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)
secrets
module was added to Python standard library in version 3.6 python.org/dev/peps/pep-0506
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([])
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.
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)
random.choices
is with replacement while random.sample
is without replacement.
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
secrets
module from other answers such as the one from Pēteris Caune! And a working link to documentation for numpy.random.choice
: numpy.org/doc/stable/reference/random/generated/…
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:)
random.choice(self, seq)
is return seq[int(self.random() * len(seq))]
.
randrange()
which means e.g. random.SystemRandom().randrange(3<<51)
exhibits significant bias. Sigh...
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.)
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.
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.
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
I usually use the random module for working with lists and randomization
import random
foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))
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
AttributeError: 'module' object has no attribute 'seed'
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)
random.choice()
and random.randrange()
?
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.
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.
random.choice
?
random.randint
has inclusive boundaries, so len(foo)
is a possible value, which would raise an IndexError
random.randrange
was added). I am just simply aware of the semantics of the random
module, and how it would work here. Or perhaps the intention of this code is to produce an IndexError
?
random.choice
. So no, I will not be modifying my perceptions.
Success story sharing
random.choice(foo)
return two different results?random.sample(lst, n)
Standard pseudo-random generators are not suitable for security/cryptographic purposes.
ref