Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
No | 279 | 24 | 2 | 12.0000 |
Em | 266 | 33 | 3 | 11.0000 |
Na | 98 | 11 | 1 | 11.0000 |
3. | 127 | 10 | 1 | 10.0000 |
Kong | 293 | 10 | 1 | 10.0000 |
Os | 295 | 28 | 3 | 9.3333 |
O | 1135 | 92 | 11 | 8.3636 |
A | 932 | 83 | 11 | 7.5455 |
cento | 414 | 15 | 2 | 7.5000 |
Legislativa | 124 | 7 | 1 | 7.0000 |
numa | 70 | 6 | 1 | 6.0000 |
América | 28 | 6 | 1 | 6.0000 |
mas | 127 | 6 | 1 | 6.0000 |
prisão | 50 | 6 | 1 | 6.0000 |
teve | 34 | 5 | 1 | 5.0000 |
através | 74 | 5 | 1 | 5.0000 |
Patacas | 96 | 5 | 1 | 5.0000 |
sendo | 91 | 5 | 1 | 5.0000 |
Sul | 39 | 5 | 1 | 5.0000 |
informática | 62 | 5 | 1 | 5.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
milhões | 656 | 2 | 27 | 0.0741 |
Hong | 313 | 1 | 10 | 0.1000 |
andar | 70 | 1 | 8 | 0.1250 |
cerca | 165 | 1 | 8 | 0.1250 |
Chefe | 669 | 1 | 7 | 0.1429 |
escalão | 102 | 1 | 7 | 0.1429 |
Universidade | 130 | 1 | 6 | 0.1667 |
anual | 47 | 1 | 6 | 0.1667 |
Direcção | 719 | 1 | 5 | 0.2000 |
prova | 36 | 1 | 5 | 0.2000 |
CCAC | 34 | 1 | 5 | 0.2000 |
cerimónia | 30 | 1 | 5 | 0.2000 |
financeiro | 126 | 1 | 5 | 0.2000 |
Secretário | 400 | 1 | 5 | 0.2000 |
acordo | 181 | 1 | 5 | 0.2000 |
áreas | 41 | 1 | 5 | 0.2000 |
máximo | 35 | 1 | 5 | 0.2000 |
vagas | 249 | 2 | 9 | 0.2222 |
Dr | 89 | 1 | 4 | 0.2500 |
exposição | 28 | 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