Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Le | 894 | 73 | 5 | 14.6000 |
Il | 391 | 27 | 2 | 13.5000 |
Les | 677 | 37 | 3 | 12.3333 |
En | 308 | 20 | 2 | 10.0000 |
mais | 291 | 19 | 2 | 9.5000 |
La | 671 | 53 | 6 | 8.8333 |
C'est | 97 | 8 | 1 | 8.0000 |
vitesse | 86 | 8 | 1 | 8.0000 |
close | 174 | 7 | 1 | 7.0000 |
C’est | 84 | 6 | 1 | 6.0000 |
la | 5283 | 450 | 80 | 5.6250 |
autour | 39 | 5 | 1 | 5.0000 |
selon | 89 | 5 | 1 | 5.0000 |
Cette | 140 | 5 | 1 | 5.0000 |
Sur | 45 | 5 | 1 | 5.0000 |
max | 27 | 5 | 1 | 5.0000 |
euros | 34 | 4 | 1 | 4.0000 |
Votre | 58 | 4 | 1 | 4.0000 |
bénéficier | 18 | 4 | 1 | 4.0000 |
car | 50 | 4 | 1 | 4.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
être | 348 | 2 | 25 | 0.0800 |
taux | 64 | 1 | 7 | 0.1429 |
réseau | 53 | 1 | 7 | 0.1429 |
nombre | 88 | 1 | 7 | 0.1429 |
Aug | 85 | 1 | 7 | 0.1429 |
maison | 63 | 1 | 6 | 0.1667 |
Date | 115 | 1 | 6 | 0.1667 |
pouvoir | 56 | 1 | 6 | 0.1667 |
voir | 62 | 1 | 6 | 0.1667 |
couleurs | 33 | 1 | 5 | 0.2000 |
véhicules | 30 | 1 | 5 | 0.2000 |
traitement | 42 | 1 | 5 | 0.2000 |
Ici | 69 | 1 | 5 | 0.2000 |
but | 37 | 1 | 5 | 0.2000 |
plusieurs | 57 | 1 | 5 | 0.2000 |
projet | 119 | 2 | 10 | 0.2000 |
pu | 32 | 1 | 5 | 0.2000 |
découvrir | 32 | 1 | 5 | 0.2000 |
ceux | 48 | 1 | 5 | 0.2000 |
version | 35 | 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