Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Die | 11132 | 682 | 8 | 85.2500 |
Der | 5966 | 396 | 5 | 79.2000 |
Am | 943 | 76 | 1 | 76.0000 |
Und | 2230 | 71 | 1 | 71.0000 |
Das | 5019 | 274 | 4 | 68.5000 |
Er | 1914 | 128 | 2 | 64.0000 |
Nach | 1369 | 53 | 1 | 53.0000 |
In | 2847 | 105 | 2 | 52.5000 |
Aber | 1282 | 49 | 1 | 49.0000 |
Im | 1617 | 98 | 2 | 49.0000 |
Ich | 1822 | 77 | 2 | 38.5000 |
Auch | 1466 | 37 | 1 | 37.0000 |
Denn | 697 | 35 | 1 | 35.0000 |
Damit | 454 | 34 | 1 | 34.0000 |
Wir | 1577 | 65 | 2 | 32.5000 |
Dies | 397 | 31 | 1 | 31.0000 |
Es | 2353 | 88 | 3 | 29.3333 |
sondern | 1047 | 29 | 1 | 29.0000 |
Eine | 1069 | 51 | 2 | 25.5000 |
So | 1153 | 74 | 3 | 24.6667 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
jene | 132 | 1 | 13 | 0.0769 |
so genannten | 136 | 1 | 10 | 0.1000 |
worden | 1378 | 14 | 138 | 0.1014 |
schweren | 82 | 1 | 9 | 0.1111 |
so genannte | 134 | 1 | 8 | 0.1250 |
höheren | 69 | 1 | 8 | 0.1250 |
DFB | 74 | 1 | 8 | 0.1250 |
historischen | 91 | 1 | 8 | 0.1250 |
Kölner | 73 | 1 | 8 | 0.1250 |
Brandenburger | 96 | 1 | 7 | 0.1429 |
schöne | 68 | 1 | 7 | 0.1429 |
zentralen | 49 | 1 | 7 | 0.1429 |
damaligen | 66 | 1 | 7 | 0.1429 |
möglichen | 92 | 1 | 7 | 0.1429 |
übrigen | 89 | 1 | 7 | 0.1429 |
wichtigste | 56 | 1 | 7 | 0.1429 |
werden | 6060 | 63 | 404 | 0.1559 |
Cottbus | 55 | 1 | 6 | 0.1667 |
Anwalt | 101 | 1 | 6 | 0.1667 |
Alten | 52 | 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