Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Ma | 577 | 38 | 1 | 38.0000 |
Questo | 432 | 22 | 1 | 22.0000 |
Si | 414 | 21 | 1 | 21.0000 |
ma | 1679 | 62 | 3 | 20.6667 |
La | 2002 | 184 | 10 | 18.4000 |
Mi | 220 | 18 | 1 | 18.0000 |
Un | 466 | 35 | 2 | 17.5000 |
Il | 1989 | 157 | 10 | 15.7000 |
Con | 209 | 15 | 1 | 15.0000 |
In | 860 | 59 | 4 | 14.7500 |
Non | 689 | 44 | 3 | 14.6667 |
Se | 417 | 29 | 2 | 14.5000 |
Di | 165 | 14 | 1 | 14.0000 |
I | 596 | 42 | 3 | 14.0000 |
Ho | 263 | 26 | 2 | 13.0000 |
Le | 526 | 38 | 3 | 12.6667 |
Nel | 324 | 23 | 2 | 11.5000 |
È | 320 | 21 | 2 | 10.5000 |
E | 1093 | 41 | 4 | 10.2500 |
190 | 10 | 1 | 10.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
ciò | 337 | 1 | 14 | 0.0714 |
tipo | 189 | 1 | 12 | 0.0833 |
Benedetto | 122 | 1 | 8 | 0.1250 |
capacità | 107 | 1 | 8 | 0.1250 |
on | 93 | 1 | 8 | 0.1250 |
cervello | 44 | 1 | 7 | 0.1429 |
materiali | 50 | 1 | 7 | 0.1429 |
bisogno | 167 | 2 | 14 | 0.1429 |
mancanza | 58 | 1 | 7 | 0.1429 |
domande | 65 | 1 | 7 | 0.1429 |
sera | 53 | 1 | 7 | 0.1429 |
gruppo | 118 | 1 | 6 | 0.1667 |
periodo | 125 | 2 | 12 | 0.1667 |
studi | 66 | 1 | 6 | 0.1667 |
pagina | 56 | 1 | 6 | 0.1667 |
concetto | 44 | 1 | 6 | 0.1667 |
voglia | 69 | 1 | 6 | 0.1667 |
crisi | 57 | 1 | 6 | 0.1667 |
60 | 40 | 1 | 6 | 0.1667 |
coloro | 127 | 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