Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
In | 852 | 49 | 1 | 49.0000 |
Hij | 415 | 34 | 1 | 34.0000 |
Dat | 1068 | 53 | 2 | 26.5000 |
Het | 2015 | 111 | 5 | 22.2000 |
Er | 436 | 21 | 1 | 21.0000 |
Dan | 195 | 19 | 1 | 19.0000 |
Maar | 753 | 37 | 2 | 18.5000 |
En | 787 | 35 | 2 | 17.5000 |
De | 3121 | 180 | 11 | 16.3636 |
Ik | 1119 | 65 | 4 | 16.2500 |
Wat | 317 | 16 | 1 | 16.0000 |
zodat | 105 | 12 | 1 | 12.0000 |
Ze | 372 | 24 | 2 | 12.0000 |
Die | 393 | 12 | 1 | 12.0000 |
We | 353 | 23 | 2 | 11.5000 |
Met | 267 | 11 | 1 | 11.0000 |
Een | 643 | 32 | 3 | 10.6667 |
Ik ben | 100 | 10 | 1 | 10.0000 |
Daar | 141 | 10 | 1 | 10.0000 |
Of | 168 | 9 | 1 | 9.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
ja | 98 | 1 | 10 | 0.1000 |
kun | 152 | 1 | 7 | 0.1429 |
uiteindelijk | 60 | 1 | 7 | 0.1429 |
oude | 91 | 1 | 6 | 0.1667 |
hoofd | 76 | 1 | 6 | 0.1667 |
gezin | 51 | 1 | 6 | 0.1667 |
daarom | 98 | 1 | 6 | 0.1667 |
volledig | 94 | 1 | 6 | 0.1667 |
uw | 100 | 1 | 6 | 0.1667 |
weekend | 50 | 1 | 6 | 0.1667 |
ronde | 34 | 1 | 5 | 0.2000 |
totale | 40 | 1 | 5 | 0.2000 |
augustus | 54 | 1 | 5 | 0.2000 |
jouw | 64 | 1 | 5 | 0.2000 |
zo’n | 176 | 2 | 9 | 0.2222 |
basis | 93 | 1 | 4 | 0.2500 |
nee | 43 | 1 | 4 | 0.2500 |
hoge | 57 | 1 | 4 | 0.2500 |
kop | 45 | 1 | 4 | 0.2500 |
leven | 226 | 3 | 12 | 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