Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Das | 1398 | 88 | 3 | 29.3333 |
Die | 3268 | 165 | 6 | 27.5000 |
Und | 317 | 23 | 1 | 23.0000 |
Im | 495 | 42 | 2 | 21.0000 |
Der | 1420 | 76 | 4 | 19.0000 |
In | 813 | 38 | 2 | 19.0000 |
dass | 1682 | 42 | 3 | 14.0000 |
Am | 292 | 26 | 2 | 13.0000 |
Auf | 232 | 12 | 1 | 12.0000 |
Wie | 297 | 21 | 2 | 10.5000 |
Wir | 769 | 50 | 5 | 10.0000 |
Da | 147 | 10 | 1 | 10.0000 |
Auch | 348 | 10 | 1 | 10.0000 |
sondern | 268 | 10 | 1 | 10.0000 |
Er | 325 | 19 | 2 | 9.5000 |
Es | 723 | 37 | 4 | 9.2500 |
Um | 159 | 9 | 1 | 9.0000 |
Durch | 152 | 9 | 1 | 9.0000 |
So | 242 | 17 | 2 | 8.5000 |
GmbH | 263 | 8 | 1 | 8.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
später | 113 | 1 | 15 | 0.0667 |
weitere | 165 | 1 | 13 | 0.0769 |
lang | 76 | 1 | 11 | 0.0909 |
2014 | 65 | 1 | 9 | 0.1111 |
Tagen | 86 | 1 | 9 | 0.1111 |
luxemburgischen | 92 | 1 | 7 | 0.1429 |
gleichen | 51 | 1 | 7 | 0.1429 |
damals | 55 | 1 | 7 | 0.1429 |
jedem | 71 | 1 | 7 | 0.1429 |
neuen | 316 | 4 | 25 | 0.1600 |
Gemeinde | 90 | 1 | 6 | 0.1667 |
größte | 39 | 1 | 6 | 0.1667 |
mehreren | 59 | 1 | 6 | 0.1667 |
Person | 81 | 1 | 6 | 0.1667 |
58 | 1 | 5 | 0.2000 | |
Jahres | 60 | 1 | 5 | 0.2000 |
aktuellen | 57 | 1 | 5 | 0.2000 |
zuvor | 54 | 1 | 5 | 0.2000 |
60 | 49 | 1 | 5 | 0.2000 |
deutschen | 73 | 1 | 5 | 0.2000 |
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