Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
baina | 1896 | 90 | 3 | 30.0000 |
Azken | 222 | 19 | 1 | 19.0000 |
Herriko | 257 | 17 | 1 | 17.0000 |
Gaur | 172 | 12 | 1 | 12.0000 |
Duela | 76 | 11 | 1 | 11.0000 |
Goierri | 202 | 21 | 2 | 10.5000 |
Datorren | 77 | 10 | 1 | 10.0000 |
zein | 213 | 10 | 1 | 10.0000 |
Lehen | 166 | 10 | 1 | 10.0000 |
festa | 122 | 9 | 1 | 9.0000 |
Beste | 139 | 9 | 1 | 9.0000 |
Zer | 77 | 8 | 1 | 8.0000 |
noiz | 81 | 8 | 1 | 8.0000 |
urriaren | 81 | 8 | 1 | 8.0000 |
Gipuzkoako | 351 | 24 | 3 | 8.0000 |
Hala | 315 | 8 | 1 | 8.0000 |
Orain | 230 | 8 | 1 | 8.0000 |
Koronabirusaren | 93 | 8 | 1 | 8.0000 |
Baina | 384 | 15 | 2 | 7.5000 |
Zumarragako | 260 | 15 | 2 | 7.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
arren | 210 | 1 | 18 | 0.0556 |
duenez | 194 | 1 | 11 | 0.0909 |
arabera | 363 | 2 | 21 | 0.0952 |
zioten | 89 | 1 | 10 | 0.1000 |
zaie | 116 | 1 | 10 | 0.1000 |
daitezke | 153 | 2 | 16 | 0.1250 |
dutelako | 66 | 1 | 7 | 0.1429 |
asteetan | 57 | 1 | 6 | 0.1667 |
gain | 401 | 3 | 18 | 0.1667 |
dauka | 46 | 1 | 6 | 0.1667 |
bateko | 69 | 1 | 6 | 0.1667 |
zion | 74 | 1 | 6 | 0.1667 |
taldea | 123 | 1 | 6 | 0.1667 |
genuen | 376 | 5 | 27 | 0.1852 |
bezperan | 28 | 1 | 5 | 0.2000 |
zirela | 57 | 1 | 5 | 0.2000 |
osoan | 107 | 2 | 10 | 0.2000 |
5ean | 24 | 1 | 5 | 0.2000 |
erabilera | 68 | 1 | 5 | 0.2000 |
barruan | 165 | 3 | 15 | 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