Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
który | 2352 | 128 | 1 | 128.0000 |
Na | 2684 | 151 | 2 | 75.5000 |
W | 7089 | 373 | 5 | 74.6000 |
która | 1348 | 70 | 1 | 70.0000 |
A | 1477 | 66 | 1 | 66.0000 |
Od | 631 | 62 | 1 | 62.0000 |
Z | 1216 | 60 | 1 | 60.0000 |
Czy | 914 | 49 | 1 | 49.0000 |
Co | 861 | 45 | 1 | 45.0000 |
Ale | 811 | 45 | 1 | 45.0000 |
bo | 1733 | 73 | 2 | 36.5000 |
Do | 1289 | 73 | 2 | 36.5000 |
Jak | 1555 | 101 | 3 | 33.6667 |
Za | 505 | 33 | 1 | 33.0000 |
Jest | 527 | 26 | 1 | 26.0000 |
a | 5299 | 172 | 7 | 24.5714 |
To | 2304 | 122 | 5 | 24.4000 |
Nie | 2061 | 116 | 5 | 23.2000 |
Prezydent | 182 | 22 | 1 | 22.0000 |
Tak | 469 | 20 | 1 | 20.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
przede | 266 | 1 | 14 | 0.0714 |
fakt | 128 | 1 | 12 | 0.0833 |
miejscowości | 260 | 1 | 12 | 0.0833 |
odbędzie | 286 | 1 | 11 | 0.0909 |
mocno | 160 | 1 | 10 | 0.1000 |
tygodnia | 96 | 1 | 10 | 0.1000 |
moment | 83 | 1 | 9 | 0.1111 |
spotkaniu | 107 | 1 | 9 | 0.1111 |
reprezentacji | 100 | 1 | 9 | 0.1111 |
wsi | 113 | 1 | 9 | 0.1111 |
rolę | 102 | 1 | 8 | 0.1250 |
proces | 112 | 1 | 8 | 0.1250 |
miejskiego | 57 | 1 | 8 | 0.1250 |
skorzystać | 98 | 1 | 8 | 0.1250 |
kraj | 67 | 1 | 8 | 0.1250 |
chcieli | 98 | 1 | 7 | 0.1429 |
długości | 72 | 1 | 7 | 0.1429 |
was | 101 | 1 | 7 | 0.1429 |
zauważyć | 46 | 1 | 7 | 0.1429 |
szanse | 84 | 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