Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Во | 2036 | 154 | 1 | 154.0000 |
а | 3159 | 123 | 1 | 123.0000 |
но | 1381 | 56 | 1 | 56.0000 |
Од | 726 | 43 | 1 | 43.0000 |
Се | 418 | 31 | 1 | 31.0000 |
На | 807 | 58 | 2 | 29.0000 |
Со | 381 | 17 | 1 | 17.0000 |
За | 591 | 34 | 2 | 17.0000 |
По | 263 | 17 | 1 | 17.0000 |
Тие | 259 | 14 | 1 | 14.0000 |
Ова | 403 | 14 | 1 | 14.0000 |
Тоа | 377 | 13 | 1 | 13.0000 |
бидејќи | 432 | 25 | 2 | 12.5000 |
Ако | 224 | 12 | 1 | 12.0000 |
Да | 206 | 11 | 1 | 11.0000 |
односно | 333 | 10 | 1 | 10.0000 |
Како | 360 | 10 | 1 | 10.0000 |
И | 491 | 30 | 3 | 10.0000 |
Дали | 197 | 10 | 1 | 10.0000 |
па | 413 | 20 | 2 | 10.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
дел | 556 | 2 | 36 | 0.0556 |
треба | 1052 | 3 | 50 | 0.0600 |
министер | 152 | 1 | 12 | 0.0833 |
сака | 163 | 1 | 11 | 0.0909 |
сакаат | 161 | 1 | 10 | 0.1000 |
2020 | 189 | 2 | 18 | 0.1111 |
страна | 407 | 2 | 18 | 0.1111 |
директор | 76 | 1 | 8 | 0.1250 |
Министерството | 305 | 1 | 8 | 0.1250 |
бројот | 374 | 2 | 16 | 0.1250 |
ширењето | 93 | 1 | 8 | 0.1250 |
дава | 59 | 1 | 8 | 0.1250 |
контакт | 104 | 1 | 8 | 0.1250 |
можеме | 104 | 1 | 8 | 0.1250 |
Законот | 100 | 1 | 8 | 0.1250 |
2019 | 98 | 1 | 7 | 0.1429 |
следи | 81 | 1 | 7 | 0.1429 |
имате | 78 | 1 | 7 | 0.1429 |
директорот | 77 | 1 | 7 | 0.1429 |
ме | 98 | 1 | 7 | 0.1429 |
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