statistics <- apply(boot.samples, 1, mean)
boot.statistics <- rowMeans(boot.samples)


3.
The first confidence interval computation seems awkward. I had never 
seen this way of computing a CI95.
Moreover, it's plain common sense to keep the results with the returned 
decimals and round for display purposes only.
And the normal intervals are computed in a more usual way later in the 
text, see sections 1.2 and 1.3.


me <- ceiling(10 * 2 * time.se)/10
round(time.mean, 1) + c(-1, 1) * me

# Straightforward.
normal_ci95 <- time.mean + c(-1, 1) * 2 * time.se
normal_ci95
round(normal_ci95, 1)

# section 1.2 , function boot.mean
interval = mean(x) + c(-1,1)*2*se

# section 1.3
with(students, mean(Height) + c(-1, 1) * 2 * sd(result))



4.
In section 1.2 there is a bootstrap function boot.mean().
The function could be improved to let users pass a conf.level of their 
choice.
And why force the function user to always have the plot displayed? Base 
functions hist() and barplot() have na argument 'plot' with default TRUE 
allowing the user to choose.

The following seems more user friendly.


boot.mean <- function(x, B, binwidth = NULL, conf.level = 0.95, plot = 
TRUE) {
   require(ggplot2)
   n <- length(x)
   boot.samples <- matrix( sample(x, size = n*B, replace = TRUE), B, n)
   boot.statistics <- rowMeans(boot.samples)
   se <- sd(boot.statistics)
   if ( is.null(binwidth) )
     binwidth <- diff(range(boot.statistics))/30

   p <- ggplot(data.frame(x = boot.statistics), aes(x = x)) +
     geom_histogram(aes(y = ..density..),binwidth = binwidth) +
     geom_density(color = "red")

   alpha <- 1 - (1 - conf.level)/2
   interval <- mean(x) + c(-1, 1) * qnorm(alpha) * se

   if(plot) {
     plot(p)
   }

   list(boot.statistics = boot.statistics, interval = interval, se = se, 
plot = p)
}


Hope this helps,

Rui Barradas



Às 17:28 de 15/03/2025, Kevin Zembower via R-help escreveu:
> Hi, Richard, thanks for replying. I should have mentioned the third
> edition, which we're using. The data file didn't change between the
> second and third editions, and the data on Body Mass Gain was the same
> as in the first edition, although the first edition data file contained
> additional variables.
> 
> According to my text, the BMGain was measured in grams. Thanks for
> pointing out that my statement of the problem lacked crucial
> information.
> 
> The matrix in my example comes from an example in
> https://pages.stat.wisc.edu/~larget/stat302/chap3.pdf, where the author
> created a bootstrap example with a matrix that consisted of one row for
> every sample in the bootstrap, and one column for each mean in the
> original data. This allowed him to find the mean for each row to create
> the bootstrap statistics.
> 
> The only need for the tidyverse is to use the read_csv() function. I'm
> regrettably lazy in not determining which of the multiple functions in
> the tidyverse library loads read_csv(), and just using that one.
> 
> Thanks, again, for helping me to further understand R and this problem.
> 
> -Kevin
> 
> On Sat, 2025-03-15 at 12:00 +0100, r-help-request using r-project.org wrote:
>> Not having the book (and which of the three editions are you using?),
>> I downloaded the data and played with it for a bit.
>> dotchart() showed the Dark and Light conditions looked quite
>> different, but also showed that there are not very many cases.
>> After trying t.test, it occurred to me that I did not know whether
>> "BMGain" means gain in *grams* or gain in *percent*.
>> Reflection told me that for a growth experiment, percent made more
>> sense, which reminded my of one of my first
>> student advising experiences, where I said "never give the computer
>> percentages; let IT calculate the percentages
>> from the baseline and outcome, because once you've thrown away
>> information, the computer can't magically get it back."
>> In particular, in the real world I'd be worried about the possibility
>> that there was some confounding going on, so I would
>> much rather have initial weight and final weight as variables.
>> If BMGain is an absolute measure, the p value for a t test is teeny
>> tiny.
>> If BMGain is a percentage, the p value for a sensible t test is about
>> 0.03.
>>
>> A permutation test went like this.
>> is.light <- d$Group == "Light"
>> is.dark <- d$Group == "Dark"
>> score <- function (g) mean(g[is.light]) - mean(g[is.dark])
>> base.score <- score(d$BMGain)
>> perm.scores <- sapply(1:997, function (i) score(sample(d$BMGain)))
>> sum(perm.scores >= base.score) / length(perm.scores)
>>
>> I don't actually see where matrix() comes into it, still less
>> anything
>> in the tidyverse.
>>
> 
> ______________________________________________
> R-help using 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.


-- 
Este e-mail foi analisado pelo software antivírus AVG para verificar a presença de vírus.
www.avg.com