Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
kad | 5432 | 215 | 3 | 71.6667 |
o | 1845 | 65 | 1 | 65.0000 |
bet | 1283 | 56 | 1 | 56.0000 |
O | 449 | 31 | 1 | 31.0000 |
tačiau | 712 | 29 | 1 | 29.0000 |
nes | 601 | 28 | 1 | 28.0000 |
Tai | 580 | 27 | 1 | 27.0000 |
jog | 565 | 22 | 1 | 22.0000 |
Po | 211 | 14 | 1 | 14.0000 |
Kai | 213 | 10 | 1 | 10.0000 |
Nuo | 143 | 9 | 1 | 9.0000 |
Į | 173 | 9 | 1 | 9.0000 |
kuri | 247 | 8 | 1 | 8.0000 |
teisės | 86 | 8 | 1 | 8.0000 |
Čia | 115 | 7 | 1 | 7.0000 |
pas | 121 | 7 | 1 | 7.0000 |
The | 43 | 7 | 1 | 7.0000 |
kurie | 582 | 20 | 3 | 6.6667 |
kuris | 349 | 13 | 2 | 6.5000 |
25 | 66 | 6 | 1 | 6.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
proc | 260 | 1 | 24 | 0.0417 |
tūkst | 271 | 1 | 23 | 0.0435 |
val | 167 | 1 | 18 | 0.0556 |
duomenimis | 135 | 1 | 14 | 0.0714 |
mln | 167 | 1 | 12 | 0.0833 |
mėnesį | 183 | 2 | 20 | 0.1000 |
pradžioje | 91 | 1 | 9 | 0.1111 |
vakarą | 93 | 2 | 15 | 0.1333 |
kuriuos | 124 | 1 | 7 | 0.1429 |
išvengti | 65 | 1 | 7 | 0.1429 |
laiką | 181 | 2 | 13 | 0.1538 |
tuomet | 97 | 1 | 6 | 0.1667 |
trys | 81 | 1 | 6 | 0.1667 |
padeda | 83 | 1 | 6 | 0.1667 |
viską | 146 | 1 | 6 | 0.1667 |
suprasti | 70 | 1 | 6 | 0.1667 |
didelis | 72 | 1 | 6 | 0.1667 |
dienas | 65 | 1 | 6 | 0.1667 |
departamentas | 34 | 1 | 6 | 0.1667 |
dienomis | 80 | 1 | 6 | 0.1667 |
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