Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
В | 715 | 53 | 1 | 53.0000 |
но | 345 | 12 | 1 | 12.0000 |
Как | 127 | 9 | 1 | 9.0000 |
во | 220 | 16 | 2 | 8.0000 |
а | 488 | 23 | 3 | 7.6667 |
со | 162 | 15 | 2 | 7.5000 |
который | 163 | 7 | 1 | 7.0000 |
что | 1432 | 61 | 9 | 6.7778 |
Мы | 112 | 6 | 1 | 6.0000 |
Я | 123 | 6 | 1 | 6.0000 |
чтобы | 217 | 11 | 2 | 5.5000 |
до | 294 | 16 | 3 | 5.3333 |
под | 127 | 5 | 1 | 5.0000 |
2015 | 52 | 5 | 1 | 5.0000 |
которые | 228 | 9 | 2 | 4.5000 |
еще | 151 | 9 | 2 | 4.5000 |
тысяч | 49 | 4 | 1 | 4.0000 |
детей | 53 | 4 | 1 | 4.0000 |
между | 78 | 4 | 1 | 4.0000 |
ни | 110 | 8 | 2 | 4.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
значит | 43 | 1 | 8 | 0.1250 |
день | 98 | 1 | 7 | 0.1429 |
бы | 161 | 2 | 13 | 0.1538 |
года | 252 | 4 | 24 | 0.1667 |
участие | 52 | 1 | 6 | 0.1667 |
власти | 63 | 1 | 6 | 0.1667 |
же | 280 | 5 | 27 | 0.1852 |
год | 67 | 2 | 10 | 0.2000 |
году | 163 | 3 | 14 | 0.2143 |
них | 100 | 2 | 9 | 0.2222 |
самым | 34 | 1 | 4 | 0.2500 |
этому | 44 | 1 | 4 | 0.2500 |
иметь | 32 | 1 | 4 | 0.2500 |
сейчас | 70 | 1 | 4 | 0.2500 |
себе | 66 | 1 | 4 | 0.2500 |
другой | 39 | 1 | 4 | 0.2500 |
минут | 36 | 1 | 4 | 0.2500 |
имя | 40 | 1 | 4 | 0.2500 |
образом | 63 | 1 | 4 | 0.2500 |
мира | 40 | 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