Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
De | 1734 | 65 | 2 | 32.5000 |
Se | 208 | 12 | 1 | 12.0000 |
hen | 56 | 6 | 1 | 6.0000 |
torüch | 68 | 5 | 1 | 5.0000 |
wiest | 49 | 5 | 1 | 5.0000 |
Bahnlien | 45 | 5 | 1 | 5.0000 |
wesen | 57 | 5 | 1 | 5.0000 |
de | 10153 | 434 | 94 | 4.6170 |
woneem | 33 | 4 | 1 | 4.0000 |
kunn | 77 | 4 | 1 | 4.0000 |
wählt | 36 | 4 | 1 | 4.0000 |
künnt | 70 | 4 | 1 | 4.0000 |
kriggt | 41 | 4 | 1 | 4.0000 |
anfungen | 29 | 4 | 1 | 4.0000 |
kamen | 84 | 4 | 1 | 4.0000 |
Stand | 149 | 4 | 1 | 4.0000 |
wiel | 66 | 4 | 1 | 4.0000 |
an’n | 258 | 35 | 9 | 3.8889 |
in’n | 523 | 34 | 11 | 3.0909 |
speelt | 83 | 6 | 2 | 3.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
März | 68 | 1 | 8 | 0.1250 |
worrn | 551 | 5 | 34 | 0.1471 |
Land | 93 | 1 | 6 | 0.1667 |
hör | 151 | 2 | 11 | 0.1818 |
September | 51 | 1 | 5 | 0.2000 |
wedder | 201 | 3 | 13 | 0.2308 |
Januar | 88 | 3 | 13 | 0.2308 |
Juni | 56 | 1 | 4 | 0.2500 |
Westen | 31 | 1 | 4 | 0.2500 |
Fro | 39 | 1 | 4 | 0.2500 |
United | 26 | 1 | 4 | 0.2500 |
gröttste | 37 | 1 | 4 | 0.2500 |
Dood | 66 | 1 | 4 | 0.2500 |
keen | 95 | 1 | 4 | 0.2500 |
ene | 53 | 1 | 4 | 0.2500 |
vun | 4097 | 46 | 173 | 0.2659 |
Johrhunnert | 65 | 2 | 7 | 0.2857 |
Kinner | 79 | 2 | 7 | 0.2857 |
ehr | 160 | 3 | 10 | 0.3000 |
Kilometer | 77 | 3 | 10 | 0.3000 |
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