改组对象列表

问题:改组对象列表

我有一个对象列表,我想对其进行洗牌。我以为可以使用该random.shuffle方法,但是当列表中包含对象时,这似乎失败了。是否有一种用于改组对象的方法或解决此问题的另一种方法?

import random

class A:
    foo = "bar"

a1 = a()
a2 = a()
b = [a1, a2]

print(random.shuffle(b))

这将失败。

I have a list of objects and I want to shuffle them. I thought I could use the random.shuffle method, but this seems to fail when the list is of objects. Is there a method for shuffling objects or another way around this?

import random

class A:
    foo = "bar"

a1 = a()
a2 = a()
b = [a1, a2]

print(random.shuffle(b))

This will fail.


回答 0

random.shuffle应该管用。这是一个示例,其中对象是列表:

from random import shuffle
x = [[i] for i in range(10)]
shuffle(x)

# print(x)  gives  [[9], [2], [7], [0], [4], [5], [3], [1], [8], [6]]
# of course your results will vary

请注意,随机播放在适当的地方起作用,并返回None。

random.shuffle should work. Here’s an example, where the objects are lists:

from random import shuffle
x = [[i] for i in range(10)]
shuffle(x)

# print(x)  gives  [[9], [2], [7], [0], [4], [5], [3], [1], [8], [6]]
# of course your results will vary

Note that shuffle works in place, and returns None.


回答 1

当您了解到就地改组就是问题所在。我也经常遇到问题,而且似乎也常常忘记如何复制列表。使用sample(a, len(a))是解决方案,使用len(a)作为样本量。有关Python文档,请参见https://docs.python.org/3.6/library/random.html#random.sample

这是使用的简单版本random.sample(),它将经过改组的结果作为新列表返回。

import random

a = range(5)
b = random.sample(a, len(a))
print a, b, "two list same:", a == b
# print: [0, 1, 2, 3, 4] [2, 1, 3, 4, 0] two list same: False

# The function sample allows no duplicates.
# Result can be smaller but not larger than the input.
a = range(555)
b = random.sample(a, len(a))
print "no duplicates:", a == list(set(b))

try:
    random.sample(a, len(a) + 1)
except ValueError as e:
    print "Nope!", e

# print: no duplicates: True
# print: Nope! sample larger than population

As you learned the in-place shuffling was the problem. I also have problem frequently, and often seem to forget how to copy a list, too. Using sample(a, len(a)) is the solution, using len(a) as the sample size. See https://docs.python.org/3.6/library/random.html#random.sample for the Python documentation.

Here’s a simple version using random.sample() that returns the shuffled result as a new list.

import random

a = range(5)
b = random.sample(a, len(a))
print a, b, "two list same:", a == b
# print: [0, 1, 2, 3, 4] [2, 1, 3, 4, 0] two list same: False

# The function sample allows no duplicates.
# Result can be smaller but not larger than the input.
a = range(555)
b = random.sample(a, len(a))
print "no duplicates:", a == list(set(b))

try:
    random.sample(a, len(a) + 1)
except ValueError as e:
    print "Nope!", e

# print: no duplicates: True
# print: Nope! sample larger than population

回答 2

我也花了一些时间来做到这一点。但是洗牌的文档非常清楚:

列表随机排列x ; 不返回。

所以你不应该print(random.shuffle(b))。相反random.shuffle(b),然后print(b)

It took me some time to get that too. But the documentation for shuffle is very clear:

shuffle list x in place; return None.

So you shouldn’t print(random.shuffle(b)). Instead do random.shuffle(b) and then print(b).


回答 3

#!/usr/bin/python3

import random

s=list(range(5))
random.shuffle(s) # << shuffle before print or assignment
print(s)

# print: [2, 4, 1, 3, 0]
#!/usr/bin/python3

import random

s=list(range(5))
random.shuffle(s) # << shuffle before print or assignment
print(s)

# print: [2, 4, 1, 3, 0]

回答 4

如果您碰巧已经使用numpy(在科学和金融应用中非常流行),则可以节省导入时间。

import numpy as np    
np.random.shuffle(b)
print(b)

http://docs.scipy.org/doc/numpy/reference/generation/numpy.random.shuffle.html

If you happen to be using numpy already (very popular for scientific and financial applications) you can save yourself an import.

import numpy as np    
np.random.shuffle(b)
print(b)

http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.shuffle.html


回答 5

>>> import random
>>> a = ['hi','world','cat','dog']
>>> random.shuffle(a,random.random)
>>> a
['hi', 'cat', 'dog', 'world']

这对我来说可以。确保设置随机方法。

>>> import random
>>> a = ['hi','world','cat','dog']
>>> random.shuffle(a,random.random)
>>> a
['hi', 'cat', 'dog', 'world']

It works fine for me. Make sure to set the random method.


