Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
щоб | 2731 | 190 | 2 | 95.0000 |
але | 2370 | 90 | 1 | 90.0000 |
яка | 1406 | 84 | 1 | 84.0000 |
які | 3964 | 241 | 3 | 80.3333 |
що | 18459 | 598 | 10 | 59.8000 |
У | 5384 | 318 | 6 | 53.0000 |
Він | 1584 | 83 | 2 | 41.5000 |
який | 2075 | 122 | 3 | 40.6667 |
а | 4087 | 106 | 3 | 35.3333 |
Як | 1550 | 68 | 2 | 34.0000 |
Також | 1511 | 68 | 2 | 34.0000 |
На | 2846 | 175 | 6 | 29.1667 |
В | 1730 | 124 | 5 | 24.8000 |
Президент | 386 | 24 | 1 | 24.0000 |
адже | 414 | 22 | 1 | 22.0000 |
Раніше | 543 | 22 | 1 | 22.0000 |
аби | 390 | 22 | 1 | 22.0000 |
Вони | 946 | 59 | 3 | 19.6667 |
Саме | 266 | 19 | 1 | 19.0000 |
Наразі | 352 | 19 | 1 | 19.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
видно | 173 | 1 | 18 | 0.0556 |
закордонних | 230 | 1 | 12 | 0.0833 |
населених | 501 | 3 | 35 | 0.0857 |
залежить | 153 | 1 | 11 | 0.0909 |
розуміють | 122 | 1 | 11 | 0.0909 |
2022 | 752 | 6 | 63 | 0.0952 |
необхідності | 74 | 1 | 10 | 0.1000 |
цін | 135 | 1 | 10 | 0.1000 |
населені | 102 | 1 | 9 | 0.1111 |
критичної | 118 | 1 | 9 | 0.1111 |
Російської | 303 | 3 | 27 | 0.1111 |
Збройних | 307 | 3 | 25 | 0.1200 |
дійсно | 160 | 1 | 8 | 0.1250 |
протиповітряної | 161 | 2 | 16 | 0.1250 |
сезоні | 67 | 1 | 8 | 0.1250 |
легко | 114 | 1 | 8 | 0.1250 |
переконаний | 101 | 1 | 8 | 0.1250 |
відомо | 417 | 2 | 16 | 0.1250 |
змогу | 64 | 1 | 8 | 0.1250 |
факт | 86 | 1 | 8 | 0.1250 |
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