Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
які | 11035 | 607 | 6 | 101.1667 |
Ця | 866 | 75 | 1 | 75.0000 |
яка | 4554 | 289 | 4 | 72.2500 |
який | 5549 | 337 | 5 | 67.4000 |
але | 8291 | 193 | 3 | 64.3333 |
Це | 5139 | 240 | 4 | 60.0000 |
Він | 3533 | 219 | 4 | 54.7500 |
а | 18009 | 310 | 6 | 51.6667 |
У | 14853 | 747 | 24 | 31.1250 |
що | 49829 | 1618 | 56 | 28.8929 |
Тільки | 567 | 28 | 1 | 28.0000 |
Там | 540 | 28 | 1 | 28.0000 |
Дуже | 466 | 27 | 1 | 27.0000 |
Всі | 697 | 27 | 1 | 27.0000 |
Така | 399 | 26 | 1 | 26.0000 |
Тут | 1045 | 49 | 2 | 24.5000 |
щоб | 6173 | 286 | 12 | 23.8333 |
Я | 4140 | 214 | 9 | 23.7778 |
Та | 1556 | 68 | 3 | 22.6667 |
бо | 2968 | 110 | 5 | 22.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
р | 5018 | 1 | 334 | 0.0030 |
ст | 3040 | 1 | 98 | 0.0102 |
зв | 1933 | 2 | 65 | 0.0308 |
обов | 1420 | 2 | 62 | 0.0323 |
тис | 952 | 2 | 53 | 0.0377 |
відповідно | 1242 | 1 | 26 | 0.0385 |
рр | 678 | 1 | 25 | 0.0400 |
Стр | 197 | 1 | 24 | 0.0417 |
здоров | 726 | 2 | 39 | 0.0513 |
комп | 449 | 1 | 17 | 0.0588 |
зобов | 719 | 2 | 33 | 0.0606 |
т | 505 | 1 | 16 | 0.0625 |
ім | 797 | 3 | 48 | 0.0625 |
св | 252 | 1 | 16 | 0.0625 |
пов | 1572 | 3 | 44 | 0.0682 |
п | 1541 | 3 | 43 | 0.0698 |
знов | 245 | 1 | 14 | 0.0714 |
Верховної | 275 | 2 | 27 | 0.0741 |
дозвіл | 139 | 1 | 13 | 0.0769 |
пам | 1129 | 3 | 38 | 0.0789 |
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