Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
पढें | 455 | 36 | 1 | 36.0000 |
लिए | 11350 | 481 | 23 | 20.9130 |
खिलाफ | 2249 | 157 | 9 | 17.4444 |
बीच | 2232 | 114 | 7 | 16.2857 |
दौरान | 2021 | 107 | 7 | 15.2857 |
आदि | 227 | 13 | 1 | 13.0000 |
अनुसार | 1233 | 60 | 5 | 12.0000 |
मुताबिक | 1452 | 58 | 5 | 11.6000 |
पालन | 100 | 11 | 1 | 11.0000 |
साथ | 5638 | 185 | 17 | 10.8824 |
अलावा | 963 | 39 | 4 | 9.7500 |
हालांकि | 1306 | 47 | 5 | 9.4000 |
तेंदुलकर | 341 | 9 | 1 | 9.0000 |
तहत | 720 | 35 | 4 | 8.7500 |
बावजूद | 498 | 25 | 3 | 8.3333 |
जाया | 87 | 8 | 1 | 8.0000 |
बढ़ावा | 76 | 8 | 1 | 8.0000 |
दौरा | 159 | 8 | 1 | 8.0000 |
अध्यक्षता | 131 | 8 | 1 | 8.0000 |
कायदा | 85 | 7 | 1 | 7.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
हैं। | 8091 | 2 | 204 | 0.0098 |
था। | 3430 | 1 | 92 | 0.0109 |
थे। | 1983 | 1 | 70 | 0.0143 |
तरीके | 317 | 1 | 22 | 0.0455 |
। | 313 | 1 | 18 | 0.0556 |
लोगों | 3720 | 9 | 157 | 0.0573 |
तौर | 864 | 2 | 33 | 0.0606 |
सुप्रीम | 361 | 1 | 14 | 0.0714 |
करने | 6782 | 34 | 416 | 0.0817 |
रहा | 4149 | 11 | 112 | 0.0982 |
मुद्दों | 220 | 2 | 20 | 0.1000 |
गेंदों | 253 | 3 | 30 | 0.1000 |
ढंग | 127 | 1 | 10 | 0.1000 |
जगहों | 101 | 1 | 9 | 0.1111 |
किया | 7567 | 37 | 316 | 0.1171 |
सकते | 1210 | 7 | 55 | 0.1273 |
मैचों | 354 | 4 | 29 | 0.1379 |
खूब | 98 | 1 | 7 | 0.1429 |
डा | 89 | 1 | 7 | 0.1429 |
मौके | 514 | 3 | 21 | 0.1429 |
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