Average sentence length |
---|
18.6388 |
sentence length | percentage |
---|---|
2 | 0.0200 |
3 | 0.1700 |
4 | 0.7300 |
5 | 1.3600 |
6 | 2.1200 |
7 | 2.8800 |
8 | 3.8300 |
9 | 4.5500 |
10 | 4.3600 |
11 | 4.6200 |
12 | 4.8000 |
13 | 4.9300 |
14 | 4.6100 |
15 | 4.2000 |
16 | 4.6000 |
17 | 4.8000 |
18 | 4.1500 |
19 | 4.5300 |
20 | 3.5800 |
21 | 3.1100 |
22 | 3.1800 |
23 | 2.8500 |
24 | 2.7100 |
25 | 2.3100 |
26 | 1.9300 |
27 | 2.0600 |
28 | 1.9400 |
29 | 1.7700 |
30 | 1.4300 |
31 | 1.4000 |
32 | 1.2000 |
33 | 1.0400 |
34 | 0.9100 |
35 | 0.9800 |
36 | 0.8500 |
37 | 0.8100 |
38 | 0.7100 |
39 | 0.5200 |
40 | 0.5100 |
41 | 0.4600 |
42 | 0.4100 |
43 | 0.4000 |
44 | 0.3700 |
45 | 0.4300 |
46 | 0.2800 |
47 | 0.1500 |
48 | 0.1500 |
49 | 0.1700 |
50 | 0.1200 |
Next we measure the length of a sentence by its number of words. Again, we are interested in average length and length distribution.
See above.
For simplicity, the number of words in a sentence is calculated as 1+(Number of blanks in the sentence).
Average length:
select avg(1+char_length(sentence)-char_length( replace(sentence," ",""))) from sentences;
Distribution:
SELECT @all:=count(*) from sentences;
select 1+char_length(sentence)-char_length( replace(sentence," ","")), 100*count(*)/@all from sentences where 50>=1+char_length(sentence)-char_length(replace(sentence," ",""))group by 1+char_length(sentence)-char_length( replace( sentence," ",""));
4.2.1 Length of sentences in characters