Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Ә | 2177 | 75 | 1 | 75.0000 |
Бу | 4330 | 224 | 3 | 74.6667 |
Алар | 1433 | 62 | 1 | 62.0000 |
Әмма | 999 | 54 | 1 | 54.0000 |
Менә | 885 | 46 | 1 | 46.0000 |
Ләкин | 811 | 45 | 1 | 45.0000 |
Аның | 1182 | 44 | 1 | 44.0000 |
Ул | 2419 | 76 | 2 | 38.0000 |
Бүген | 559 | 30 | 1 | 30.0000 |
Бер | 566 | 29 | 1 | 29.0000 |
әмма | 684 | 28 | 1 | 28.0000 |
Һәм | 430 | 27 | 1 | 27.0000 |
ә | 1791 | 79 | 3 | 26.3333 |
Шул | 861 | 25 | 1 | 25.0000 |
Мин | 1498 | 74 | 3 | 24.6667 |
Әле | 294 | 24 | 1 | 24.0000 |
Тик | 417 | 22 | 1 | 22.0000 |
Без | 1254 | 57 | 3 | 19.0000 |
17 | 135 | 16 | 1 | 16.0000 |
Беренче | 247 | 15 | 1 | 15.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
чакырды | 260 | 1 | 26 | 0.0385 |
әйтүенчә | 482 | 1 | 23 | 0.0435 |
сүзләренчә | 537 | 1 | 15 | 0.0667 |
ителде | 249 | 1 | 15 | 0.0667 |
белдерүенчә | 210 | 1 | 15 | 0.0667 |
әйтте | 723 | 3 | 41 | 0.0732 |
барысы | 333 | 1 | 13 | 0.0769 |
иде | 5665 | 20 | 254 | 0.0787 |
итте | 955 | 3 | 37 | 0.0811 |
әйткәндә | 128 | 1 | 12 | 0.0833 |
белдерде | 1232 | 5 | 56 | 0.0893 |
катнашты | 273 | 1 | 11 | 0.0909 |
майда | 78 | 1 | 10 | 0.1000 |
фикеренчә | 281 | 1 | 10 | 0.1000 |
янындагы | 98 | 1 | 10 | 0.1000 |
чыкты | 554 | 3 | 28 | 0.1071 |
үтте | 219 | 2 | 17 | 0.1176 |
июньдә | 55 | 1 | 8 | 0.1250 |
tä | 71 | 1 | 8 | 0.1250 |
татарларына | 63 | 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