Python pick 20 random results from list [duplicate]

Question

I would like to get 20 random results from the following list:

coordinates = [
  [20, 140], [40, 140], [60, 140], [80, 140], [100, 140], [120, 140],
  [20, 120], [40, 120], [60, 120], [80, 120], [100, 120], [120, 120],
  [20, 100], [40, 100], [60, 100], [80, 100], [100, 100], [120, 100],
  [20, 80], [40, 80], [60, 80], [80, 80], [100, 80], [120, 80],
  [20, 60], [40, 60], [60, 60], [80, 60], [100, 60], [120, 60],
  [20, 40], [40, 40], [60, 40], [80, 40], [100, 40], [120, 40]
]

I tried random.shuffle but it returns None.

Answer 1

If you want 20 unique values in random order, use random.sample():

random.sample(coordinates, 20)

random.sample(population, k)¶

Return a k length list of unique elements chosen from the population sequence or set. Used for random sampling without replacement.

>>> random.sample(coordinates, 20)
[[80, 60], [40, 100], [80, 100], [60, 80], [60, 100], [40, 60], [40, 80], [80, 120], [120, 140], [120, 100], [100, 80], [40, 120], [80, 140], [100, 140], [20, 80], [120, 80], [100, 100], [20, 40], [120, 120], [100, 120]]

---

You could use random.choice() 20 times, but this will not be "unique"—elements may be duplicated, because one is randomly selected each time:

>>> [random.choice(coordinates) for _ in range(20)]
[[80, 80], [40, 140], [80, 140], [60, 60], [120, 100], [20, 120], [100, 80], [120, 100], [20, 60], [100, 120], [100, 40], [80, 80], [100, 80], [80, 120], [20, 40], [100, 80], [60, 80], [80, 140], [40, 40], [120, 40]]