Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
а | 1243 | 64 | 1 | 64.0000 |
У | 823 | 62 | 3 | 20.6667 |
али | 444 | 17 | 1 | 17.0000 |
јер | 228 | 13 | 1 | 13.0000 |
па | 172 | 9 | 1 | 9.0000 |
Општина | 31 | 7 | 1 | 7.0000 |
Да | 80 | 6 | 1 | 6.0000 |
часа | 82 | 6 | 1 | 6.0000 |
И | 121 | 5 | 1 | 5.0000 |
првог | 45 | 5 | 1 | 5.0000 |
даље | 85 | 5 | 1 | 5.0000 |
кроз | 131 | 5 | 1 | 5.0000 |
воде | 48 | 5 | 1 | 5.0000 |
кише | 28 | 4 | 1 | 4.0000 |
ученика | 52 | 4 | 1 | 4.0000 |
30. | 21 | 4 | 1 | 4.0000 |
Милош | 41 | 4 | 1 | 4.0000 |
људе | 35 | 4 | 1 | 4.0000 |
колико | 67 | 4 | 1 | 4.0000 |
болнице | 126 | 4 | 1 | 4.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Опште | 119 | 1 | 6 | 0.1667 |
тек | 59 | 1 | 6 | 0.1667 |
2022 | 103 | 2 | 12 | 0.1667 |
тада | 54 | 1 | 6 | 0.1667 |
нас | 134 | 2 | 12 | 0.1667 |
потрошачи | 40 | 1 | 6 | 0.1667 |
периоду | 61 | 1 | 6 | 0.1667 |
довољно | 36 | 1 | 5 | 0.2000 |
могли | 48 | 1 | 5 | 0.2000 |
тренутно | 61 | 1 | 5 | 0.2000 |
стране | 64 | 1 | 5 | 0.2000 |
округа | 41 | 1 | 5 | 0.2000 |
могуће | 31 | 1 | 5 | 0.2000 |
управе | 32 | 1 | 5 | 0.2000 |
друге | 59 | 1 | 5 | 0.2000 |
година | 285 | 8 | 35 | 0.2286 |
часова | 136 | 3 | 13 | 0.2308 |
примило | 26 | 1 | 4 | 0.2500 |
њега | 37 | 1 | 4 | 0.2500 |
грмљавином | 51 | 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