Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Gli | 1010 | 95 | 1 | 95.0000 |
Il | 7572 | 529 | 8 | 66.1250 |
La | 6860 | 527 | 8 | 65.8750 |
Si | 1145 | 57 | 1 | 57.0000 |
Le | 2617 | 215 | 4 | 53.7500 |
Nel | 1371 | 96 | 2 | 48.0000 |
I | 3170 | 231 | 5 | 46.2000 |
Lo | 661 | 42 | 1 | 42.0000 |
Dal | 436 | 39 | 1 | 39.0000 |
Con | 1300 | 36 | 1 | 36.0000 |
Per | 2983 | 127 | 5 | 25.4000 |
Una | 1063 | 74 | 3 | 24.6667 |
In | 3359 | 144 | 6 | 24.0000 |
Di | 459 | 24 | 1 | 24.0000 |
Al | 528 | 24 | 1 | 24.0000 |
Un | 1446 | 115 | 5 | 23.0000 |
È | 1651 | 87 | 4 | 21.7500 |
Questa | 637 | 43 | 2 | 21.5000 |
Nella | 497 | 42 | 2 | 21.0000 |
Ciò | 350 | 19 | 1 | 19.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
dobbiamo | 124 | 1 | 11 | 0.0909 |
milioni | 488 | 4 | 42 | 0.0952 |
tipi | 203 | 1 | 10 | 0.1000 |
coloro | 181 | 1 | 10 | 0.1000 |
datore | 136 | 1 | 10 | 0.1000 |
lezione | 107 | 1 | 9 | 0.1111 |
istruttore | 83 | 1 | 9 | 0.1111 |
caratteristica | 63 | 1 | 8 | 0.1250 |
tipo | 561 | 3 | 24 | 0.1250 |
salita | 67 | 1 | 8 | 0.1250 |
regime | 54 | 1 | 8 | 0.1250 |
mancanza | 111 | 1 | 8 | 0.1250 |
vecchia | 93 | 1 | 7 | 0.1429 |
portafoglio | 47 | 1 | 7 | 0.1429 |
opportuno | 57 | 1 | 7 | 0.1429 |
margine | 56 | 1 | 7 | 0.1429 |
facciamo | 70 | 1 | 7 | 0.1429 |
dose | 49 | 1 | 7 | 0.1429 |
entrambe | 77 | 1 | 7 | 0.1429 |
incontrato | 42 | 1 | 7 | 0.1429 |
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