Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
а | 2242 | 119 | 3 | 39.6667 |
јер | 564 | 27 | 1 | 27.0000 |
па | 349 | 19 | 1 | 19.0000 |
У | 1894 | 130 | 7 | 18.5714 |
На | 811 | 49 | 3 | 16.3333 |
Лука | 401 | 14 | 1 | 14.0000 |
Херцеговине | 484 | 14 | 1 | 14.0000 |
али | 904 | 40 | 3 | 13.3333 |
И | 343 | 23 | 2 | 11.5000 |
close | 156 | 11 | 1 | 11.0000 |
штампу | 588 | 11 | 1 | 11.0000 |
Циљ | 106 | 10 | 1 | 10.0000 |
Овај | 149 | 9 | 1 | 9.0000 |
Па | 79 | 8 | 1 | 8.0000 |
Херцеговини | 278 | 8 | 1 | 8.0000 |
За | 382 | 23 | 3 | 7.6667 |
оквиру | 245 | 15 | 2 | 7.5000 |
Предсједник | 154 | 15 | 2 | 7.5000 |
Као | 158 | 7 | 1 | 7.0000 |
издавање | 95 | 7 | 1 | 7.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
електричне | 234 | 1 | 23 | 0.0435 |
Босне | 517 | 2 | 33 | 0.0606 |
2010 | 155 | 1 | 15 | 0.0667 |
2011 | 245 | 2 | 29 | 0.0690 |
2009 | 128 | 1 | 12 | 0.0833 |
Универзитета | 163 | 1 | 11 | 0.0909 |
Кодекса | 132 | 1 | 11 | 0.0909 |
2013 | 349 | 3 | 30 | 0.1000 |
2014 | 537 | 5 | 48 | 0.1042 |
2008 | 120 | 1 | 9 | 0.1111 |
Републике | 828 | 7 | 61 | 0.1148 |
Општих | 52 | 1 | 8 | 0.1250 |
Социјалистичке | 160 | 1 | 8 | 0.1250 |
Бање | 92 | 1 | 8 | 0.1250 |
Представничког | 92 | 1 | 8 | 0.1250 |
2012 | 247 | 3 | 22 | 0.1364 |
финансија | 96 | 1 | 7 | 0.1429 |
мало | 124 | 1 | 7 | 0.1429 |
сад | 79 | 1 | 7 | 0.1429 |
ступа | 44 | 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