Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
kad | 54328 | 1439 | 4 | 359.7500 |
tačiau | 7286 | 256 | 1 | 256.0000 |
nes | 6377 | 242 | 1 | 242.0000 |
jog | 5792 | 236 | 1 | 236.0000 |
o | 17386 | 441 | 3 | 147.0000 |
Ji | 1809 | 110 | 1 | 110.0000 |
tad | 2096 | 105 | 1 | 105.0000 |
Į | 1743 | 103 | 1 | 103.0000 |
Jie | 1653 | 102 | 1 | 102.0000 |
Nors | 2072 | 94 | 1 | 94.0000 |
Šis | 1164 | 80 | 1 | 80.0000 |
kurie | 5848 | 316 | 4 | 79.0000 |
Kaip | 3195 | 152 | 2 | 76.0000 |
Po | 2218 | 148 | 2 | 74.0000 |
kuri | 2664 | 147 | 2 | 73.5000 |
Ir | 3581 | 138 | 2 | 69.0000 |
Ši | 925 | 69 | 1 | 69.0000 |
O | 4472 | 136 | 2 | 68.0000 |
Jau | 1018 | 65 | 1 | 65.0000 |
Tad | 1101 | 60 | 1 | 60.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
teigimu | 1831 | 1 | 90 | 0.0111 |
tūkst | 2682 | 2 | 148 | 0.0135 |
proc | 2531 | 2 | 137 | 0.0146 |
mln | 1779 | 2 | 126 | 0.0159 |
val | 1447 | 1 | 63 | 0.0159 |
mlrd | 442 | 1 | 45 | 0.0222 |
šiek | 837 | 1 | 31 | 0.0323 |
duomenimis | 1105 | 3 | 81 | 0.0370 |
min | 235 | 1 | 20 | 0.0500 |
nustatė | 242 | 1 | 19 | 0.0526 |
dr | 332 | 1 | 18 | 0.0556 |
1-ąją | 117 | 1 | 17 | 0.0588 |
manau | 567 | 1 | 16 | 0.0625 |
d | 246 | 2 | 30 | 0.0667 |
pašnekovė | 188 | 1 | 15 | 0.0667 |
nuomone | 639 | 2 | 29 | 0.0690 |
manymu | 363 | 1 | 14 | 0.0714 |
atkreipia | 197 | 1 | 14 | 0.0714 |
pašnekovas | 197 | 1 | 14 | 0.0714 |
akivaizdu | 173 | 1 | 14 | 0.0714 |
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