Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Die | 33561 | 1781 | 9 | 197.8889 |
Der | 17309 | 1102 | 6 | 183.6667 |
Damit | 1393 | 87 | 1 | 87.0000 |
Das | 16161 | 680 | 8 | 85.0000 |
Denn | 1810 | 73 | 1 | 73.0000 |
Und | 5415 | 127 | 2 | 63.5000 |
Ein | 4900 | 298 | 5 | 59.6000 |
Er | 4434 | 236 | 4 | 59.0000 |
Außerdem | 836 | 56 | 1 | 56.0000 |
Seit | 1620 | 108 | 2 | 54.0000 |
In | 9226 | 254 | 5 | 50.8000 |
Im | 5291 | 281 | 6 | 46.8333 |
Insgesamt | 712 | 45 | 1 | 45.0000 |
Aber | 3121 | 86 | 2 | 43.0000 |
Sein | 529 | 37 | 1 | 37.0000 |
So | 2821 | 143 | 4 | 35.7500 |
Demnach | 366 | 35 | 1 | 35.0000 |
Eine | 3355 | 208 | 6 | 34.6667 |
Nun | 1230 | 69 | 2 | 34.5000 |
Diese | 2048 | 130 | 4 | 32.5000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
worden | 3901 | 15 | 328 | 0.0457 |
mitteilte | 622 | 2 | 29 | 0.0690 |
ereignete | 115 | 1 | 12 | 0.0833 |
belässt | 71 | 1 | 12 | 0.0833 |
hieß | 707 | 1 | 11 | 0.0909 |
schnellen | 145 | 1 | 11 | 0.0909 |
schöne | 138 | 1 | 10 | 0.1000 |
eigentlichen | 79 | 1 | 10 | 0.1000 |
erneuten | 79 | 1 | 10 | 0.1000 |
bestehenden | 148 | 1 | 9 | 0.1111 |
Verbraucher | 137 | 1 | 9 | 0.1111 |
Einheit | 113 | 1 | 9 | 0.1111 |
wahre | 69 | 1 | 9 | 0.1111 |
werden | 18872 | 121 | 988 | 0.1225 |
Dow | 151 | 1 | 8 | 0.1250 |
jüngste | 110 | 1 | 8 | 0.1250 |
damalige | 106 | 1 | 8 | 0.1250 |
Kroatien | 97 | 1 | 8 | 0.1250 |
diesjährigen | 95 | 1 | 8 | 0.1250 |
50.000 | 81 | 1 | 8 | 0.1250 |
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