Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Det | 617 | 28 | 1 | 28.0000 |
I | 668 | 45 | 2 | 22.5000 |
Den | 455 | 30 | 2 | 15.0000 |
En | 312 | 15 | 1 | 15.0000 |
Da | 167 | 8 | 1 | 8.0000 |
mens | 142 | 7 | 1 | 7.0000 |
men | 413 | 20 | 3 | 6.6667 |
hvor | 125 | 6 | 1 | 6.0000 |
Flere | 72 | 5 | 1 | 5.0000 |
skriver | 70 | 5 | 1 | 5.0000 |
Til | 65 | 5 | 1 | 5.0000 |
fordi | 58 | 5 | 1 | 5.0000 |
York | 30 | 4 | 1 | 4.0000 |
igjen | 95 | 4 | 1 | 4.0000 |
finansminister | 15 | 4 | 1 | 4.0000 |
Stavanger | 33 | 4 | 1 | 4.0000 |
møttes | 36 | 4 | 1 | 4.0000 |
hjemme | 35 | 4 | 1 | 4.0000 |
Her | 73 | 4 | 1 | 4.0000 |
fly | 32 | 4 | 1 | 4.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
norske | 131 | 1 | 7 | 0.1429 |
2018 | 107 | 2 | 11 | 0.1818 |
år | 680 | 12 | 63 | 0.1905 |
rammet | 68 | 1 | 5 | 0.2000 |
ingen | 46 | 1 | 5 | 0.2000 |
gamle | 65 | 1 | 5 | 0.2000 |
norsk | 76 | 1 | 5 | 0.2000 |
begge | 51 | 1 | 5 | 0.2000 |
både | 104 | 1 | 5 | 0.2000 |
slutten | 71 | 1 | 5 | 0.2000 |
høyeste | 32 | 1 | 4 | 0.2500 |
imidlertid | 38 | 1 | 4 | 0.2500 |
partiet | 45 | 1 | 4 | 0.2500 |
Times | 20 | 1 | 4 | 0.2500 |
være | 109 | 2 | 8 | 0.2500 |
begynnelsen | 35 | 1 | 4 | 0.2500 |
annen | 42 | 1 | 4 | 0.2500 |
juli | 71 | 2 | 8 | 0.2500 |
bli | 106 | 2 | 7 | 0.2857 |
største | 121 | 2 | 7 | 0.2857 |
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