Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Die | 10819 | 633 | 5 | 126.6000 |
Der | 4335 | 291 | 4 | 72.7500 |
Im | 2154 | 119 | 2 | 59.5000 |
Seit | 506 | 56 | 1 | 56.0000 |
Das | 4905 | 252 | 5 | 50.4000 |
Er | 858 | 47 | 1 | 47.0000 |
Bei | 1226 | 40 | 1 | 40.0000 |
Am | 883 | 76 | 2 | 38.0000 |
Nach | 907 | 37 | 1 | 37.0000 |
Eine | 1279 | 73 | 2 | 36.5000 |
Auch | 1139 | 30 | 1 | 30.0000 |
Da | 547 | 30 | 1 | 30.0000 |
Aber | 626 | 29 | 1 | 29.0000 |
So | 868 | 58 | 2 | 29.0000 |
In | 2907 | 86 | 3 | 28.6667 |
Ein | 1588 | 84 | 3 | 28.0000 |
Als | 804 | 27 | 1 | 27.0000 |
Hier | 583 | 27 | 1 | 27.0000 |
Denn | 408 | 26 | 1 | 26.0000 |
sondern | 1050 | 25 | 1 | 25.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
klassischen | 120 | 1 | 12 | 0.0833 |
zentralen | 76 | 1 | 11 | 0.0909 |
speziellen | 88 | 1 | 9 | 0.1111 |
allgemeinen | 66 | 1 | 9 | 0.1111 |
Großen | 54 | 1 | 8 | 0.1250 |
Betroffenen | 86 | 1 | 8 | 0.1250 |
besseren | 56 | 1 | 8 | 0.1250 |
erfolgreichen | 93 | 1 | 8 | 0.1250 |
positiven | 63 | 1 | 8 | 0.1250 |
private | 72 | 1 | 8 | 0.1250 |
eigenes | 104 | 1 | 8 | 0.1250 |
150 | 70 | 1 | 8 | 0.1250 |
möglichen | 77 | 1 | 8 | 0.1250 |
Geist | 65 | 1 | 7 | 0.1429 |
Deutschen | 86 | 1 | 7 | 0.1429 |
steirische | 64 | 1 | 7 | 0.1429 |
traditionellen | 59 | 1 | 7 | 0.1429 |
klassische | 64 | 1 | 7 | 0.1429 |
Semester | 63 | 1 | 7 | 0.1429 |
Seminar | 59 | 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