Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Den | 9665 | 731 | 4 | 182.7500 |
Dat | 9876 | 221 | 2 | 110.5000 |
Déi | 7133 | 313 | 3 | 104.3333 |
De | 14965 | 1116 | 12 | 93.0000 |
Eng | 3923 | 250 | 3 | 83.3333 |
Hie | 1182 | 78 | 1 | 78.0000 |
Am | 5517 | 305 | 4 | 76.2500 |
Mat | 2687 | 73 | 1 | 73.0000 |
Och | 3791 | 64 | 1 | 64.0000 |
Als | 1028 | 59 | 1 | 59.0000 |
Mä | 2098 | 57 | 1 | 57.0000 |
Hei | 2553 | 110 | 2 | 55.0000 |
Dëst | 1234 | 53 | 1 | 53.0000 |
An | 13562 | 147 | 3 | 49.0000 |
Ma | 1339 | 48 | 1 | 48.0000 |
Nom | 507 | 48 | 1 | 48.0000 |
Si | 3448 | 184 | 4 | 46.0000 |
Do | 2425 | 124 | 3 | 41.3333 |
E | 3343 | 162 | 4 | 40.5000 |
Daat | 678 | 40 | 1 | 40.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
ginn | 30637 | 49 | 1268 | 0.0386 |
Aart | 528 | 2 | 27 | 0.0741 |
Etienne | 289 | 1 | 13 | 0.0769 |
Schouljoer | 127 | 1 | 12 | 0.0833 |
lang | 103 | 1 | 11 | 0.0909 |
Guiden | 482 | 1 | 9 | 0.1111 |
historesch | 159 | 1 | 9 | 0.1111 |
dramatesch | 115 | 1 | 9 | 0.1111 |
Duo | 99 | 1 | 9 | 0.1111 |
Gusty | 81 | 1 | 9 | 0.1111 |
Zentralbank | 57 | 1 | 9 | 0.1111 |
aaneren | 182 | 2 | 16 | 0.1250 |
ëmsou | 136 | 1 | 8 | 0.1250 |
Paulette | 91 | 1 | 8 | 0.1250 |
méigleche | 91 | 1 | 8 | 0.1250 |
Wollef | 85 | 1 | 8 | 0.1250 |
Aneren | 61 | 1 | 8 | 0.1250 |
Milliarden | 415 | 6 | 46 | 0.1304 |
gaangen | 1465 | 11 | 84 | 0.1310 |
Säiten | 838 | 2 | 15 | 0.1333 |
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