1972

Assume I have the following list:

foo = ['a', 'b', 'c', 'd', 'e']

What is the simplest way to retrieve an item at random from this list?

martineau
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Ray
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    Before you post a new answer, consider there are already 10+ answers for this question. Please, make sure that your answer contributes information that is not among existing answers. – Nine Tails Oct 01 '20 at 07:04

15 Answers15

2961

Use random.choice()

import random

foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))

For cryptographically secure random choices (e.g. for generating a passphrase from a wordlist) use secrets.choice()

import secrets

foo = ['battery', 'correct', 'horse', 'staple']
print(secrets.choice(foo))

secrets is new in Python 3.6, on older versions of Python you can use the random.SystemRandom class:

import random

secure_random = random.SystemRandom()
print(secure_random.choice(foo))
Boris
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Pēteris Caune
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    Does making two consecutive calls of `random.choice(foo)` return two different results? – Eduardo Pignatelli Dec 07 '18 at 17:13
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    @EduardoPignatelli Each choice is random, so it *can* return two different results, but depending on the start seed, it's not guaranteed. If you want to select *n* distinct random elements from a list *lst*, use [`random.sample(lst, n)`](https://docs.python.org/library/random.html#random.sample) – Graham Dec 23 '18 at 17:02
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    on a related note, `Standard pseudo-random generators are not suitable for security/cryptographic purposes.` [ref](https://bandit.readthedocs.io/en/latest/blacklists/blacklist_calls.html#b311-random) – XoXo Jan 30 '19 at 19:46
231

If you want to randomly select more than one item from a list, or select an item from a set, I'd recommend using random.sample instead.

import random
group_of_items = {1, 2, 3, 4}               # a sequence or set will work here.
num_to_select = 2                           # set the number to select here.
list_of_random_items = random.sample(group_of_items, num_to_select)
first_random_item = list_of_random_items[0]
second_random_item = list_of_random_items[1] 

If you're only pulling a single item from a list though, choice is less clunky, as using sample would have the syntax random.sample(some_list, 1)[0] instead of random.choice(some_list).

Unfortunately though, choice only works for a single output from sequences (such as lists or tuples). Though random.choice(tuple(some_set)) may be an option for getting a single item from a set.

EDIT: Using Secrets

As many have pointed out, if you require more secure pseudorandom samples, you should use the secrets module:

import secrets                              # imports secure module.
secure_random = secrets.SystemRandom()      # creates a secure random object.
group_of_items = {1, 2, 3, 4}               # a sequence or set will work here.
num_to_select = 2                           # set the number to select here.
list_of_random_items = secure_random.sample(group_of_items, num_to_select)
first_random_item = list_of_random_items[0]
second_random_item = list_of_random_items[1]

EDIT: Pythonic One-Liner

If you want a more pythonic one-liner for selecting multiple items, you can use unpacking.

import random
first_random_item, second_random_item = random.sample(group_of_items, 2)
Paul
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    BTW `secrets` module was added to Python standard library in version 3.6 https://www.python.org/dev/peps/pep-0506/ – and1er Dec 13 '19 at 06:53
170

If you also need the index, use random.randrange

from random import randrange
random_index = randrange(len(foo))
print(foo[random_index])
Boris
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Juampi
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46

As of Python 3.6 you can use the secrets module, which is preferable to the random module for cryptography or security uses.

To print a random element from a list:

import secrets
foo = ['a', 'b', 'c', 'd', 'e']
print(secrets.choice(foo))

To print a random index:

print(secrets.randbelow(len(foo)))

For details, see PEP 506.

