Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
et | 2428 | 96 | 4 | 24.0000 |
kes | 431 | 18 | 1 | 18.0000 |
kuid | 288 | 14 | 1 | 14.0000 |
mis | 579 | 27 | 2 | 13.5000 |
See | 270 | 11 | 1 | 11.0000 |
kus | 266 | 10 | 1 | 10.0000 |
ole | 398 | 20 | 2 | 10.0000 |
Euroopa | 170 | 10 | 1 | 10.0000 |
saa | 123 | 8 | 1 | 8.0000 |
Kas | 88 | 7 | 1 | 7.0000 |
Aga | 131 | 7 | 1 | 7.0000 |
mille | 129 | 7 | 1 | 7.0000 |
Kui | 336 | 14 | 2 | 7.0000 |
Ta | 104 | 7 | 1 | 7.0000 |
mida | 275 | 14 | 2 | 7.0000 |
Me | 87 | 6 | 1 | 6.0000 |
teise | 61 | 5 | 1 | 5.0000 |
sest | 228 | 5 | 1 | 5.0000 |
kaasa | 101 | 5 | 1 | 5.0000 |
üle | 263 | 13 | 3 | 4.3333 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
aastal | 169 | 1 | 17 | 0.0588 |
mõttes | 45 | 1 | 6 | 0.1667 |
jooksul | 111 | 2 | 11 | 0.1818 |
saanud | 124 | 1 | 5 | 0.2000 |
võimalik | 109 | 1 | 5 | 0.2000 |
olukord | 47 | 1 | 5 | 0.2000 |
vähem | 67 | 1 | 5 | 0.2000 |
tegemist | 64 | 1 | 4 | 0.2500 |
Eestis | 149 | 1 | 4 | 0.2500 |
valitsus | 54 | 1 | 4 | 0.2500 |
sellele | 82 | 1 | 4 | 0.2500 |
praegu | 167 | 3 | 11 | 0.2727 |
ajal | 170 | 4 | 13 | 0.3077 |
siin | 76 | 1 | 3 | 0.3333 |
liige | 28 | 1 | 3 | 0.3333 |
täna | 98 | 1 | 3 | 0.3333 |
nädalat | 32 | 1 | 3 | 0.3333 |
valmis | 72 | 2 | 6 | 0.3333 |
sellega | 68 | 1 | 3 | 0.3333 |
mõned | 43 | 1 | 3 | 0.3333 |
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