Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
která | 2128 | 86 | 1 | 86.0000 |
kteří | 1306 | 55 | 1 | 55.0000 |
který | 2847 | 102 | 2 | 51.0000 |
Po | 1240 | 92 | 2 | 46.0000 |
Pro | 1171 | 89 | 2 | 44.5000 |
Je | 1523 | 71 | 2 | 35.5000 |
S | 538 | 33 | 1 | 33.0000 |
Z | 729 | 32 | 1 | 32.0000 |
Tato | 439 | 30 | 1 | 30.0000 |
Od | 522 | 30 | 1 | 30.0000 |
V | 5132 | 227 | 9 | 25.2222 |
Do | 525 | 25 | 1 | 25.0000 |
Ve | 692 | 47 | 2 | 23.5000 |
takže | 505 | 19 | 1 | 19.0000 |
Na | 2301 | 131 | 7 | 18.7143 |
Ale | 540 | 37 | 2 | 18.5000 |
A | 1445 | 71 | 4 | 17.7500 |
Není | 198 | 17 | 1 | 17.0000 |
Jako | 303 | 16 | 1 | 16.0000 |
Všechny | 277 | 15 | 1 | 15.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
sv | 212 | 1 | 16 | 0.0625 |
vzniku | 162 | 1 | 13 | 0.0769 |
naopak | 230 | 1 | 12 | 0.0833 |
životě | 176 | 1 | 11 | 0.0909 |
např | 685 | 1 | 11 | 0.0909 |
týmu | 165 | 1 | 10 | 0.1000 |
tzv | 349 | 1 | 9 | 0.1111 |
připojení | 138 | 1 | 9 | 0.1111 |
jediný | 98 | 1 | 9 | 0.1111 |
vývoji | 75 | 1 | 8 | 0.1250 |
pravděpodobně | 133 | 1 | 8 | 0.1250 |
založení | 100 | 1 | 8 | 0.1250 |
ploše | 72 | 1 | 8 | 0.1250 |
typů | 124 | 1 | 8 | 0.1250 |
jinou | 79 | 1 | 8 | 0.1250 |
velká | 152 | 1 | 8 | 0.1250 |
s.r.o | 236 | 1 | 7 | 0.1429 |
provoz | 150 | 1 | 7 | 0.1429 |
sezóně | 57 | 1 | 7 | 0.1429 |
č | 308 | 2 | 14 | 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