Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Im | 179 | 13 | 1 | 13.0000 |
Am | 117 | 12 | 1 | 12.0000 |
In | 298 | 12 | 1 | 12.0000 |
Auch | 165 | 9 | 1 | 9.0000 |
Die | 1116 | 42 | 5 | 8.4000 |
Aber | 111 | 8 | 1 | 8.0000 |
weil | 85 | 8 | 1 | 8.0000 |
Wie | 116 | 7 | 1 | 7.0000 |
sowie | 102 | 7 | 1 | 7.0000 |
Ein | 177 | 7 | 1 | 7.0000 |
werde | 88 | 7 | 1 | 7.0000 |
ab | 142 | 7 | 1 | 7.0000 |
wegen | 111 | 6 | 1 | 6.0000 |
Das | 534 | 24 | 4 | 6.0000 |
Auf | 84 | 6 | 1 | 6.0000 |
Er | 143 | 11 | 2 | 5.5000 |
dass | 659 | 22 | 4 | 5.5000 |
Der | 560 | 16 | 3 | 5.3333 |
waren | 128 | 5 | 1 | 5.0000 |
Bei | 108 | 5 | 1 | 5.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Wochen | 76 | 1 | 8 | 0.1250 |
Tag | 67 | 1 | 8 | 0.1250 |
Tagen | 57 | 1 | 8 | 0.1250 |
neue | 125 | 1 | 7 | 0.1429 |
mich | 66 | 1 | 6 | 0.1667 |
ihrem | 49 | 1 | 6 | 0.1667 |
andere | 71 | 1 | 6 | 0.1667 |
August | 37 | 1 | 6 | 0.1667 |
diesen | 54 | 1 | 6 | 0.1667 |
neuen | 130 | 2 | 11 | 0.1818 |
anderen | 111 | 2 | 11 | 0.1818 |
mir | 47 | 1 | 5 | 0.2000 |
eigenen | 39 | 1 | 5 | 0.2000 |
September | 35 | 1 | 5 | 0.2000 |
weiteren | 40 | 1 | 5 | 0.2000 |
ihr | 120 | 1 | 4 | 0.2500 |
ihren | 87 | 2 | 8 | 0.2500 |
März | 67 | 2 | 8 | 0.2500 |
alle | 140 | 2 | 8 | 0.2500 |
wenig | 51 | 1 | 4 | 0.2500 |
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