Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
A | 275 | 20 | 1 | 20.0000 |
En | 356 | 20 | 1 | 20.0000 |
Pour | 337 | 18 | 1 | 18.0000 |
La | 566 | 51 | 3 | 17.0000 |
Il | 651 | 32 | 2 | 16.0000 |
Le | 920 | 76 | 5 | 15.2000 |
C’est | 399 | 25 | 2 | 12.5000 |
Ce | 217 | 10 | 1 | 10.0000 |
Dans | 144 | 10 | 1 | 10.0000 |
Les | 546 | 37 | 4 | 9.2500 |
sous | 135 | 8 | 1 | 8.0000 |
Au | 169 | 15 | 2 | 7.5000 |
mais | 275 | 14 | 2 | 7.0000 |
la | 6878 | 496 | 73 | 6.7945 |
Faso | 412 | 13 | 2 | 6.5000 |
On | 152 | 12 | 2 | 6.0000 |
pendant | 71 | 5 | 1 | 5.0000 |
C'est | 45 | 5 | 1 | 5.0000 |
technique | 55 | 5 | 1 | 5.0000 |
travers | 114 | 10 | 2 | 5.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
autre | 84 | 1 | 8 | 0.1250 |
pouvoir | 124 | 1 | 8 | 0.1250 |
être | 270 | 3 | 21 | 0.1429 |
d’autres | 76 | 1 | 6 | 0.1667 |
clubs | 23 | 1 | 5 | 0.2000 |
point | 80 | 1 | 5 | 0.2000 |
état | 31 | 1 | 5 | 0.2000 |
régions | 39 | 1 | 5 | 0.2000 |
sûr | 27 | 1 | 5 | 0.2000 |
longtemps | 28 | 1 | 5 | 0.2000 |
football | 56 | 1 | 5 | 0.2000 |
seule | 36 | 1 | 4 | 0.2500 |
d’être | 74 | 1 | 4 | 0.2500 |
mon | 60 | 1 | 4 | 0.2500 |
durable | 41 | 1 | 4 | 0.2500 |
capacités | 36 | 1 | 4 | 0.2500 |
yeux | 19 | 1 | 4 | 0.2500 |
objectifs | 39 | 1 | 4 | 0.2500 |
conditions | 78 | 2 | 8 | 0.2500 |
me | 61 | 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