Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
El | 1120 | 85 | 4 | 21.2500 |
La | 908 | 80 | 4 | 20.0000 |
A | 197 | 19 | 1 | 19.0000 |
En | 663 | 39 | 3 | 13.0000 |
Y | 181 | 12 | 1 | 12.0000 |
pero | 416 | 24 | 2 | 12.0000 |
Con | 123 | 10 | 1 | 10.0000 |
según | 118 | 9 | 1 | 9.0000 |
Los | 334 | 18 | 2 | 9.0000 |
Un | 103 | 8 | 1 | 8.0000 |
la | 8328 | 683 | 94 | 7.2660 |
Según | 96 | 7 | 1 | 7.0000 |
No | 231 | 19 | 3 | 6.3333 |
los | 3383 | 286 | 47 | 6.0851 |
sociales | 106 | 6 | 1 | 6.0000 |
Así | 94 | 6 | 1 | 6.0000 |
Desde | 84 | 6 | 1 | 6.0000 |
tras | 137 | 6 | 1 | 6.0000 |
expresó | 39 | 6 | 1 | 6.0000 |
las | 2486 | 192 | 34 | 5.6471 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
total | 118 | 1 | 8 | 0.1250 |
tipo | 88 | 1 | 7 | 0.1429 |
estar | 74 | 1 | 7 | 0.1429 |
siendo | 56 | 1 | 6 | 0.1667 |
va | 126 | 1 | 6 | 0.1667 |
poder | 118 | 1 | 6 | 0.1667 |
uno | 157 | 1 | 6 | 0.1667 |
parte | 259 | 3 | 18 | 0.1667 |
años | 402 | 9 | 47 | 0.1915 |
respuesta | 44 | 1 | 5 | 0.2000 |
empresa | 86 | 1 | 5 | 0.2000 |
Estados | 98 | 1 | 5 | 0.2000 |
número | 86 | 1 | 5 | 0.2000 |
nuestra | 74 | 1 | 5 | 0.2000 |
millones | 245 | 3 | 15 | 0.2000 |
servicio | 59 | 1 | 5 | 0.2000 |
Donald | 25 | 1 | 5 | 0.2000 |
mayores | 69 | 1 | 5 | 0.2000 |
hecho | 138 | 2 | 9 | 0.2222 |
fin | 87 | 2 | 9 | 0.2222 |
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