Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Il | 532 | 33 | 1 | 33.0000 |
mais | 342 | 19 | 1 | 19.0000 |
Les | 735 | 53 | 4 | 13.2500 |
Cette | 175 | 13 | 1 | 13.0000 |
Le | 915 | 74 | 6 | 12.3333 |
A | 115 | 12 | 1 | 12.0000 |
Pour | 229 | 11 | 1 | 11.0000 |
La | 773 | 72 | 7 | 10.2857 |
Au | 100 | 10 | 1 | 10.0000 |
En | 355 | 30 | 3 | 10.0000 |
depuis | 150 | 9 | 1 | 9.0000 |
Nous | 218 | 15 | 2 | 7.5000 |
la | 6278 | 490 | 78 | 6.2821 |
car | 67 | 5 | 1 | 5.0000 |
selon | 69 | 5 | 1 | 5.0000 |
particulier | 54 | 5 | 1 | 5.0000 |
permettant | 32 | 5 | 1 | 5.0000 |
qu'il | 89 | 9 | 2 | 4.5000 |
C'est | 104 | 8 | 2 | 4.0000 |
Si | 127 | 8 | 2 | 4.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
processus | 63 | 1 | 8 | 0.1250 |
parti | 63 | 1 | 7 | 0.1429 |
décision | 58 | 1 | 6 | 0.1667 |
nombre | 66 | 1 | 6 | 0.1667 |
forme | 59 | 1 | 6 | 0.1667 |
fort | 39 | 1 | 5 | 0.2000 |
problèmes | 38 | 1 | 5 | 0.2000 |
pratiques | 52 | 1 | 5 | 0.2000 |
seul | 59 | 1 | 5 | 0.2000 |
jeu | 39 | 1 | 5 | 0.2000 |
partenariat | 41 | 1 | 5 | 0.2000 |
principe | 44 | 1 | 5 | 0.2000 |
taux | 42 | 1 | 5 | 0.2000 |
présent | 38 | 1 | 5 | 0.2000 |
texte | 42 | 1 | 5 | 0.2000 |
modèle | 35 | 1 | 5 | 0.2000 |
titre | 34 | 1 | 5 | 0.2000 |
seule | 48 | 1 | 5 | 0.2000 |
différentes | 36 | 1 | 4 | 0.2500 |
changement | 45 | 1 | 4 | 0.2500 |
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