Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Die | 3354 | 199 | 3 | 66.3333 |
Und | 669 | 43 | 1 | 43.0000 |
Der | 1770 | 103 | 3 | 34.3333 |
Das | 1486 | 93 | 3 | 31.0000 |
Doch | 382 | 28 | 1 | 28.0000 |
Dabei | 151 | 21 | 1 | 21.0000 |
sondern | 316 | 16 | 1 | 16.0000 |
daß | 1963 | 43 | 3 | 14.3333 |
Im | 478 | 42 | 3 | 14.0000 |
Es | 630 | 42 | 3 | 14.0000 |
Aber | 382 | 23 | 2 | 11.5000 |
Am | 238 | 23 | 2 | 11.5000 |
Er | 567 | 34 | 3 | 11.3333 |
Schon | 95 | 11 | 1 | 11.0000 |
Ich | 491 | 32 | 3 | 10.6667 |
In | 920 | 31 | 3 | 10.3333 |
Wie | 238 | 20 | 2 | 10.0000 |
Außerdem | 119 | 10 | 1 | 10.0000 |
Nach | 398 | 20 | 2 | 10.0000 |
allem | 313 | 18 | 2 | 9.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Tagen | 86 | 1 | 13 | 0.0769 |
weitere | 129 | 1 | 9 | 0.1111 |
Hamburger | 76 | 1 | 9 | 0.1111 |
solche | 107 | 1 | 8 | 0.1250 |
Jahres | 125 | 1 | 7 | 0.1429 |
solchen | 55 | 1 | 7 | 0.1429 |
Monaten | 90 | 2 | 14 | 0.1429 |
dessen | 146 | 1 | 7 | 0.1429 |
Amt | 47 | 1 | 7 | 0.1429 |
Reihe | 48 | 1 | 6 | 0.1667 |
hohen | 52 | 1 | 6 | 0.1667 |
Dezember | 64 | 1 | 6 | 0.1667 |
Start | 37 | 1 | 6 | 0.1667 |
Westen | 61 | 1 | 6 | 0.1667 |
Markt | 84 | 1 | 6 | 0.1667 |
europäischen | 61 | 1 | 6 | 0.1667 |
gemeinsamen | 37 | 1 | 6 | 0.1667 |
französischen | 48 | 1 | 6 | 0.1667 |
bald | 64 | 1 | 6 | 0.1667 |
weniger | 158 | 2 | 11 | 0.1818 |
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