Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
It | 333 | 17 | 1 | 17.0000 |
The | 1788 | 86 | 8 | 10.7500 |
A | 210 | 10 | 1 | 10.0000 |
but | 356 | 20 | 2 | 10.0000 |
This | 469 | 28 | 3 | 9.3333 |
And | 100 | 9 | 1 | 9.0000 |
We | 271 | 16 | 2 | 8.0000 |
They | 121 | 7 | 1 | 7.0000 |
How | 48 | 6 | 1 | 6.0000 |
questions | 52 | 6 | 1 | 6.0000 |
experience | 96 | 6 | 1 | 6.0000 |
tools | 51 | 6 | 1 | 6.0000 |
What | 73 | 6 | 1 | 6.0000 |
technology | 95 | 6 | 1 | 6.0000 |
If | 151 | 6 | 1 | 6.0000 |
You | 157 | 10 | 2 | 5.0000 |
There | 118 | 5 | 1 | 5.0000 |
standards | 40 | 5 | 1 | 5.0000 |
too | 71 | 5 | 1 | 5.0000 |
down | 69 | 5 | 1 | 5.0000 |
Word | Frequency | Number of right neighbors | Number of left neighbors | Ratio |
---|---|---|---|---|
her | 123 | 1 | 8 | 0.1250 |
own | 95 | 1 | 7 | 0.1429 |
able | 74 | 1 | 6 | 0.1667 |
total | 47 | 1 | 5 | 0.2000 |
Jul | 72 | 1 | 5 | 0.2000 |
further | 88 | 1 | 5 | 0.2000 |
often | 65 | 1 | 5 | 0.2000 |
always | 46 | 1 | 5 | 0.2000 |
want | 81 | 2 | 9 | 0.2222 |
such | 195 | 2 | 9 | 0.2222 |
big | 54 | 1 | 4 | 0.2500 |
general | 65 | 1 | 4 | 0.2500 |
certain | 38 | 1 | 4 | 0.2500 |
gas | 30 | 1 | 4 | 0.2500 |
third | 36 | 1 | 4 | 0.2500 |
actually | 40 | 1 | 4 | 0.2500 |
article | 45 | 1 | 4 | 0.2500 |
recent | 36 | 1 | 4 | 0.2500 |
months | 46 | 2 | 7 | 0.2857 |
part | 142 | 3 | 10 | 0.3000 |
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