Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
За | 2445 | 120 | 1 | 120.0000 |
які | 4809 | 310 | 3 | 103.3333 |
яка | 1660 | 101 | 1 | 101.0000 |
З | 1380 | 97 | 1 | 97.0000 |
У | 5949 | 288 | 3 | 96.0000 |
На | 3056 | 177 | 2 | 88.5000 |
Він | 864 | 63 | 1 | 63.0000 |
В | 2390 | 165 | 3 | 55.0000 |
Для | 849 | 55 | 1 | 55.0000 |
який | 2307 | 149 | 3 | 49.6667 |
що | 15750 | 733 | 15 | 48.8667 |
щоб | 1564 | 113 | 3 | 37.6667 |
а | 4598 | 107 | 3 | 35.6667 |
Ми | 968 | 70 | 2 | 35.0000 |
Але | 692 | 34 | 1 | 34.0000 |
але | 1523 | 62 | 2 | 31.0000 |
Після | 569 | 30 | 1 | 30.0000 |
Це | 1350 | 76 | 3 | 25.3333 |
До | 730 | 44 | 2 | 22.0000 |
Вона | 406 | 21 | 1 | 21.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
ім | 381 | 1 | 31 | 0.0323 |
вул | 1637 | 1 | 24 | 0.0417 |
Юлії | 305 | 1 | 22 | 0.0455 |
доступ | 242 | 1 | 19 | 0.0526 |
р | 260 | 1 | 17 | 0.0588 |
існування | 127 | 1 | 11 | 0.0909 |
Чернівецької | 712 | 5 | 50 | 0.1000 |
необхідності | 117 | 1 | 10 | 0.1000 |
визначити | 84 | 1 | 10 | 0.1000 |
ст | 88 | 1 | 9 | 0.1111 |
мокрий | 167 | 1 | 9 | 0.1111 |
склали | 96 | 1 | 9 | 0.1111 |
Верховної | 250 | 2 | 18 | 0.1111 |
дощ | 788 | 2 | 17 | 0.1176 |
дозволу | 101 | 1 | 8 | 0.1250 |
пропонують | 109 | 1 | 8 | 0.1250 |
Кетрін | 105 | 1 | 8 | 0.1250 |
житлово-комунального | 86 | 1 | 8 | 0.1250 |
села | 116 | 1 | 8 | 0.1250 |
заробітної | 125 | 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