Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Et | 468 | 31 | 1 | 31.0000 |
Il | 1305 | 63 | 3 | 21.0000 |
De | 318 | 20 | 1 | 20.0000 |
C'est | 305 | 19 | 1 | 19.0000 |
close | 202 | 18 | 1 | 18.0000 |
C’est | 264 | 18 | 1 | 18.0000 |
En | 959 | 70 | 4 | 17.5000 |
A | 497 | 34 | 2 | 17.0000 |
Le | 2329 | 192 | 12 | 16.0000 |
La | 1953 | 175 | 11 | 15.9091 |
Nous | 605 | 30 | 2 | 15.0000 |
mais | 1109 | 42 | 3 | 14.0000 |
Ce | 595 | 41 | 3 | 13.6667 |
Comme | 131 | 13 | 1 | 13.0000 |
Cette | 371 | 26 | 2 | 13.0000 |
Je | 645 | 37 | 3 | 12.3333 |
Tout | 125 | 12 | 1 | 12.0000 |
Les | 1897 | 143 | 12 | 11.9167 |
Si | 367 | 22 | 2 | 11.0000 |
On | 377 | 21 | 2 | 10.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Sep | 185 | 1 | 15 | 0.0667 |
processus | 89 | 1 | 9 | 0.1111 |
mode | 76 | 1 | 8 | 0.1250 |
maladie | 80 | 1 | 8 | 0.1250 |
gros | 64 | 1 | 7 | 0.1429 |
Connection | 224 | 1 | 6 | 0.1667 |
délai | 41 | 1 | 6 | 0.1667 |
marque | 110 | 1 | 6 | 0.1667 |
parcours | 72 | 1 | 6 | 0.1667 |
type | 110 | 1 | 6 | 0.1667 |
phase | 45 | 1 | 6 | 0.1667 |
série | 76 | 1 | 6 | 0.1667 |
filles | 43 | 1 | 6 | 0.1667 |
programme | 142 | 2 | 11 | 0.1818 |
faudrait | 35 | 1 | 5 | 0.2000 |
modèles | 66 | 1 | 5 | 0.2000 |
dossier | 50 | 1 | 5 | 0.2000 |
véhicule | 57 | 1 | 5 | 0.2000 |
grandes | 73 | 1 | 5 | 0.2000 |
consultation | 31 | 1 | 5 | 0.2000 |
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