Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Il | 9901 | 365 | 3 | 121.6667 |
Ce | 1708 | 110 | 1 | 110.0000 |
Les | 6683 | 548 | 5 | 109.6000 |
Le | 9539 | 607 | 8 | 75.8750 |
À | 1861 | 74 | 1 | 74.0000 |
La | 7830 | 530 | 9 | 58.8889 |
Pour | 985 | 51 | 1 | 51.0000 |
Une | 1385 | 93 | 2 | 46.5000 |
Un | 1344 | 86 | 2 | 43.0000 |
On | 1206 | 53 | 2 | 26.5000 |
C’est | 271 | 16 | 1 | 16.0000 |
afin | 718 | 24 | 2 | 12.0000 |
Tout | 210 | 12 | 1 | 12.0000 |
car | 950 | 22 | 2 | 11.0000 |
Et | 169 | 11 | 1 | 11.0000 |
Quand | 173 | 10 | 1 | 10.0000 |
puisque | 139 | 10 | 1 | 10.0000 |
jusqu'au | 137 | 10 | 1 | 10.0000 |
été | 4907 | 408 | 42 | 9.7143 |
Si | 381 | 19 | 2 | 9.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
d | 2325 | 2 | 164 | 0.0122 |
l | 6424 | 2 | 53 | 0.0377 |
d’autres | 113 | 1 | 10 | 0.1000 |
bientôt | 99 | 1 | 9 | 0.1111 |
instruments | 83 | 1 | 8 | 0.1250 |
the | 167 | 1 | 8 | 0.1250 |
Marie | 146 | 1 | 8 | 0.1250 |
documentaire | 59 | 1 | 8 | 0.1250 |
cultures | 86 | 1 | 7 | 0.1429 |
faux | 62 | 1 | 7 | 0.1429 |
exemples | 50 | 1 | 7 | 0.1429 |
cercle | 79 | 1 | 7 | 0.1429 |
d'entre | 135 | 2 | 13 | 0.1538 |
virus | 58 | 1 | 6 | 0.1667 |
difficulté | 77 | 1 | 6 | 0.1667 |
race | 71 | 1 | 6 | 0.1667 |
dépôt | 52 | 1 | 6 | 0.1667 |
bloc | 37 | 1 | 6 | 0.1667 |
décor | 68 | 1 | 6 | 0.1667 |
Centre | 141 | 1 | 6 | 0.1667 |
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