Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
We | 973 | 56 | 3 | 18.6667 |
The | 4178 | 233 | 14 | 16.6429 |
am | 334 | 25 | 2 | 12.5000 |
It | 952 | 36 | 3 | 12.0000 |
This | 1218 | 60 | 5 | 12.0000 |
They | 347 | 22 | 2 | 11.0000 |
He | 432 | 19 | 2 | 9.5000 |
Please | 189 | 19 | 2 | 9.5000 |
And | 363 | 25 | 3 | 8.3333 |
She | 166 | 8 | 1 | 8.0000 |
As | 293 | 15 | 2 | 7.5000 |
After | 168 | 7 | 1 | 7.0000 |
Now | 131 | 7 | 1 | 7.0000 |
By | 125 | 7 | 1 | 7.0000 |
Town | 217 | 7 | 1 | 7.0000 |
It’s | 118 | 7 | 1 | 7.0000 |
In | 807 | 46 | 7 | 6.5714 |
These | 244 | 13 | 2 | 6.5000 |
points | 99 | 6 | 1 | 6.0000 |
So | 188 | 12 | 2 | 6.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
wanted | 85 | 1 | 9 | 0.1111 |
able | 257 | 1 | 9 | 0.1111 |
interesting | 73 | 1 | 8 | 0.1250 |
part | 302 | 3 | 24 | 0.1250 |
trying | 92 | 1 | 7 | 0.1429 |
involved | 114 | 2 | 12 | 0.1667 |
hundred | 28 | 1 | 6 | 0.1667 |
likely | 73 | 1 | 6 | 0.1667 |
afternoon | 46 | 1 | 6 | 0.1667 |
individual | 93 | 1 | 6 | 0.1667 |
century | 35 | 1 | 6 | 0.1667 |
International | 89 | 1 | 6 | 0.1667 |
excellent | 104 | 1 | 6 | 0.1667 |
recent | 77 | 1 | 6 | 0.1667 |
perhaps | 55 | 1 | 6 | 0.1667 |
rather | 100 | 1 | 6 | 0.1667 |
commitment | 48 | 1 | 5 | 0.2000 |
tried | 55 | 1 | 5 | 0.2000 |
type | 96 | 1 | 5 | 0.2000 |
entire | 82 | 1 | 5 | 0.2000 |
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