Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
प्रशासनिक | 111 | 10 | 1 | 10.0000 |
उपजिल्लामा | 595 | 28 | 3 | 9.3333 |
यसरी | 449 | 9 | 1 | 9.0000 |
झापा | 99 | 9 | 1 | 9.0000 |
२०६५ | 55 | 8 | 1 | 8.0000 |
सन | 200 | 16 | 2 | 8.0000 |
जर्मन | 132 | 8 | 1 | 8.0000 |
सुझाव | 57 | 8 | 1 | 8.0000 |
प्रयोगमा | 93 | 8 | 1 | 8.0000 |
अन्तरले | 154 | 8 | 1 | 8.0000 |
वैदिक | 82 | 7 | 1 | 7.0000 |
२०७१ | 53 | 7 | 1 | 7.0000 |
२०५४ | 62 | 7 | 1 | 7.0000 |
अन्तरिम | 53 | 7 | 1 | 7.0000 |
शिक्षक | 85 | 7 | 1 | 7.0000 |
२०६३ | 59 | 7 | 1 | 7.0000 |
भाव | 71 | 7 | 1 | 7.0000 |
२०४८ | 55 | 7 | 1 | 7.0000 |
कलकत्ता | 50 | 6 | 1 | 6.0000 |
लोकतान्त्रिक | 66 | 6 | 1 | 6.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
। | 41394 | 2 | 591 | 0.0034 |
छिन्। | 228 | 1 | 16 | 0.0625 |
हो | 7025 | 29 | 320 | 0.0906 |
वि | 835 | 4 | 42 | 0.0952 |
गरेका | 4981 | 24 | 235 | 0.1021 |
गरिएको | 4729 | 25 | 231 | 0.1082 |
राखेर | 178 | 1 | 9 | 0.1111 |
सालदेखि | 121 | 2 | 17 | 0.1176 |
गर्नको | 216 | 2 | 16 | 0.1250 |
व्यक्तिले | 82 | 1 | 8 | 0.1250 |
मदरसाहरू | 64 | 1 | 8 | 0.1250 |
देखिन्छ | 406 | 3 | 22 | 0.1364 |
समितिले | 72 | 1 | 7 | 0.1429 |
गतिमा | 73 | 1 | 7 | 0.1429 |
मल्लले | 70 | 1 | 7 | 0.1429 |
कुराहरू | 65 | 1 | 7 | 0.1429 |
अभियानमा | 63 | 1 | 7 | 0.1429 |
हुन् | 2062 | 18 | 121 | 0.1488 |
गर्नाले | 109 | 2 | 13 | 0.1538 |
छ | 8578 | 38 | 246 | 0.1545 |
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