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Topic: Is there a way to calculate an average ranking from uneven lists?
Replies: 15   Last Post: Oct 30, 2013 12:18 PM

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Ben Bacarisse

Posts: 751
Registered: 7/4/07
Re: Is there a way to calculate an average ranking from uneven lists?
Posted: Oct 27, 2013 7:46 PM
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Jennifer Murphy <JenMurphy@jm.invalid> writes:

> On Sun, 27 Oct 2013 13:36:29 -0600, Virgil <virgil@ligriv.com> wrote:
>

>>In article <chpq69prq63kh364qqmphkmqedhgm5ti6h@4ax.com>,
>> Jennifer Murphy <JenMurphy@jm.invalid> wrote:
>>

>>> There are many lists containing rankings of great books. Some are
>>> limited to a particular genre (historical novels, biographies, science
>>> fiction). Others are more general. Some are fairly short (50-100 books).
>>> Others are much longer (1,001 books).
>>>
>>> Is there a way to "average" the data from as many of these lists as
>>> possible to get some sort of composite ranking of all of the books that
>>> appear in any of the lists?

<snip>
>>One way to compare rankings when there are different numbers of objects
>>ranked in different rankings is to scale them all over the same range,
>>such as from 0% to 100%.
>>
>>Thus in all rankings a lowest rank would rank 0% and the highest 100%,
>>and the middle one, if there were one, would rank 50%.
>>Four items with no ties would rank 0%, 33 1/3%, 66 2/3% and 100%,
>>and so on.
>>
>>For something of rank r out of n ranks use (r-1)/(n-1) times 100%.

>
> In the lists I have, the highest ranking entity is R=1, the lowest is
> R=N. For that, I think the formula is (N-R)/(N-1). No?


Here's another idea to add to the mix. Some rankings (and I think this
is one) have the property the top is more significant than the bottom.
Anyone who picks a book to be no. 1 should have carefully weighed it up
against no. 2 and no. 3. But what about no. 1001? How likely is it
that some tiny alteration in the assessment might make it no. 995 or
998? And this effect is related to the absolute length of the list, not
just the relative position within it.

Put it another way, outstanding things stand out. Once you are into the
more run of the mill. the distinction become less significant.

As a result, you might consider a negative exponential weighting -- a
ranking of R is given a value of w^(R-1) with 0 < w < 1. Thus all first
positions are "worth" 1, all second positions are worth w, and all third
positions w^2 and so on.

--
Ben.



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