Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
ATSskinner | 1313 | 77 | 1 | 77.0000 |
They | 2723 | 112 | 3 | 37.3333 |
York | 1004 | 37 | 1 | 37.0000 |
That | 1289 | 55 | 2 | 27.5000 |
So | 401 | 26 | 1 | 26.0000 |
row | 475 | 26 | 1 | 26.0000 |
That’s | 526 | 25 | 1 | 25.0000 |
He’s | 273 | 24 | 1 | 24.0000 |
Both | 498 | 21 | 1 | 21.0000 |
Orleans | 309 | 21 | 1 | 21.0000 |
But | 1253 | 58 | 3 | 19.3333 |
In | 2645 | 96 | 5 | 19.2000 |
Diego | 325 | 19 | 1 | 19.0000 |
Coach | 212 | 19 | 1 | 19.0000 |
She | 293 | 17 | 1 | 17.0000 |
cent | 289 | 15 | 1 | 15.0000 |
He | 2001 | 90 | 6 | 15.0000 |
hill | 354 | 15 | 1 | 15.0000 |
Barbuda | 1144 | 44 | 3 | 14.6667 |
The | 15465 | 790 | 55 | 14.3636 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
vs | 1280 | 2 | 68 | 0.0294 |
Play | 619 | 2 | 48 | 0.0417 |
pm | 344 | 1 | 22 | 0.0455 |
able | 1429 | 1 | 21 | 0.0476 |
kind | 376 | 1 | 19 | 0.0526 |
plenty | 610 | 2 | 29 | 0.0690 |
responsible | 266 | 1 | 14 | 0.0714 |
ahead | 355 | 2 | 24 | 0.0833 |
O/U | 2062 | 12 | 139 | 0.0863 |
erectile | 104 | 1 | 11 | 0.0909 |
failed | 207 | 1 | 11 | 0.0909 |
thanks | 335 | 2 | 22 | 0.0909 |
04 | 59 | 1 | 11 | 0.0909 |
01 | 47 | 1 | 10 | 0.1000 |
02 | 47 | 1 | 10 | 0.1000 |
willing | 97 | 1 | 10 | 0.1000 |
dealing | 114 | 1 | 10 | 0.1000 |
tonight | 5444 | 31 | 302 | 0.1026 |
03 | 46 | 1 | 9 | 0.1111 |
06 | 50 | 1 | 9 | 0.1111 |
In this subsection, we compute the ratio of the number of right neighbors and the number of left neighbors. Again, we look for words with extreme ratios:
Data for first table:
select word,w.freq,aa.cnt, bb.cnt,aa.cnt/bb.cnt as r from words w, (select w1_id,count(c.w2_id) as cnt from co_n c where w1_id>100 group by w1_id) aa, (select w2_id,count(c.w1_id) as cnt from co_n c where w2_id>100 group by w2_id) bb where w_id=aa.w1_id and aa.w1_id=bb.w2_id order by r desc limit 20;
Diagram data:
select aa.cnt, bb.cnt from (select w1_id,count(c.w2_id) as cnt from co_n c where w1_id>100 group by w1_id) aa, (select w2_id,count(c.w1_id) as cnt from co_n c where w2_id>100 group by w2_id) bb where aa.w1_id=bb.w2_id;
5.1.7.1 Number of NN co-occurrences vs. Frequency I
5.1.7.2 Number of NN co-occurrences vs. Frequency II