Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Ze | 3414 | 204 | 1 | 204.0000 |
Hij | 5068 | 246 | 2 | 123.0000 |
Maar | 5572 | 122 | 1 | 122.0000 |
En | 6103 | 118 | 1 | 118.0000 |
Zo | 2134 | 114 | 1 | 114.0000 |
Dat | 10259 | 201 | 2 | 100.5000 |
Deze | 1773 | 89 | 1 | 89.0000 |
Een | 7568 | 437 | 5 | 87.4000 |
Zij | 1180 | 84 | 1 | 84.0000 |
Dan | 1484 | 75 | 1 | 75.0000 |
Die | 3149 | 147 | 2 | 73.5000 |
n | 3335 | 212 | 3 | 70.6667 |
Het | 19898 | 715 | 12 | 59.5833 |
Zijn | 1016 | 57 | 1 | 57.0000 |
Of | 1279 | 55 | 1 | 55.0000 |
Wel | 709 | 54 | 1 | 54.0000 |
Ook | 4271 | 95 | 2 | 47.5000 |
In | 9011 | 230 | 5 | 46.0000 |
De | 44877 | 2219 | 51 | 43.5098 |
Bovendien | 717 | 43 | 1 | 43.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
kun | 886 | 2 | 38 | 0.0526 |
vorig | 1629 | 5 | 69 | 0.0725 |
z | 736 | 2 | 24 | 0.0833 |
gebrek | 236 | 1 | 11 | 0.0909 |
tal | 128 | 1 | 11 | 0.0909 |
NAC | 144 | 1 | 10 | 0.1000 |
bijv | 101 | 1 | 9 | 0.1111 |
aanhangers | 124 | 1 | 9 | 0.1111 |
gepubliceerde | 63 | 1 | 9 | 0.1111 |
RKC | 85 | 1 | 9 | 0.1111 |
sluiting | 114 | 1 | 8 | 0.1250 |
vrijlating | 103 | 1 | 8 | 0.1250 |
manieren | 104 | 1 | 8 | 0.1250 |
AC | 114 | 1 | 8 | 0.1250 |
o.a | 126 | 1 | 8 | 0.1250 |
tak | 92 | 1 | 7 | 0.1429 |
ex | 95 | 1 | 7 | 0.1429 |
ingetrokken | 71 | 1 | 7 | 0.1429 |
16.00 | 41 | 1 | 7 | 0.1429 |
verhoging | 146 | 1 | 7 | 0.1429 |
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