Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
n | 3830 | 209 | 5 | 41.8000 |
En | 800 | 41 | 2 | 20.5000 |
Die | 4237 | 230 | 12 | 19.1667 |
So | 240 | 19 | 1 | 19.0000 |
Ons | 976 | 52 | 3 | 17.3333 |
Vir | 228 | 16 | 1 | 16.0000 |
Dit | 1620 | 46 | 3 | 15.3333 |
Maar | 595 | 28 | 2 | 14.0000 |
Met | 189 | 14 | 1 | 14.0000 |
In | 756 | 34 | 3 | 11.3333 |
Toe | 188 | 11 | 1 | 11.0000 |
Op | 216 | 10 | 1 | 10.0000 |
Hier | 139 | 10 | 1 | 10.0000 |
Wanneer | 131 | 10 | 1 | 10.0000 |
As | 525 | 19 | 2 | 9.5000 |
Hulle | 460 | 19 | 2 | 9.5000 |
Nie | 86 | 9 | 1 | 9.0000 |
Hierdie | 420 | 18 | 2 | 9.0000 |
Dis | 386 | 26 | 3 | 8.6667 |
die | 31794 | 1664 | 197 | 8.4467 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
groter | 129 | 1 | 11 | 0.0909 |
pragtige | 70 | 1 | 10 | 0.1000 |
soort | 92 | 1 | 9 | 0.1111 |
saam | 585 | 3 | 22 | 0.1364 |
mnr | 66 | 1 | 7 | 0.1429 |
aantal | 58 | 1 | 7 | 0.1429 |
kop | 121 | 1 | 6 | 0.1667 |
wonderlike | 68 | 1 | 6 | 0.1667 |
daai | 120 | 1 | 6 | 0.1667 |
diegene | 55 | 1 | 6 | 0.1667 |
gewone | 67 | 1 | 6 | 0.1667 |
album | 38 | 1 | 6 | 0.1667 |
oomblik | 112 | 2 | 12 | 0.1667 |
verdere | 53 | 1 | 6 | 0.1667 |
omtrent | 71 | 1 | 5 | 0.2000 |
duidelike | 34 | 1 | 5 | 0.2000 |
2009 | 75 | 1 | 5 | 0.2000 |
Sondag | 71 | 1 | 5 | 0.2000 |
hulself | 53 | 1 | 5 | 0.2000 |
algemene | 65 | 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