Average sentence length |
---|
19.4114 |
sentence length | percentage |
---|---|
2 | 0.0100 |
3 | 0.1900 |
4 | 0.6600 |
5 | 1.2400 |
6 | 1.9700 |
7 | 2.2300 |
8 | 3.0600 |
9 | 3.5800 |
10 | 3.7100 |
11 | 3.9800 |
12 | 4.2500 |
13 | 4.7600 |
14 | 4.4000 |
15 | 4.5600 |
16 | 4.3200 |
17 | 4.1900 |
18 | 4.6300 |
19 | 4.1400 |
20 | 3.7300 |
21 | 3.5500 |
22 | 3.4800 |
23 | 3.3700 |
24 | 3.1100 |
25 | 2.9100 |
26 | 2.6000 |
27 | 2.3600 |
28 | 2.1400 |
29 | 2.1600 |
30 | 1.7500 |
31 | 1.7300 |
32 | 1.5500 |
33 | 1.3300 |
34 | 1.2900 |
35 | 1.2800 |
36 | 0.8700 |
37 | 0.8700 |
38 | 0.8500 |
39 | 0.6500 |
40 | 0.6000 |
41 | 0.5500 |
42 | 0.3900 |
43 | 0.3500 |
44 | 0.2200 |
45 | 0.2000 |
46 | 0.1000 |
47 | 0.0500 |
48 | 0.0100 |
49 | 0.0500 |
50 | 0.0200 |
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