Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Nella | 481 | 39 | 1 | 39.0000 |
close | 649 | 32 | 1 | 32.0000 |
Ma | 1572 | 63 | 2 | 31.5000 |
Anche | 703 | 30 | 1 | 30.0000 |
Da | 723 | 50 | 2 | 25.0000 |
Vi | 294 | 23 | 1 | 23.0000 |
E | 2225 | 75 | 4 | 18.7500 |
Alla | 284 | 17 | 1 | 17.0000 |
Ha | 239 | 17 | 1 | 17.0000 |
Un | 1397 | 109 | 7 | 15.5714 |
Una | 1059 | 75 | 5 | 15.0000 |
Dopo | 654 | 29 | 2 | 14.5000 |
Uniti | 253 | 14 | 1 | 14.0000 |
La | 7014 | 505 | 37 | 13.6486 |
Il | 7905 | 518 | 39 | 13.2821 |
Alberto | 192 | 26 | 2 | 13.0000 |
Si | 1262 | 73 | 6 | 12.1667 |
suo | 2251 | 205 | 17 | 12.0588 |
sua | 2299 | 204 | 17 | 12.0000 |
In | 3599 | 143 | 12 | 11.9167 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
attuali | 149 | 1 | 13 | 0.0769 |
collezione | 101 | 1 | 12 | 0.0833 |
louis | 125 | 1 | 11 | 0.0909 |
migliaia | 127 | 1 | 10 | 0.1000 |
voglia | 171 | 2 | 17 | 0.1176 |
collega | 81 | 1 | 8 | 0.1250 |
datore | 99 | 1 | 8 | 0.1250 |
pagato | 67 | 1 | 7 | 0.1429 |
centinaia | 80 | 1 | 7 | 0.1429 |
avesse | 85 | 1 | 7 | 0.1429 |
sostanza | 113 | 1 | 7 | 0.1429 |
materie | 80 | 1 | 7 | 0.1429 |
Content-Length | 395 | 1 | 7 | 0.1429 |
coloro | 188 | 1 | 7 | 0.1429 |
documentario | 58 | 1 | 7 | 0.1429 |
tutte | 1086 | 5 | 31 | 0.1613 |
palazzo | 53 | 1 | 6 | 0.1667 |
lana | 38 | 1 | 6 | 0.1667 |
raccolto | 64 | 1 | 6 | 0.1667 |
farà | 100 | 1 | 6 | 0.1667 |
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