Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
있도록 | 2515 | 117 | 1 | 117.0000 |
라고 | 4848 | 166 | 3 | 55.3333 |
이라고 | 9507 | 146 | 3 | 48.6667 |
고 | 31707 | 369 | 8 | 46.1250 |
며 | 15780 | 81 | 2 | 40.5000 |
그런데 | 941 | 40 | 1 | 40.0000 |
는 | 4800 | 193 | 5 | 38.6000 |
을 | 3021 | 164 | 5 | 32.8000 |
를 | 2897 | 164 | 5 | 32.8000 |
의 | 2576 | 93 | 3 | 31.0000 |
에 | 2863 | 92 | 3 | 30.6667 |
가 | 2917 | 121 | 4 | 30.2500 |
그러나 | 2282 | 111 | 4 | 27.7500 |
서초구 | 292 | 24 | 1 | 24.0000 |
다만 | 2756 | 95 | 4 | 23.7500 |
은 | 1846 | 71 | 3 | 23.6667 |
용산구 | 305 | 22 | 1 | 22.0000 |
막대한 | 199 | 16 | 1 | 16.0000 |
본인 | 197 | 16 | 1 | 16.0000 |
이동 | 211 | 15 | 1 | 15.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
밝혔다 | 11595 | 1 | 304 | 0.0033 |
있다 | 40146 | 6 | 1369 | 0.0044 |
했다 | 13911 | 4 | 566 | 0.0071 |
전했다 | 3387 | 1 | 133 | 0.0075 |
예정이다 | 2754 | 2 | 226 | 0.0088 |
것이다 | 5647 | 3 | 323 | 0.0093 |
한다 | 8847 | 7 | 585 | 0.0120 |
발표했다 | 1041 | 1 | 79 | 0.0127 |
계획이다 | 1867 | 2 | 148 | 0.0135 |
보도했다 | 1069 | 1 | 70 | 0.0143 |
전망이다 | 770 | 1 | 67 | 0.0149 |
때문이다 | 2052 | 2 | 103 | 0.0194 |
설명했다 | 3220 | 1 | 51 | 0.0196 |
받았다 | 1480 | 2 | 96 | 0.0208 |
상태다 | 649 | 1 | 48 | 0.0208 |
말했다 | 12522 | 1 | 46 | 0.0217 |
있었다 | 2476 | 3 | 138 | 0.0217 |
못했다 | 1152 | 2 | 91 | 0.0220 |
않았다 | 2262 | 4 | 170 | 0.0235 |
강조했다 | 2791 | 1 | 40 | 0.0250 |
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