Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Il | 1468 | 63 | 1 | 63.0000 |
Pour | 729 | 46 | 1 | 46.0000 |
En | 838 | 70 | 2 | 35.0000 |
La | 1887 | 161 | 5 | 32.2000 |
Le | 2405 | 206 | 7 | 29.4286 |
Cette | 396 | 26 | 1 | 26.0000 |
Au | 240 | 25 | 1 | 25.0000 |
Vous | 450 | 24 | 1 | 24.0000 |
Je | 663 | 39 | 2 | 19.5000 |
car | 271 | 16 | 1 | 16.0000 |
mais | 1030 | 43 | 3 | 14.3333 |
Ils | 261 | 14 | 1 | 14.0000 |
A | 521 | 28 | 2 | 14.0000 |
Les | 1813 | 124 | 9 | 13.7778 |
De | 365 | 25 | 2 | 12.5000 |
Un | 599 | 37 | 3 | 12.3333 |
Comme | 135 | 12 | 1 | 12.0000 |
Tout | 140 | 11 | 1 | 11.0000 |
C'est | 455 | 22 | 2 | 11.0000 |
Nous | 607 | 33 | 3 | 11.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
nouveaux | 109 | 1 | 8 | 0.1250 |
contrat | 76 | 1 | 7 | 0.1429 |
cheval | 41 | 1 | 7 | 0.1429 |
processus | 48 | 1 | 7 | 0.1429 |
serait | 72 | 1 | 7 | 0.1429 |
équipe | 46 | 1 | 6 | 0.1667 |
Pierre | 77 | 1 | 6 | 0.1667 |
parc | 39 | 1 | 6 | 0.1667 |
rythme | 48 | 1 | 6 | 0.1667 |
dispositif | 35 | 1 | 6 | 0.1667 |
niveaux | 42 | 1 | 6 | 0.1667 |
tenir | 44 | 1 | 6 | 0.1667 |
types | 54 | 1 | 6 | 0.1667 |
univers | 40 | 1 | 6 | 0.1667 |
gros | 68 | 1 | 6 | 0.1667 |
logement | 28 | 1 | 6 | 0.1667 |
véhicule | 40 | 1 | 5 | 0.2000 |
moyens | 83 | 1 | 5 | 0.2000 |
joli | 32 | 1 | 5 | 0.2000 |
comportement | 39 | 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