我如何洗牌对象列表?我尝试了random.shuffle:

import random

b = [object(), object()]

print(random.shuffle(b))

但它输出:

None

当前回答

""" to shuffle random, set random= True """

def shuffle(x,random=False):
     shuffled = []
     ma = x
     if random == True:
         rando = [ma[i] for i in np.random.randint(0,len(ma),len(ma))]
         return rando
     if random == False:
          for i in range(len(ma)):
          ave = len(ma)//3
          if i < ave:
             shuffled.append(ma[i+ave])
          else:
             shuffled.append(ma[i-ave])    
     return shuffled

其他回答

def shuffle(_list):
    if not _list == []:
        import random
        list2 = []
        while _list != []:
            card = random.choice(_list)
            _list.remove(card)
            list2.append(card)
        while list2 != []:
            card1 = list2[0]
            list2.remove(card1)
            _list.append(card1)
        return _list
import random
class a:
    foo = "bar"

a1 = a()
a2 = a()
b = [a1.foo,a2.foo]
random.shuffle(b)

你可以这样做:

>>> A = ['r','a','n','d','o','m']
>>> B = [1,2,3,4,5,6]
>>> import random
>>> random.sample(A+B, len(A+B))
[3, 'r', 4, 'n', 6, 5, 'm', 2, 1, 'a', 'o', 'd']

如果您想返回到两个列表,那么您可以将这个长列表拆分为两个。

你可以使用shuffle或者sample。它们都来自于随机模块。

import random
def shuffle(arr1):
    n=len(arr1)
    b=random.sample(arr1,n)
    return b

OR

import random
def shuffle(arr1):
    random.shuffle(arr1)
    return arr1

它工作得很好。我在这里尝试用函数作为列表对象:

    from random import shuffle

    def foo1():
        print "foo1",

    def foo2():
        print "foo2",

    def foo3():
        print "foo3",

    A=[foo1,foo2,foo3]

    for x in A:
        x()

    print "\r"

    shuffle(A)
    for y in A:
        y()

它打印出: Foo1 foo2 foo3 Foo2 foo3 foo1 (最后一行的傻瓜是随机排列的)