从不同长度的多个列表中获取随机整数的更有效和通用的方法

Question

What I want is to be able to have mutable lists of mutable length and draw a random number from these lists. The trick is that these lists are different sizes so if a list is smaller than the others a number from it should not come up as often as from the other bigger lists.

Here's what I have now:

 import random

 a = 1
 b = 11
 c = 111
 d = 1111
 e = 11111
 f = 111111
 g = f - e
 q = (f - e) + (d - c) + (b - a)
 r = ((f - e) / q ) * g
 s = g - ((b - a) / q ) * g

 low = 1
 mid = 1
 high = 1

 i = 1
 while i < 666666:
     x = random.randint(a, b)
     y = random.randint(c, d)
     z = random.randint(e, f)
     v = random.randint(1, g)
     if v < r:
         print(z)
         high += 1
     else:
         if v > s:
             print(x)
             low += 1
         else:
             print(y)
             mid += 1
     i += 1

 print(low, mid, high)



  ...
  72027
  81188
  57 6579 660032

It works, but it's very crude and requires me to know which list is the longest, etc. Thanks.

Answer 1

Make a list of all the input lists. Then pick a random number from 0 to the combined length of the lists. Then loop over the incremental lengths of the lists until you get to the one where the index points, and select the element of that list.

l1 = [1, 2, 3, 4, 5]
l2 = [10, 20, 30]
l3 = [100, 101, 105, 107, 111, 125, 130]
l4 = [200, 202, 204, 300, 303, 306, 400, 404, 408]
list_of_lists = [l1, l2, l3, l4]
index = random.randint(0, sum(len(l) for l in list_of_lists))
listnum = 0
prev_length = 0
running_length = len(list_of_lists[listnum])
while index >= running_length:
    prev_length = running_length
    listnum += 1
    running_length += len(list_of_lists[listnum])
result = list_of_lists[listnum][index - prev_length]

Answer 2

Just pick from the concatenaded lists, possibly without reinventing the wheel and avoiding indexing:

import itertools as it
import random as ra

l1 = [1, 2, 3, 4, 5]
l2 = [10, 20, 30]
l3 = [100, 101, 105, 107, 111, 125, 130]
l4 = [200, 202, 204, 300, 303, 306, 400, 404, 408]

res = ra.choice(list(it.chain(l1, l2, l3, l4)))
print(res)

Note: if the number of lists varies, make a list of them and pass it to chain.from_iterable