Chris_Rands
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35

I propose a script for removing randomly picked up items off a list until it is empty:

Maintain a set and remove randomly picked up element (with choice) until list is empty.

s=set(range(1,6))
import random

while len(s)>0:
  s.remove(random.choice(list(s)))
  print(s)

Three runs give three different answers:

>>> 
set([1, 3, 4, 5])
set([3, 4, 5])
set([3, 4])
set([4])
set([])
>>> 
set([1, 2, 3, 5])
set([2, 3, 5])
set([2, 3])
set([2])
set([])

>>> 
set([1, 2, 3, 5])
set([1, 2, 3])
set([1, 2])
set([1])
set([])
Community
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kiriloff
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    Or you could just `random.shuffle` the `list` once and either iterate it or pop it to produce results. Either would result in a perfectly adequate "select randomly with no repeats" stream, it's just that the randomness would be introduced at the beginning. – ShadowRanger Dec 25 '15 at 03:23
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    Theoretically you can use the [pop()](https://docs.python.org/3.4/library/stdtypes.html#set.pop) method of a set to remove an arbitrary element from a set and return it, but it's probably not random enough. – Joubarc Jun 17 '16 at 07:07
22
foo = ['a', 'b', 'c', 'd', 'e']
number_of_samples = 1

In python 2:

random_items = random.sample(population=foo, k=number_of_samples)

In python 3:

random_items = random.choices(population=foo, k=number_of_samples)
Fardin
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18

numpy solution: numpy.random.choice

For this question, it works the same as the accepted answer (import random; random.choice()), but I added it because the programmer may have imported numpy already (like me) & also there are some differences between the two methods that may concern your actual use case.

import numpy as np    
np.random.choice(foo) # randomly selects a single item

For reproducibility, you can do:

np.random.seed(123)
np.random.choice(foo) # first call will always return 'c'

For samples of one or more items, returned as an array, pass the size argument:

np.random.choice(foo, 5)          # sample with replacement (default)
np.random.choice(foo, 5, False)   # sample without replacement
C8H10N4O2
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11

if you need the index just use:

import random
foo = ['a', 'b', 'c', 'd', 'e']
print int(random.random() * len(foo))
print foo[int(random.random() * len(foo))]

random.choice does the same:)

Janek Olszak
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    @tc. Actually, it does do essentially the same. Implementation of `random.choice(self, seq)` is `return seq[int(self.random() * len(seq))]`. – wim Apr 12 '13 at 04:56
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    @wim That's a little disappointing, but the *very* disappointing thing is that's also the definition of `randrange()` which means e.g. `random.SystemRandom().randrange(3<<51)` exhibits significant bias. *Sigh...* – tc. Apr 13 '13 at 23:55
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    @kevinsa5 Ultimately it's because a `float` (an IEEE double) can only take a finite number of values in [0,1). `Random.random()` generates its output in the traditional way: pick a random integer in `[0, 2**53)` and divide by `2**53` (53 is the number of bits in a double). So `random()` returns 2**53 equiprobable doubles, and you can divide this evenly into N outputs only if N is a power of 2. The bias is small for small N, but see `collections.Counter(random.SystemRandom().randrange(3<<51)%6 for i in range(100000)).most_common()`. (Java's Random.nextInt() avoids such bias.) – tc. Jan 18 '14 at 16:37
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    @tc. I suppose anything less than about `2**40`, (which is 1099511627776), would be small enough for the bias to not matter in practice? This should really be pointed out in the documentation, because if somebody is not meticulous, they might not expect problems to come from this part of their code. – Evgeni Sergeev May 29 '15 at 01:28
  • @tc.: Actually, `random` uses `getrandbits` to get an adequate number of bits to generate a result for larger `randrange`s (`random.choice` is also using that). This is true on both 2.7 and 3.5. It only uses `self.random() * len(seq)` when `getrandbits` is not available. It's not doing the stupid thing you think it is. – ShadowRanger Dec 25 '15 at 03:32
11

How to randomly select an item from a list?

Assume I have the following list:

foo = ['a', 'b', 'c', 'd', 'e']  

What is the simplest way to retrieve an item at random from this list?

