Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
År | 2215 | 171 | 1 | 171.0000 |
Hon | 3418 | 148 | 1 | 148.0000 |
Han | 12064 | 441 | 3 | 147.0000 |
Efter | 3937 | 124 | 1 | 124.0000 |
Vid | 2462 | 113 | 1 | 113.0000 |
Denna | 1985 | 106 | 1 | 106.0000 |
När | 3006 | 87 | 1 | 87.0000 |
På | 3689 | 164 | 2 | 82.0000 |
Som | 1363 | 76 | 1 | 76.0000 |
Sedan | 1218 | 73 | 1 | 73.0000 |
Från | 1431 | 71 | 1 | 71.0000 |
Dessa | 1454 | 69 | 1 | 69.0000 |
För | 2681 | 61 | 1 | 61.0000 |
Där | 726 | 47 | 1 | 47.0000 |
Under | 4563 | 177 | 4 | 44.2500 |
Många | 965 | 43 | 1 | 43.0000 |
Med | 1393 | 43 | 1 | 43.0000 |
Då | 1111 | 43 | 1 | 43.0000 |
Till | 1491 | 43 | 1 | 43.0000 |
Den | 12616 | 550 | 13 | 42.3077 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
bl.a | 1641 | 1 | 53 | 0.0189 |
sig. | 628 | 1 | 51 | 0.0196 |
f.Kr | 716 | 2 | 63 | 0.0317 |
med. | 301 | 1 | 23 | 0.0435 |
rollen | 339 | 1 | 19 | 0.0526 |
resten | 316 | 1 | 17 | 0.0588 |
intresserad | 155 | 1 | 17 | 0.0588 |
beroende | 640 | 2 | 31 | 0.0645 |
kommun | 3618 | 19 | 292 | 0.0651 |
del. | 143 | 1 | 15 | 0.0667 |
medlem | 776 | 2 | 28 | 0.0714 |
bestående | 436 | 1 | 13 | 0.0769 |
delarna | 377 | 2 | 25 | 0.0800 |
s.k | 612 | 1 | 12 | 0.0833 |
tvungen | 221 | 1 | 12 | 0.0833 |
ledamot | 477 | 2 | 24 | 0.0833 |
tillsammans | 3051 | 8 | 95 | 0.0842 |
visserligen | 142 | 1 | 10 | 0.1000 |
typer | 448 | 2 | 20 | 0.1000 |
av. | 159 | 1 | 10 | 0.1000 |
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