The number of headers on a page influences how likely someone viewing the page with LHS ToC will be able to see a definition without scrolling. I analyzed the traffic stats for 2009-08 which included around 880 of the most common main namespace pages. From that I graphed the cumulative distribution of page views per number of page headers. This indicates the percentage of page views on pages with at most a certain number of headers. For instance, on a common setup (1280x800 monitor with Firefox) the ToC for a page with about 12 headers will push the definitions off the first screen. Looking at the graph we can see that 75% of views are of pages with have at most 11 headers. This means that about 25% of views are of pages that, with high probability, will not have a definition on the initial screen. Using the graph one can make determinations about modified viewer experiences. I think it's unacceptably high that around a quarter of page views will not have a definition on the "landing" screen and is the main reason why almost any of the options are preferable to the default. This is especially true since as time goes on, pages get more and more headers. --Bequw → τ 04:32, 24 February 2010 (UTC)
Why isn't there an option for "table of contents being above the entries, uncollapsed, on the top-left, showing only language headings, with the numbers being hidden"? --Vahagn Petrosyan 07:49, 25 February 2010 (UTC)
Using CSS, there seems to be know way to make the choice of option page-dependent (within the main namespace). Am I correct? (Something with many languages, each with only one part of speech, might benefit from a languages-only TOC, while something with a lot of English and nothing else might benefit from a TOC showing POS.)—msh210℠ 17:00, 25 February 2010 (UTC)
Does option 6 mean "keep TOCs as they are now, but collapsed by default"? It doesn't specify.—msh210℠ 18:56, 25 February 2010 (UTC)
To get an idea of the result of the vote, I have used the following online calculator: Ranked-ballot voting calculator, Washington University, St. Louis.
Without the use of a tiebreaker, Schulze method gives a tie: "a, d, e and g lose no beatpath comparisons" (a=1,d=4,e=5,g=7). Given "the ranking d>f>e>a>g>c>b was used as a random-ballot tiebreaker", the calculator has determined that the winner according Schulze method is option 4, labeled "d" in the data below.
The data I have fed into the calculator (11 votes):
1:c>e>f>g>d>b>a 1:e>d>c>g>b>f>a 1:b>a>f>d>c>g>e 1:a>g>d>e>c>b>f 1:a>b=d>f>g>c=e 1:g>e>d>f>a>c>b 1:d>f>e>a>g>c>b 1:a>c>g>b=d=e=f 1:d>e>c>b>g>a>f 1:e>a=b=c=d=e=g 1:f>a=g>c=e>b=d
I have obtained this data by mapping 1 to a, 2 to b, ..., 7 to g, from
3>5>6>7>4>2>1 5>4>3>7>2>6>1 2>1>6>4>3>7>5 1>7>4>5>3>2>6 1>2=4>6>7>3=5 7>5>4>6>1>3>2 4>6>5>1>7>3>2 1>3>7>2=4=5=6 4>5>3>2>7>1>6 5>1=2=3=4=5=7 6>1=7>3=5>2=4
The pairwise matrix:
against | ||||||||
---|---|---|---|---|---|---|---|---|
a | b | c | d | e | f | g | ||
for | a | 6½ | 7½ | 5½ | 5 | 6 | 6 | |
b | 4½ | 2½ | 3 | 2½ | 6½ | 3½ | ||
c | 3½ | 8½ | 3½ | 4 | 6 | 5½ | ||
d | 5½ | 8 | 7½ | 5½ | 7½ | 5½ | ||
e | 6 | 8½ | 7 | 5½ | 6½ | 5 | ||
f | 5 | 4½ | 5 | 3½ | 4½ | 5 | ||
g | 5 | 7½ | 5½ | 5½ | 6 | 6 |