Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
Харин | 1815 | 94 | 1 | 94.0000 |
Тэр | 1467 | 57 | 1 | 57.0000 |
Энэ | 5664 | 170 | 3 | 56.6667 |
Одоо | 857 | 45 | 1 | 45.0000 |
Манай | 975 | 75 | 2 | 37.5000 |
мэндийн | 1308 | 71 | 2 | 35.5000 |
Ийм | 665 | 33 | 1 | 33.0000 |
Зарим | 306 | 19 | 1 | 19.0000 |
Та | 605 | 35 | 2 | 17.5000 |
Бид | 1039 | 52 | 3 | 17.3333 |
Нэг | 710 | 48 | 3 | 16.0000 |
Өнгөрсөн | 471 | 31 | 2 | 15.5000 |
Би | 1211 | 60 | 4 | 15.0000 |
өдөртөө | 318 | 44 | 3 | 14.6667 |
Тийм | 591 | 14 | 1 | 14.0000 |
Өнөөдөр | 535 | 40 | 3 | 13.3333 |
Ирэх | 159 | 13 | 1 | 13.0000 |
Ямар | 331 | 13 | 1 | 13.0000 |
Хотын | 161 | 12 | 1 | 12.0000 |
Шүүх | 114 | 12 | 1 | 12.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
байна | 18751 | 34 | 818 | 0.0416 |
гарчээ | 144 | 1 | 17 | 0.0588 |
болжээ | 405 | 2 | 34 | 0.0588 |
гэлээ | 534 | 1 | 17 | 0.0588 |
гэв | 478 | 1 | 17 | 0.0588 |
билээ | 1254 | 3 | 46 | 0.0652 |
ярилцлаа | 444 | 1 | 15 | 0.0667 |
улиралд | 117 | 1 | 14 | 0.0714 |
жилээр | 79 | 1 | 12 | 0.0833 |
өндөржүүлсэн | 127 | 1 | 11 | 0.0909 |
2020 | 1030 | 7 | 73 | 0.0959 |
байлаа | 1074 | 6 | 62 | 0.0968 |
шүү | 1946 | 8 | 82 | 0.0976 |
байв | 602 | 4 | 41 | 0.0976 |
бодоход | 88 | 1 | 10 | 0.1000 |
доо | 176 | 1 | 9 | 0.1111 |
авчээ | 95 | 1 | 9 | 0.1111 |
мөнгөөр | 88 | 1 | 9 | 0.1111 |
хэрээр | 80 | 1 | 9 | 0.1111 |
танилцууллаа | 109 | 1 | 9 | 0.1111 |
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