Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Si | 1661 | 71 | 1 | 71.0000 |
Nella | 731 | 60 | 1 | 60.0000 |
Questa | 605 | 51 | 1 | 51.0000 |
Lo | 517 | 38 | 1 | 38.0000 |
close | 1144 | 35 | 1 | 35.0000 |
Se | 1358 | 58 | 2 | 29.0000 |
Gli | 1250 | 84 | 3 | 28.0000 |
trova | 3087 | 55 | 2 | 27.5000 |
Siamo | 251 | 23 | 1 | 23.0000 |
Vi | 685 | 45 | 2 | 22.5000 |
Un | 710 | 66 | 3 | 22.0000 |
Una | 716 | 44 | 2 | 22.0000 |
trovano | 902 | 44 | 2 | 22.0000 |
Quando | 374 | 21 | 1 | 21.0000 |
Le | 2679 | 206 | 10 | 20.6000 |
Ci | 444 | 19 | 1 | 19.0000 |
Questo | 992 | 55 | 3 | 18.3333 |
Ma | 279 | 18 | 1 | 18.0000 |
Qui | 313 | 17 | 1 | 17.0000 |
Con | 942 | 33 | 2 | 16.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Tipologia | 136 | 1 | 14 | 0.0714 |
sec | 152 | 1 | 13 | 0.0769 |
m⊃2 | 225 | 1 | 12 | 0.0833 |
m2 | 1483 | 10 | 110 | 0.0909 |
coloro | 289 | 1 | 11 | 0.0909 |
conseguenze | 70 | 1 | 9 | 0.1111 |
seduta | 159 | 2 | 18 | 0.1111 |
dispone | 759 | 4 | 34 | 0.1176 |
Smjestaj | 80 | 2 | 17 | 0.1176 |
Canale | 72 | 1 | 8 | 0.1250 |
1947 | 33 | 1 | 7 | 0.1429 |
brevi | 81 | 1 | 7 | 0.1429 |
sostanza | 49 | 1 | 7 | 0.1429 |
insediamenti | 50 | 1 | 7 | 0.1429 |
Sv | 108 | 1 | 7 | 0.1429 |
disturbo | 38 | 1 | 7 | 0.1429 |
corridoio | 189 | 1 | 7 | 0.1429 |
prof | 83 | 1 | 7 | 0.1429 |
mucosa | 25 | 1 | 6 | 0.1667 |
Spirito | 58 | 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