如何使用numpy在分段间隔上生成随机数

Question

I am using numpy module in python to generate random numbers. When I need to generate random numbers in a continuous interval such as [a,b], I will use

(b-a)*np.random.rand(1)+a

but now I Need to generate a uniform random number in the interval [a, b] and [c, d], what should I do?

I want to generate a random number that is uniform over the length of all the intervals. I do not select an interval with equal probability, and then generate a random number inside the interval. If [a, b] and [c, d] are equal in length, There is no problem with this use, but when the lengths of the intervals are not equal, the random numbers generated by this method are not completely uniform.

Answer 1

You could do something like

a,b,c,d = 1,2,7,9
N = 10
r = np.random.uniform(a-b,d-c,N)
r += np.where(r<0,b,c)
r
# array([7.30557415, 7.42185479, 1.48986144, 7.95916547, 1.30422703,
#        8.79749665, 8.19329762, 8.72669862, 1.88426196, 8.33789181])

Answer 2

You can use

np.random.uniform(a,b)

for your random numbers between a and b (including a but excluding b)

So for random number in [a,b] and [c,d], you can use

np.random.choice( [np.random.uniform(a,b) , np.random.uniform(c,d)] )

Answer 3

Here's a recipe:

def random_multiinterval(*intervals, shape=(1,)):
    # FIXME assert intervals are valid and non-overlapping
    size = sum(i[1] - i[0] for i in intervals)
    v = size * np.random.rand(*shape)
    res = np.zeros_like(v)
    for i in intervals:
        res += (0 < v) * (v < (i[1] - i[0])) * (i[0] + v)
        v -= i[1] - i[0]
    return res

In [11]: random_multiinterval((1, 2), (3, 4))
Out[11]: array([1.34391171])

In [12]: random_multiinterval((1, 2), (3, 4), shape=(3, 3))
Out[12]:
array([[1.42936024, 3.30961893, 1.01379663],
       [3.19310627, 1.05386192, 1.11334538],
       [3.2837065 , 1.89239373, 3.35785566]])

Note: This is uniformly distributed over N (non-overlapping) intervals, even if they have different sizes.

Answer 4

You can just assign a probability for how likely it will be [a,b] or [c,d] and then generate accordingly:

import numpy as np
import random

random_roll = random.random()
a = 1
b = 5
c = 7
d = 10
if random_roll > .5: # half the time we will use [a,b]
    my_num = (b - a) * np.random.rand(1) + a
else: # the other half we will use [c,d]
    my_num = (d - c) * np.random.rand(1) + c
print(my_num)