How can I 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?

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))

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 = {'a', 'b', 'c', 'd', 'e'}  # 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 = {'a', 'b', 'c', 'd', 'e'}  # 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({'a', 'b', 'c', 'd', 'e'}, 2)

If you also need the index, use random.randrange

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

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.

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([])

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)

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)

And 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

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:)

> ## 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

In short, use random.sample method

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
lst = ['a', 'b', 'c', 'd', 'e']
random.seed(0)  # remove this line, if you want different results for each run
rand_lst = random.sample(lst,3)  # 3 is the number of sample you want to retrieve
print(rand_lst)

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

here is a running code https://onecompiler.com/python/3xem5jjvz

I usually use the random module for working with lists and randomization

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

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)

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

You could just:

from random import randint

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

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

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

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

The recommended numpy way is now to use an explicit RNG:

from numpy.random import default_rng

rng = default_rng()
rng.choice(foo)

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)

You can randomly print/pick an element from the given dictionary by randomising the slicing arguments using the slicing method. But make sure you have imported the random module, thats the big boy here.

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

Here we are setting a random number in the slice argument. Any number given in the foo[] would return the element attaining the index similar to the number given in slice. Just setting a random number there would do you job.