Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Как | 1929 | 88 | 1 | 88.0000 |
которые | 2515 | 162 | 2 | 81.0000 |
который | 1966 | 116 | 2 | 58.0000 |
А | 1492 | 58 | 1 | 58.0000 |
которая | 1175 | 56 | 1 | 56.0000 |
Это | 1705 | 91 | 2 | 45.5000 |
но | 3087 | 81 | 2 | 40.5000 |
Но | 1808 | 77 | 2 | 38.5000 |
чтобы | 2258 | 149 | 4 | 37.2500 |
И | 1658 | 73 | 2 | 36.5000 |
что | 17753 | 585 | 18 | 32.5000 |
На | 2082 | 127 | 4 | 31.7500 |
В | 8614 | 404 | 13 | 31.0769 |
С | 931 | 73 | 3 | 24.3333 |
а | 4522 | 111 | 5 | 22.2000 |
которое | 531 | 22 | 1 | 22.0000 |
При | 764 | 22 | 1 | 22.0000 |
Мы | 1311 | 84 | 4 | 21.0000 |
Я | 1443 | 79 | 4 | 19.7500 |
рамках | 635 | 39 | 2 | 19.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
сельского | 123 | 1 | 12 | 0.0833 |
случаях | 96 | 1 | 12 | 0.0833 |
Бидзина | 151 | 1 | 10 | 0.1000 |
команда | 146 | 1 | 9 | 0.1111 |
корр | 91 | 1 | 9 | 0.1111 |
хотели | 87 | 1 | 8 | 0.1250 |
сын | 92 | 1 | 8 | 0.1250 |
2010 | 419 | 5 | 36 | 0.1389 |
утверждает | 129 | 1 | 7 | 0.1429 |
г | 71 | 1 | 7 | 0.1429 |
перспективе | 58 | 1 | 7 | 0.1429 |
отсутствие | 80 | 1 | 7 | 0.1429 |
кажется | 180 | 1 | 7 | 0.1429 |
точки | 265 | 2 | 13 | 0.1538 |
2013 | 362 | 5 | 32 | 0.1563 |
оба | 69 | 1 | 6 | 0.1667 |
гибели | 70 | 1 | 6 | 0.1667 |
2011 | 401 | 5 | 30 | 0.1667 |
важную | 38 | 1 | 6 | 0.1667 |
Бидзины | 97 | 1 | 6 | 0.1667 |
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