The `ordertrans()` function in the `hyper2` package
In hyper2: The Hyperdirichlet Distribution, Mark 2

set.seed(0)
library("hyper2")
options(rmarkdown.html_vignette.check_title = FALSE)
knitr::opts_chunk$set(echo = TRUE)
knitr::opts_chunk$set(fig.width=6, fig.height=6)
knit_print.function <- function(x, ...){dput(x)}
registerS3method(
  "knit_print", "function", knit_print.function,
  envir = asNamespace("knitr")
)

knitr::include_graphics(system.file("help/figures/hyper2.png", package = "hyper2"))

ordertrans

To cite the hyper2 package in publications, please use @hankin2017_rmd. The ordertrans() function can be difficult to understand and this short document provides some sensible use-cases. The manpage provides a very simple example, but here we are going to use an even simpler example:

x <- c(d=2,a=3,b=1,c=4)
x

In the above, object x is a named vector with elements seq_along(x) in some order. It means that competitor d came second, competitor a came third, b came first and c came fourth. Technically x is an order vector because it answers the question "where did a particular competitor come?" However, it might equally be a rank vector because it answers the question "who came first? who came second?" (see also the discussion at rrank.Rd).

But note that x is not helpfully structured to answer either of these questions: you have to go searching through the names or the ranks respectively---both of which may appear in any order---to find the competitor or rank you are interested in. To find the rank vector (that is, who came first, second etc), one would use sort():

sort(x)

But this is suboptimal to find the order vector ["where did a particular competitor come?"]. For the order vector, we want to rearrange the elements of x so that the names are in alphabetical order. That way we can see straightaway where competitor a placed (in this case, third). This nontrivial task is accomplished by function ordertrans():

o <- ordertrans(x) # by default, sorts names() into alphabetical order
o

Observe that objects x and o are equal in the sense that they are a rearrangement of one another:

identical(x, o[c(4,1,2,3)])
identical(o, x[c(2,3,4,1)])

One consequence of this is that the resulting Plackett-Luce support functions will be the same:

(Sx <- ordervec2supp(x))
(So <- ordervec2supp(o))

Note carefully that the two support functions above can be mathematically identical but not formally identical() because neither the order of the brackets, nor the order of terms within a bracket, is defined. They look the same on my system but YMMV. It is possible that the two support functions might appear to be different even though they are mathematically the same. We can verify that the two are mathematically identical using package idiom ==:

Sx==So

Another use-case

How about this:

(x <- c(d=2, a=3, c=4, b=3, e=6))
(y <- c(e=3, c=2, a=4, b=5, d=1))
x+y

Above, observe that R is behaving as intended $\ldots$ but is potentially confusing. Take the first element. This has value 2+5=7 and name d from the name of element 1 of x. But it would be reasonable to ask for the first element to be the sum of element d of x and element d of y, which would be 2+1=3. This can be done with ordertrans():

ordertrans(x) + ordertrans(y)

See how ordertrans() has rearranged both x and y so that the names are in alphabetical order.

But this is not perfect. Consider:

(z <- c(f=3, g=2, h=4, a=5, b=1))  # names NOT letters[1:5]
ordertrans(x) + ordertrans(z)  # arguably not well-defined

In this case the result is arguably incorrect.

Skating

Let us consider the skating dataset and use ordertrans() to study it (note that the skating dataset is analysed in more depth in skating.Rmd).

skating_table

We might ask how judges J1 and J2 compare to one another? We need to create vectors like x and y above:

j1 <- skating_table[,1]  # column 1 is judge number 1
names(j1) <- rownames(skating_table)
j2 <- skating_table[,2]  # column 2 is judge number 2
names(j2) <- rownames(skating_table)
j1
j2
cbind(j1,j2)

In the above, see how objects j1 and j2 have identical names, in the same order. Observe that hughes is ranked 1 (that is, first) by J1, and 4 (that is, 4th) by J2. This makes it sensible to plot j1 against j2:

par(pty='s')  # forces plot to be square
plot(j1,j2,asp=1,pty='s',xlim=c(0,25),ylim=c(0,25),pch=16,xlab='judge 1',ylab='judge 2')
abline(0,1)  # diagonal line
for(i in seq_along(j1)){text(j1[i],j2[i],names(j1)[i],pos=4,col='gray',cex=0.7)}

In figure \@ref(fig:j1vsj2), we see general agreement but differences in detail. For example, hughes is ranked first by judge 1 and fourth by judge 2. However, other problems are not so easy. Suppose we wish to compare the ranks according to likelihood with the ranks according to some points system.

mL <- skating_maxp  # predefined; use maxp(skating) to calculate ab initio
mL

Note that in the above, the competitors' names are in alphabetical order. We first need to convert strengths to ranks:

mL[] <- rank(-mL)  # minus because ranks orders from weak to strong
mL

(note that the names are in the same order as before, alphabetical). In the above we see that slutskya ranks first, hughes second, and so on. Another way of ranking the skaters is to use a Borda-type system: essentially the rowsums of the table (and, of course, the lowest score wins):

mP <- rowSums(skating_table)  # 'P' for Points
mP[] <- rank(mP,ties='first') # positive sign here
mP

It is not at all obvious how to compare mP and mL. For example, we might be interested in hegel. It takes some effort to find that her likelihood rank is 20 and her Borda rank is 19. Function ordertrans() facilitates this:

ordertrans(mP,names(mL))

See above how ordertrans() shows the points-based ranks but in alphabetical order, to facilitate comparison with mL. We can now plot these against one another:

plot(mL,ordertrans(mP,names(mL)))

However, figure \@ref(fig:crapplot) is a bit crude. Function ordertransplot() gives a more visually pleasing output, see figure \@ref(fig:showoffordertransplot).

ordertransplot(mL,mP,xlab="likelihood rank",ylab="Borda rank")

So now we may compare judge 1 against likelihood:

ordertransplot(mL,j1,xlab="likelihood rank",ylab="Judge 1 rank")

In figure \@ref(fig:transplotjudge1), looking at the lower-left corner, we see (reading horizontally) that the likelihood method placed Slutskya first, then Hughes second, then Kwan third; while (reading vertically) judge 1 placed Hughes first, then Kwan, then Slutskya.