The first random number genera
The first random number generator comes from Python itself.
To use it you need to import the Python package. Then call therandom() method. See below.
import random
print (random.random())
It is important to know that a random number generatorreally is random. If so, it should have uniform distributionbetween 0 and 1. Test the random number generators in Python foruniformity.
Answer:
What you are asking for is not random; indeed, it isfairly structured. The point of random data is that it isn’tstructured: you can’t say anything about what the values will be,only what they will be on average.
You can still do what you want, though: just specify the rangesyou want your numbers to fall into.
>>> def random_function(*args):... n = sum(args)... bottom = 0... for m in args:... for _ in range(m):... yield random.uniform(bottom, bottom + m/n)... bottom += m/n...>>> list(random_function(6,3,1))[0.1849778317803791, 0.2779140519434712, 0.08380168147928498, 0.5477412922676888, 0.5158697440011519, 0.5535466918038039, 0.8046993690361345, 0.714614514522802, 0.7102988180048052, 0.9608096335125095]>>> list(random_function(6,3,1))[0.29313403848522546, 0.5543469551407342, 0.14208842652528347, 0.3444640248041187, 0.3168266508239002, 0.5956620829410604, 0.673021479097414, 0.7141779120687652, 0.7783099010964684, 0.9103924993423906]
Explanation:
Given (6,3,1), work out their sum (10). We want six numbersbetween 0 and 6/10, three numbers between 6/10 and 6/10 + 3/10, andone number between 6/10 + 3/10 and 6/10 + 3/10 + 1/10. We cangenerate these by calling
random.uniform(0, 6/10) six times,random.uniform(6/10, 9/10) three times, andrandom.uniform(9/10, 1) once.
Another approach could be
As part of the random module (import random), you can userandom.uniform(a,b) instead of random.random() where a to b is therange in floating point.
But second approach may or may not have significant effect onuniformity. So, its better to go with first approach.
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