Average sentence length |
---|
17.4939 |
sentence length | percentage |
---|---|
2 | 0.0100 |
3 | 0.4200 |
4 | 1.2000 |
5 | 1.9700 |
6 | 2.7400 |
7 | 3.0300 |
8 | 3.9000 |
9 | 4.0800 |
10 | 4.2800 |
11 | 5.1500 |
12 | 5.2300 |
13 | 5.0400 |
14 | 4.6400 |
15 | 4.6100 |
16 | 4.4300 |
17 | 4.8100 |
18 | 4.2400 |
19 | 4.0000 |
20 | 3.6900 |
21 | 3.5700 |
22 | 3.3800 |
23 | 2.9700 |
24 | 2.7400 |
25 | 2.6100 |
26 | 2.2500 |
27 | 1.8100 |
28 | 2.0000 |
29 | 1.7100 |
30 | 1.5900 |
31 | 1.2200 |
32 | 1.1100 |
33 | 1.0000 |
34 | 0.7700 |
35 | 0.8400 |
36 | 0.4900 |
37 | 0.4900 |
38 | 0.4200 |
39 | 0.3800 |
40 | 0.3100 |
41 | 0.3000 |
42 | 0.2200 |
43 | 0.1100 |
44 | 0.1300 |
45 | 0.0400 |
46 | 0.0200 |
47 | 0.0300 |
48 | 0.0100 |
50 | 0.0100 |
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