Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
který | 3907 | 135 | 1 | 135.0000 |
Podle | 1890 | 125 | 1 | 125.0000 |
která | 2643 | 105 | 1 | 105.0000 |
V | 4607 | 276 | 3 | 92.0000 |
kde | 1890 | 66 | 1 | 66.0000 |
že | 15188 | 388 | 6 | 64.6667 |
kteří | 1458 | 58 | 1 | 58.0000 |
Po | 1069 | 95 | 2 | 47.5000 |
Ale | 809 | 60 | 2 | 30.0000 |
takže | 449 | 26 | 1 | 26.0000 |
A | 2188 | 88 | 4 | 22.0000 |
Na | 2641 | 148 | 7 | 21.1429 |
Ten | 404 | 21 | 1 | 21.0000 |
Před | 313 | 21 | 1 | 21.0000 |
aby | 2512 | 81 | 4 | 20.2500 |
protože | 1198 | 59 | 3 | 19.6667 |
Ve | 887 | 58 | 3 | 19.3333 |
První | 317 | 18 | 1 | 18.0000 |
které | 4048 | 137 | 8 | 17.1250 |
Pokud | 737 | 34 | 2 | 17.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
údajně | 216 | 1 | 16 | 0.0625 |
léta | 167 | 1 | 14 | 0.0714 |
play | 131 | 1 | 12 | 0.0833 |
°C | 141 | 2 | 24 | 0.0833 |
návštěvě | 79 | 1 | 10 | 0.1000 |
Pásma | 97 | 1 | 10 | 0.1000 |
života | 285 | 3 | 25 | 0.1200 |
chvíle | 93 | 1 | 8 | 0.1250 |
ředitele | 110 | 1 | 8 | 0.1250 |
týdnech | 105 | 1 | 8 | 0.1250 |
podpoře | 68 | 1 | 8 | 0.1250 |
skupině | 110 | 1 | 8 | 0.1250 |
fakt | 147 | 1 | 8 | 0.1250 |
dovolit | 56 | 1 | 8 | 0.1250 |
dobou | 61 | 1 | 8 | 0.1250 |
zajistit | 105 | 1 | 8 | 0.1250 |
míry | 87 | 1 | 7 | 0.1429 |
reprezentace | 84 | 1 | 7 | 0.1429 |
týdnu | 96 | 1 | 7 | 0.1429 |
Andreje | 96 | 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