Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
В | 387 | 28 | 1 | 28.0000 |
но | 459 | 23 | 1 | 23.0000 |
Това | 291 | 17 | 1 | 17.0000 |
Не | 200 | 11 | 1 | 11.0000 |
По | 129 | 8 | 1 | 8.0000 |
през | 300 | 26 | 4 | 6.5000 |
а | 462 | 13 | 2 | 6.5000 |
също | 106 | 6 | 1 | 6.0000 |
във | 206 | 17 | 3 | 5.6667 |
Ако | 137 | 5 | 1 | 5.0000 |
със | 194 | 5 | 1 | 5.0000 |
Ще | 70 | 4 | 1 | 4.0000 |
предаде | 19 | 4 | 1 | 4.0000 |
Какво | 39 | 4 | 1 | 4.0000 |
където | 88 | 4 | 1 | 4.0000 |
че | 1532 | 48 | 14 | 3.4286 |
подкрепа | 29 | 3 | 1 | 3.0000 |
държави | 63 | 6 | 2 | 3.0000 |
проект | 25 | 3 | 1 | 3.0000 |
Много | 41 | 3 | 1 | 3.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
края | 69 | 1 | 7 | 0.1429 |
трябва | 289 | 2 | 14 | 0.1429 |
сте | 88 | 1 | 7 | 0.1429 |
време | 192 | 2 | 13 | 0.1538 |
началото | 67 | 1 | 6 | 0.1667 |
ни | 218 | 1 | 6 | 0.1667 |
част | 122 | 1 | 6 | 0.1667 |
мен | 49 | 1 | 5 | 0.2000 |
защо | 70 | 1 | 5 | 0.2000 |
могат | 125 | 1 | 5 | 0.2000 |
целия | 32 | 1 | 5 | 0.2000 |
седмици | 37 | 1 | 5 | 0.2000 |
беше | 201 | 3 | 14 | 0.2143 |
години | 211 | 4 | 18 | 0.2222 |
теб | 28 | 1 | 4 | 0.2500 |
точно | 76 | 1 | 4 | 0.2500 |
иска | 55 | 1 | 4 | 0.2500 |
вас | 30 | 1 | 4 | 0.2500 |
Министерството | 24 | 1 | 4 | 0.2500 |
Турция | 37 | 1 | 4 | 0.2500 |
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