If you want close to truly random, then I suggest secrets.choice from the standard library (New in Python 3.6.):

>>> from secrets import choice         # Python 3 only
>>> choice(list('abcde'))
'c'

The above is equivalent to my former recommendation, using a SystemRandom object from the random module with the choice method - available earlier in Python 2:

>>> import random                      # Python 2 compatible
>>> sr = random.SystemRandom()
>>> foo = list('abcde')
>>> foo
['a', 'b', 'c', 'd', 'e']

And now:

>>> sr.choice(foo)
'd'
>>> sr.choice(foo)
'e'
>>> sr.choice(foo)
'a'
>>> sr.choice(foo)
'b'
>>> sr.choice(foo)
'a'
>>> sr.choice(foo)
'c'
>>> sr.choice(foo)
'c'

If you want a deterministic pseudorandom selection, use the choice function (which is actually a bound method on a Random object):

>>> random.choice
<bound method Random.choice of <random.Random object at 0x800c1034>>

It seems random, but it's actually not, which we can see if we reseed it repeatedly:

>>> random.seed(42); random.choice(foo), random.choice(foo), random.choice(foo)
('d', 'a', 'b')
>>> random.seed(42); random.choice(foo), random.choice(foo), random.choice(foo)
('d', 'a', 'b')
>>> random.seed(42); random.choice(foo), random.choice(foo), random.choice(foo)
('d', 'a', 'b')
>>> random.seed(42); random.choice(foo), random.choice(foo), random.choice(foo)
('d', 'a', 'b')
>>> random.seed(42); random.choice(foo), random.choice(foo), random.choice(foo)
('d', 'a', 'b')

A comment:

This is not about whether random.choice is truly random or not. If you fix the seed, you will get the reproducible results -- and that's what seed is designed for. You can pass a seed to SystemRandom, too. sr = random.SystemRandom(42)

Well, yes you can pass it a "seed" argument, but you'll see that the SystemRandom object simply ignores it:

def seed(self, *args, **kwds):
    "Stub method.  Not used for a system random number generator."
    return None
Aaron Hall
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  • This is just pedantic, but secrets isn't truly random, it's cryptographically secure pseudorandom. – Paul Jan 10 '21 at 22:07
8

This is the code with a variable that defines the random index:

import random

foo = ['a', 'b', 'c', 'd', 'e']
randomindex = random.randint(0,len(foo)-1) 
print (foo[randomindex])
## print (randomindex)

This is the code without the variable:

import random

foo = ['a', 'b', 'c', 'd', 'e']
print (foo[random.randint(0,len(foo)-1)])

And this is the code in the shortest and smartest way to do it:

import random

foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))

(python 2.7)

Will Dereham
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Liam
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4

Random item selection:

import random

my_list = [1, 2, 3, 4, 5]
num_selections = 2

new_list = random.sample(my_list, num_selections)

To preserve the order of the list, you could do:

randIndex = random.sample(range(len(my_list)), n_selections)
randIndex.sort()
new_list = [my_list[i] for i in randIndex]

Duplicate of https://stackoverflow.com/a/49682832/4383027

Solomon Vimal
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The following code demonstrates if you need to produce the same items. You can also specify how many samples you want to extract.
The sample method returns a new list containing elements from the population while leaving the original population unchanged. The resulting list is in selection order so that all sub-slices will also be valid random samples.

import random as random
random.seed(0)  # don't use seed function, if you want different results in each run
print(random.sample(foo,3))  # 3 is the number of sample you want to retrieve

Output:['d', 'e', 'a']
Memin
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3

You could just:

from random import randint

foo = ["a", "b", "c", "d", "e"]

print(foo[randint(0,4)])
1

This may already be an answer but you could use random.shuffle. Example:

import random
foo = ['a', 'b', 'c', 'd', 'e']
random.shuffle(foo)
Scott
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Jax
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-2

We can also do this using randint.

from random import randint
l= ['a','b','c']

def get_rand_element(l):
    if l:
        return l[randint(0,len(l)-1)]
    else:
        return None

get_rand_element(l)
Liam
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Abdul Majeed
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    Why on earth would you do it this way, when there's `random.choice()` and `random.randrange()`? – alexis Mar 19 '16 at 15:10
  • "random.choice()" will give you "IndexError: list index out of range" on empty list. – Abdul Majeed Sep 05 '17 at 05:45
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    As it should: that's what exceptions are for. Choosing from an empty list is an error. Returning `None` just kicks the can to some random later point where the invalid "element" triggers an exception; or worse yet, you get an incorrect program instead of an exception, and you don't even know it. – alexis Sep 05 '17 at 09:27