Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
A | 7090 | 324 | 7 | 46.2857 |
Ha | 578 | 29 | 1 | 29.0000 |
Az | 2656 | 142 | 6 | 23.6667 |
hogy | 6093 | 142 | 7 | 20.2857 |
Ez | 677 | 20 | 1 | 20.0000 |
de | 1606 | 70 | 4 | 17.5000 |
mert | 746 | 31 | 2 | 15.5000 |
Egy | 329 | 15 | 1 | 15.0000 |
Nagyon | 131 | 11 | 1 | 11.0000 |
Nem | 518 | 31 | 3 | 10.3333 |
amit | 439 | 19 | 2 | 9.5000 |
ami | 530 | 19 | 2 | 9.5000 |
Még | 130 | 8 | 1 | 8.0000 |
Úgy | 106 | 8 | 1 | 8.0000 |
minél | 76 | 8 | 1 | 8.0000 |
hiszen | 235 | 8 | 1 | 8.0000 |
ugyanis | 143 | 7 | 1 | 7.0000 |
akik | 357 | 12 | 2 | 6.0000 |
amely | 440 | 12 | 2 | 6.0000 |
amelyek | 181 | 6 | 1 | 6.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
módon | 144 | 1 | 9 | 0.1111 |
években | 74 | 1 | 7 | 0.1429 |
tud | 112 | 1 | 7 | 0.1429 |
legnagyobb | 138 | 1 | 6 | 0.1667 |
ismert | 92 | 1 | 6 | 0.1667 |
sokat | 89 | 1 | 6 | 0.1667 |
kevés | 73 | 1 | 6 | 0.1667 |
keresztül | 144 | 1 | 6 | 0.1667 |
belül | 158 | 2 | 11 | 0.1818 |
kedves | 77 | 1 | 5 | 0.2000 |
mindent | 111 | 1 | 5 | 0.2000 |
más | 366 | 4 | 20 | 0.2000 |
szépen | 58 | 1 | 5 | 0.2000 |
múlva | 56 | 1 | 5 | 0.2000 |
tudjuk | 92 | 1 | 5 | 0.2000 |
további | 110 | 1 | 5 | 0.2000 |
legjobb | 106 | 1 | 5 | 0.2000 |
késõbb | 58 | 1 | 5 | 0.2000 |
éve | 98 | 2 | 9 | 0.2222 |
tudja | 134 | 2 | 9 | 0.2222 |
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