如何创建一个包含 50 个随机数的列表，其中偶数是奇数的两倍？

Question

can you help me with this problem please? I have to create a list of 50 random numbers from 1 to 10 where the odd numbers frequency is aproximately twice the even number frequency.

Answer 1

Try two list, one with 33 random odd numbers and the other with 17 random even numbers, and then merge the two lists.

Answer 2

Solution as simplified

import random
odd=[item for item in range(1,11,2)]
even=[item for item in range(2,11,2)]
# twice chance 50/3 ~17   odd=33  even=17
random_even=[random.choice(even) for _ in range(17)]
random_odd=[random.choice(odd) for _ in range(33)]
random_number=random_even+random_odd # sum 
random.shuffle(random_number) # messing up order
print("even",len(random_even),"odd",len(random_odd)) #check len
print("len list  : ",len(random_number))#check len
print(random_number)#show list

Answer 3

random.choices has an argument weights , which can makes the odd number roughly twice as many as the even number:

>>> nums = range(1, 11)
>>> random.choices(nums, weights=[2 / 3, 1 / 3] * 5, k=50)
[7, 4, 4, 2, 1, 1, 7, 9, 6, 9, 3, 9, 9, 5, 9, 3, 2, 1, 9, 10, 9, 3, 7, 5, 3, 8, 8, 10, 9, 5, 2, 9, 9, 6, 5, 5, 10, 5, 6, 5, 1, 3, 3, 2, 5, 6, 5, 5, 5, 2]
>>> sum(i % 2 == 0 for i in _)
16
>>> (50 - 16) / 16
2.125

If you have strict requirements on their quantitative relationship, you can generate two lists, concatenate and shuffle:

>>> n_even = 50 // 3
>>> n_odd = 50 - n_even
>>> rand = random.choices(nums[::2], k=n_odd) + random.choices(nums[1::2], k=n_even)
>>> random.shuffle(rand)
>>> rand
[2, 10, 5, 8, 7, 10, 3, 5, 1, 7, 8, 10, 7, 5, 3, 9, 3, 10, 3, 2, 1, 10, 9, 8, 7, 6, 1, 1, 8, 1, 5, 4, 9, 9, 2, 1, 5, 1, 4, 7, 1, 2, 5, 7, 7, 1, 7, 1, 1, 5]