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

[Richard O'Keefe]

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.


On Sat, 15 Mar 2025 at 07:52, Kevin Zembower via R-help
<r-help using r-project.org> wrote:
>
> Thank you all, very much, for your kind and detailed explanations. I
> didn't understand, mainly, that the matrix() call only called its
> parameters once. I was certain that this was a bug with sample()
> getting seeded with a constant value, and giving the same permutation.
>
> I think I need to make my MWE a little less minimal to continue
> learning. If you're familiar with the Lock5 statistics textbook, I'm
> working on the Light and Dark mice example, where groups of mice were
> exposed or not to light at night, then measured for weight gain. The
> statistic is mean difference in weight gain between the two groups.
>
> My understanding of how I'm supposed to construct a randomized
> distribution is to join the weight gains of the 10 mice exposed to
> light at night to the 8 mice not exposed to light at night. After
> shuffling this data, I arbitrarily group the first 10 values into the
> 'light' group, and the last 8 into the 'dark' group, and find the
> difference in their means.
>
> I think I can do this correctly with:
> ===================
> ## Less-minimal working example
> library(tidyverse)
>
> library(Lock5Data)
> data(LightatNight)
> str(LightatNight)
>
> ## Or, if you don't have the Lock5Data library:
> (d <-
> read_csv("https://www.lock5stat.com/datasets3e/LigthtatNight.csv"))
>
> (lt <- d$BMGain[d$Group == "Light"])
> (dk <- d$BMGain[d$Group == "Dark"])
> (n_lt <- length(lt))
> (n_dk <- length(dk))
>
> (data <- c(lt, dk))
>
> B <- 10 #Will be 1000
> n <- length(data)
>
> random.samples <- matrix(NA, B, n)
> random.statistics <- rep(NA, B)
>
> for(i in 1:B) {
>     random.samples[i,] <- sample(data)
>     random.statistics[i] <- mean(random.samples[i, 1:n_lt]) -
>         mean(random.samples[i, (n_lt + 1):(n_lt + n_dk)])
> }
> random.samples
> random.statistics
>
> ## Trying to do it without a for(), using Peter's suggestion:
> (random.samples <- matrix(replicate(B, sample(data)), B, n,
> byrow=TRUE))
> compute.diff.means <- function(x) {
>     return(mean(x[1:n_lt]) - mean(x[(n_lt+1):(n_lt+n_dk)]))
> }
> (random.statistics <- apply(random.samples, 1, compute.diff.means))
> =======================
>
> I think both of these methods give me the data I'm trying for. Any
> suggestions on my R coding techniques are welcome.
>
> Thank you all, again, for taking the time and effort to help me. Your
> help is greatly appreciated.
>
> -Kevin
>
> On Thu, 2025-03-13 at 17:00 -0400, Kevin Zembower wrote:
> > Hello, all,
> >
> > I'm learning to do randomized distributions in my Stats 101 class*. I
> > thought I could do it with a call to sample() inside a matrix(),
> > like:
> >
> > > matrix(sample(1:10, replace=TRUE), 5, 10, byrow=TRUE)
> >      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> > [1,]    8    2    3    1    8    2    8    8    9     8
> > [2,]    8    2    3    1    8    2    8    8    9     8
> > [3,]    8    2    3    1    8    2    8    8    9     8
> > [4,]    8    2    3    1    8    2    8    8    9     8
> > [5,]    8    2    3    1    8    2    8    8    9     8
> > >
> >
> > Imagine my surprise to learn that all the rows were the same
> > permutation. I thought each time sample() was called inside the
> > matrix,
> > it would generate a different permutation.
> >
> > I modeled this after the bootstrap sample techniques in
> > https://pages.stat.wisc.edu/~larget/stat302/chap3.pdf. I don't
> > understand why it works in bootstrap samples (with replace=TRUE), but
> > not in randomized distributions (with replace=FALSE).
> >
> > Thanks for any insight you can share with me, and any suggestions for
> > getting rows in a matrix with different permutations.
> >
> > -Kevin
> >
> > *No, this isn't a homework problem. We're using Lock5 as the text in
> > class, along with its StatKey web application. I'm just trying to get
> > more out of the class by also solving our problems using R, for which
> > I'm not receiving any class credit.
>
>
>
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