问题:如何从列表中随机选择一个项目?
假设我有以下列表:
foo = ['a', 'b', 'c', 'd', 'e']
从此列表中随机检索项目的最简单方法是什么?
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?
回答 0
采用 random.choice()
import random
foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))
对于密码安全的随机选择(例如,用于从单词列表生成密码短语),请使用secrets.choice()
import secrets
foo = ['battery', 'correct', 'horse', 'staple']
print(secrets.choice(foo))
secrets
是Python 3.6中的新功能,在旧版本的Python上,您可以使用random.SystemRandom
此类:
import random
secure_random = random.SystemRandom()
print(secure_random.choice(foo))
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))
回答 1
如果您想从列表中随机选择一个以上的项目,或者从一组中选择一个项目,则建议random.sample
改用。
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]
如果您只是从列表中拉出一个项目,那么选择就不会那么笨拙,因为使用sample的语法将random.sample(some_list, 1)[0]
是random.choice(some_list)
。
但是不幸的是,选择仅适用于序列(例如列表或元组)中的单个输出。虽然random.choice(tuple(some_set))
可能是从集合中获取单个项目的选项。
编辑:使用秘密
正如许多人指出的那样,如果需要更安全的伪随机样本,则应使用secrets模块:
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]
编辑:Pythonic一线
如果您希望使用更具Python风格的单行代码来选择多个项目,则可以使用拆包。
import random
first_random_item, second_random_item = random.sample(group_of_items, 2)
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)
回答 2
如果您还需要索引,请使用 random.randrange
from random import randrange
random_index = randrange(len(foo))
print(foo[random_index])
If you also need the index, use random.randrange
from random import randrange
random_index = randrange(len(foo))
print(foo[random_index])
回答 3
从Python 3.6开始,您可以使用该secrets
模块,该random
模块比加密或安全用途的模块更好。
要从列表中打印随机元素:
import secrets
foo = ['a', 'b', 'c', 'd', 'e']
print(secrets.choice(foo))
要打印随机索引:
print(secrets.randbelow(len(foo)))
有关详细信息,请参阅PEP 506。
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.
回答 4
我提出了一个脚本,用于从列表中删除随机拾取的项目,直到它为空:
维持set
并删除随机拾取的元素(带有choice
),直到列表为空。
s=set(range(1,6))
import random
while len(s)>0:
s.remove(random.choice(list(s)))
print(s)
三个运行给出三个不同的答案:
>>>
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([])
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([])
回答 5
foo = ['a', 'b', 'c', 'd', 'e']
number_of_samples = 1
在python 2:
random_items = random.sample(population=foo, k=number_of_samples)
在python 3:
random_items = random.choices(population=foo, k=number_of_samples)
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)
回答 6
numpy
解: numpy.random.choice
对于这个问题,它的作用与接受的答案(import random; random.choice()
)相同,但是我添加了它,因为程序员可能已经导入numpy
了(像我一样),并且这两种方法之间可能存在一些差异,这可能与您的实际用例有关。
import numpy as np
np.random.choice(foo) # randomly selects a single item
为了重现性,您可以执行以下操作:
np.random.seed(123)
np.random.choice(foo) # first call will always return 'c'
对于以形式返回的一个或多个项目的样本array
,请传递size
参数:
np.random.choice(foo, 5) # sample with replacement (default)
np.random.choice(foo, 5, False) # sample 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)
& 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
回答 7
如何从列表中随机选择一个项目?
假设我有以下列表:
foo = ['a', 'b', 'c', 'd', 'e']
从此列表中随机检索项目的最简单方法是什么?
如果您想接近真正的随机性,那么我建议secrets.choice
从标准库(Python 3.6中的新增功能)中进行建议:
>>> from secrets import choice # Python 3 only
>>> choice(list('abcde'))
'c'
上面的内容等同于我以前的建议,即使用模块中的SystemRandom
对象random
和choice
方法-早于Python 2:
>>> import random # Python 2 compatible
>>> sr = random.SystemRandom()
>>> foo = list('abcde')
>>> foo
['a', 'b', 'c', 'd', 'e']
现在:
>>> 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'
如果需要确定性伪随机选择,请使用choice
函数(实际上是Random
对象上的绑定方法):
>>> random.choice
<bound method Random.choice of <random.Random object at 0x800c1034>>
看来是随机的,但实际上不是,我们可以看看是否反复播种:
>>> 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')
一条评论:
这与random.choice是否真正随机无关。如果修复种子,您将获得可重复的结果-这就是种子的设计目的。您也可以将种子传递给SystemRandom。sr = random.SystemRandom(42)
好吧,是的,您可以给它传递一个“种子”参数,但是您会看到该SystemRandom
对象只是忽略了它:
def seed(self, *args, **kwds):
"Stub method. Not used for a system random number generator."
return None
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')
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
回答 8
如果您需要索引,请使用:
import random
foo = ['a', 'b', 'c', 'd', 'e']
print int(random.random() * len(foo))
print foo[int(random.random() * len(foo))]
random.choice做同样的事情:)
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:)
回答 9
这是带有定义随机索引的变量的代码:
import random
foo = ['a', 'b', 'c', 'd', 'e']
randomindex = random.randint(0,len(foo)-1)
print (foo[randomindex])
## print (randomindex)
这是没有变量的代码:
import random
foo = ['a', 'b', 'c', 'd', 'e']
print (foo[random.randint(0,len(foo)-1)])
这是用最短和最聪明的方法实现的代码:
import random
foo = ['a', 'b', 'c', 'd', 'e']
print(random.choice(foo))
(python 2.7)
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)
回答 10
以下代码演示了是否需要生产相同的物品。您还可以指定要提取的样本数量。
该sample
方法返回一个新列表,其中包含总体中的元素,而保留原始总体不变。结果列表按选择顺序排列,因此所有子切片也将是有效的随机样本。
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']
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']
回答 11
随机项目选择:
import random
my_list = [1, 2, 3, 4, 5]
num_selections = 2
new_list = random.sample(my_list, num_selections)
要保留列表的顺序,您可以执行以下操作:
randIndex = random.sample(range(len(my_list)), n_selections)
randIndex.sort()
new_list = [my_list[i] for i in randIndex]
重复的https://stackoverflow.com/a/49682832/4383027
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
回答 12
我们也可以使用randint做到这一点。
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)
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)
回答 13
您可以:
from random import randint
foo = ["a", "b", "c", "d", "e"]
print(foo[randint(0,4)])
You could just:
from random import randint
foo = ["a", "b", "c", "d", "e"]
print(foo[randint(0,4)])