Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Under | 752 | 55 | 1 | 55.0000 |
Han | 690 | 47 | 1 | 47.0000 |
Och | 1403 | 47 | 1 | 47.0000 |
Den | 3059 | 186 | 4 | 46.5000 |
Så | 604 | 37 | 1 | 37.0000 |
Du | 1256 | 34 | 1 | 34.0000 |
Som | 594 | 33 | 1 | 33.0000 |
Det | 9198 | 159 | 5 | 31.8000 |
Men | 1816 | 61 | 2 | 30.5000 |
Med | 810 | 28 | 1 | 28.0000 |
De | 2673 | 121 | 5 | 24.2000 |
Sedan | 315 | 24 | 1 | 24.0000 |
Här | 453 | 22 | 1 | 22.0000 |
I | 4034 | 208 | 10 | 20.8000 |
På | 952 | 62 | 3 | 20.6667 |
Detta | 1614 | 59 | 3 | 19.6667 |
Enligt | 379 | 19 | 1 | 19.0000 |
Vid | 375 | 18 | 1 | 18.0000 |
Projektet | 306 | 18 | 1 | 18.0000 |
Denna | 509 | 34 | 2 | 17.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
tillsammans | 762 | 1 | 25 | 0.0400 |
typer | 197 | 1 | 12 | 0.0833 |
tillgång | 412 | 2 | 21 | 0.0952 |
möjlighet | 619 | 3 | 31 | 0.0968 |
delen | 256 | 2 | 19 | 0.1053 |
moderna | 115 | 1 | 9 | 0.1111 |
vikten | 144 | 1 | 9 | 0.1111 |
handla | 148 | 1 | 8 | 0.1250 |
vanligaste | 47 | 1 | 8 | 0.1250 |
chansen | 82 | 1 | 8 | 0.1250 |
syn | 174 | 1 | 8 | 0.1250 |
risk | 182 | 2 | 16 | 0.1250 |
grund | 755 | 2 | 16 | 0.1250 |
svårt | 417 | 3 | 22 | 0.1364 |
finska | 90 | 1 | 7 | 0.1429 |
konstatera | 62 | 1 | 7 | 0.1429 |
denne | 71 | 1 | 7 | 0.1429 |
resten | 103 | 1 | 7 | 0.1429 |
krav | 714 | 7 | 49 | 0.1429 |
fredag | 49 | 1 | 7 | 0.1429 |
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