Python pick 20 random results from list

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.

Please read http://stackoverflow.com/help/how-to-ask. Show us the code that is not working. — Alexandre Cartapanis, Jun 03 '16 at 08:04
Also with random.choice() it only returns one result, i need 20 results in a new list — Thom, Jun 03 '16 at 08:06
`random.shuffle` returns nothing. He shuffles your source parameter directly — qvpham, Jun 03 '16 at 08:08

score 11 · Accepted Answer · edited Dec 21 '19 at 02:50

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]]

If you need random choice with replacement, use [`random.choices`](https://docs.python.org/library/random.html#random.choices) (new in Python 3.6) instead of using `random.choice` in a list comprehension. — Boris Verkhovskiy, Dec 21 '19 at 02:59

score 6 · Answer 2 · edited Dec 21 '19 at 02:53

6

I think you might be looking for random.sample from the random library. You can use like this:

import random
my_new_list = random.sample(coordinates, 20)

edited Dec 21 '19 at 02:53

Boris Verkhovskiy

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79

answered Jun 03 '16 at 08:08

Leukonoe

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score 2 · Answer 3 · edited Dec 21 '19 at 03:02

2

One way of doing it is to use random.shuffle. Instead of returning a new shuffled list, shuffle modifies the list in place, which is why it wasn't returning anything.

from random import shuffle
shuffle(coordinates)
result = coordinates[:20]

edited Dec 21 '19 at 03:02

Boris Verkhovskiy

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answered Jun 03 '16 at 08:08

Matthias Schreiber

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score 2 · Answer 4 · edited Dec 21 '19 at 03:03

2

random.shuffle shuffles the input parameter directly and returns nothing.

With random.shuffle:

import random
random.shuffle(coordinates)[:20]

But it changes your source list. My opinion: it's not good.

You can use random.sample

import random
random.sample(coordinates, 20)

edited Dec 21 '19 at 03:03

Boris Verkhovskiy

10,733
7
77
79

answered Jun 03 '16 at 08:12

qvpham

1,857
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score -2 · Answer 5 · edited Jun 03 '16 at 10:23

Try the following recipe:

coordinaten = [[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]]

import random

print(random.sample(coordinaten, 20))

Python pick 20 random results from list

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