Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Den | 939 | 61 | 1 | 61.0000 |
Det | 2698 | 82 | 2 | 41.0000 |
De | 740 | 31 | 1 | 31.0000 |
I | 1263 | 91 | 3 | 30.3333 |
Detta | 540 | 25 | 1 | 25.0000 |
Du | 406 | 20 | 1 | 20.0000 |
Och | 408 | 18 | 1 | 18.0000 |
Hur | 209 | 16 | 1 | 16.0000 |
Så | 170 | 14 | 1 | 14.0000 |
Ett | 286 | 22 | 2 | 11.0000 |
När | 387 | 11 | 1 | 11.0000 |
Han | 201 | 10 | 1 | 10.0000 |
Om | 551 | 20 | 2 | 10.0000 |
Jag | 630 | 39 | 4 | 9.7500 |
Vi | 665 | 38 | 4 | 9.5000 |
Vad | 191 | 15 | 2 | 7.5000 |
Att | 283 | 20 | 3 | 6.6667 |
men | 1137 | 39 | 6 | 6.5000 |
medan | 120 | 6 | 1 | 6.0000 |
Med | 269 | 12 | 2 | 6.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
leda | 116 | 1 | 8 | 0.1250 |
tillgång | 103 | 1 | 8 | 0.1250 |
typ | 88 | 1 | 7 | 0.1429 |
nej | 66 | 1 | 7 | 0.1429 |
del | 525 | 4 | 26 | 0.1538 |
riktigt | 60 | 1 | 6 | 0.1667 |
mindre | 274 | 3 | 18 | 0.1667 |
hälften | 44 | 1 | 6 | 0.1667 |
grad | 57 | 1 | 5 | 0.2000 |
handla | 41 | 1 | 5 | 0.2000 |
amerikanska | 72 | 1 | 5 | 0.2000 |
framförallt | 50 | 1 | 5 | 0.2000 |
omfattande | 65 | 1 | 5 | 0.2000 |
övergripande | 38 | 1 | 5 | 0.2000 |
gränsöverskridande | 49 | 1 | 5 | 0.2000 |
såväl | 79 | 1 | 5 | 0.2000 |
gamla | 106 | 1 | 5 | 0.2000 |
lätt | 137 | 2 | 10 | 0.2000 |
typer | 60 | 1 | 5 | 0.2000 |
ihåg | 36 | 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