Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
있도록 | 2403 | 104 | 1 | 104.0000 |
고 | 32513 | 422 | 7 | 60.2857 |
라고 | 4872 | 179 | 3 | 59.6667 |
이라고 | 8986 | 146 | 4 | 36.5000 |
며 | 15295 | 98 | 3 | 32.6667 |
는 | 4875 | 191 | 6 | 31.8333 |
를 | 2680 | 147 | 5 | 29.4000 |
을 | 2591 | 136 | 5 | 27.2000 |
은 | 1916 | 76 | 3 | 25.3333 |
에 | 2608 | 97 | 4 | 24.2500 |
다만 | 2345 | 88 | 4 | 22.0000 |
의 | 2637 | 85 | 4 | 21.2500 |
그런데 | 892 | 42 | 2 | 21.0000 |
그러나 | 2740 | 101 | 5 | 20.2000 |
내달 | 169 | 17 | 1 | 17.0000 |
가 | 2590 | 113 | 7 | 16.1429 |
두기 | 164 | 14 | 1 | 14.0000 |
종로구 | 395 | 26 | 2 | 13.0000 |
논란 | 125 | 13 | 1 | 13.0000 |
연락을 | 166 | 12 | 1 | 12.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
밝혔다 | 12250 | 1 | 293 | 0.0034 |
있다 | 38508 | 8 | 1347 | 0.0059 |
했다 | 15916 | 4 | 600 | 0.0067 |
전했다 | 3579 | 1 | 141 | 0.0071 |
한다 | 8706 | 7 | 583 | 0.0120 |
보도했다 | 1191 | 1 | 76 | 0.0132 |
예정이다 | 2708 | 3 | 218 | 0.0138 |
계획이다 | 1641 | 2 | 144 | 0.0139 |
발표했다 | 1047 | 1 | 70 | 0.0143 |
것이다 | 5829 | 5 | 328 | 0.0152 |
전망이다 | 768 | 1 | 63 | 0.0159 |
방침이다 | 814 | 1 | 60 | 0.0167 |
상태다 | 760 | 1 | 59 | 0.0169 |
설명했다 | 3374 | 1 | 58 | 0.0172 |
올랐다 | 708 | 1 | 52 | 0.0192 |
말했다 | 11220 | 1 | 46 | 0.0217 |
중이다 | 1662 | 2 | 83 | 0.0241 |
받았다 | 2658 | 3 | 115 | 0.0261 |
나섰다 | 892 | 2 | 74 | 0.0270 |
강조했다 | 2326 | 1 | 35 | 0.0286 |
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