如果您只需要采样而不更换:

>>>随机导入>>>random.sample(范围(1, 100), 3)[77, 52, 45]

random.sample 需要一个总体和一个样本大小k 并返回 k 个随机的总体成员.

如果你必须控制k大于len(population)的情况，你需要准备捕捉ValueError>:

>>>尝试:... random.sample(range(1, 2), 3)... 除了 ValueError:... 打印('样本大小超过人口规模.')...样本量超过人口规模

I know how to generate a random number within a range in Python.

random.randint(numLow, numHigh)

And I know I can put this in a loop to generate n amount of these numbers

for x in range (0, n):
    listOfNumbers.append(random.randint(numLow, numHigh))

However, I need to make sure each number in that list is unique. Other than a load of conditional statements, is there a straightforward way of generating n number of unique random numbers?

The important thing is that each number in the list is different to the others..

So

[12, 5, 6, 1] = good

But

[12, 5, 5, 1] = bad, because the number 5 occurs twice.

If you just need sampling without replacement:

>>> import random
>>> random.sample(range(1, 100), 3)
[77, 52, 45]

random.sample takes a population and a sample size k and returns k random members of the population.

If you have to control for the case where k is larger than len(population), you need to be prepared to catch a ValueError:

>>> try:
...   random.sample(range(1, 2), 3)
... except ValueError:
...   print('Sample size exceeded population size.')
... 
Sample size exceeded population size