回答 6

如果您有多个列表,则可能要先定义排列(随机排列列表/重新排列列表中项目的方式),然后将其应用于所有列表:

import random

perm = list(range(len(list_one)))
random.shuffle(perm)
list_one = [list_one[index] for index in perm]
list_two = [list_two[index] for index in perm]

脾气暴躁

如果您的列表是numpy数组,则更为简单:

import numpy as np

perm = np.random.permutation(len(list_one))
list_one = list_one[perm]
list_two = list_two[perm]

处理器

我创建了mpu具有以下consistent_shuffle功能的小型实用程序包:

import mpu

# Necessary if you want consistent results
import random
random.seed(8)

# Define example lists
list_one = [1,2,3]
list_two = ['a', 'b', 'c']

# Call the function
list_one, list_two = mpu.consistent_shuffle(list_one, list_two)

请注意,它mpu.consistent_shuffle接受任意数量的参数。因此,您也可以使用它洗牌三个或更多列表。

If you have multiple lists, you might want to define the permutation (the way you shuffle the list / rearrange the items in the list) first and then apply it to all lists:

import random

perm = list(range(len(list_one)))
random.shuffle(perm)
list_one = [list_one[index] for index in perm]
list_two = [list_two[index] for index in perm]

Numpy / Scipy

If your lists are numpy arrays, it is simpler:

import numpy as np

perm = np.random.permutation(len(list_one))
list_one = list_one[perm]
list_two = list_two[perm]

mpu

I’ve created the small utility package mpu which has the consistent_shuffle function:

import mpu

# Necessary if you want consistent results
import random
random.seed(8)

# Define example lists
list_one = [1,2,3]
list_two = ['a', 'b', 'c']

# Call the function
list_one, list_two = mpu.consistent_shuffle(list_one, list_two)

Note that mpu.consistent_shuffle takes an arbitrary number of arguments. So you can also shuffle three or more lists with it.


回答 7

from random import random
my_list = range(10)
shuffled_list = sorted(my_list, key=lambda x: random())

对于要交换订购功能的某些应用程序,此替代方法可能很有用。

from random import random
my_list = range(10)
shuffled_list = sorted(my_list, key=lambda x: random())

This alternative may be useful for some applications where you want to swap the ordering function.


回答 8

在某些情况下,使用numpy数组时,请random.shuffle在数组中使用创建的重复数据。

另一种方法是使用numpy.random.shuffle。如果您已经在使用numpy,那么这是优于generic的首选方法random.shuffle

numpy.random.shuffle

>>> import numpy as np
>>> import random

使用random.shuffle

>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])


>>> random.shuffle(foo)
>>> foo

array([[1, 2, 3],
       [1, 2, 3],
       [4, 5, 6]])

使用numpy.random.shuffle

>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])


>>> np.random.shuffle(foo)
>>> foo

array([[1, 2, 3],
       [7, 8, 9],
       [4, 5, 6]])

In some cases when using numpy arrays, using random.shuffle created duplicate data in the array.

An alternative is to use numpy.random.shuffle. If you’re working with numpy already, this is the preferred method over the generic random.shuffle.

numpy.random.shuffle

Example

>>> import numpy as np
>>> import random

Using random.shuffle:

>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])


>>> random.shuffle(foo)
>>> foo

array([[1, 2, 3],
       [1, 2, 3],
       [4, 5, 6]])

Using numpy.random.shuffle:

>>> foo = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> foo

array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9]])


>>> np.random.shuffle(foo)
>>> foo

array([[1, 2, 3],
       [7, 8, 9],
       [4, 5, 6]])

回答 9

对于单行代码,请使用random.sample(list_to_be_shuffled, length_of_the_list)示例:

import random
random.sample(list(range(10)), 10)

输出:[2、9、7、8、3、0、4、1、6、5]

For one-liners, userandom.sample(list_to_be_shuffled, length_of_the_list) with an example:

import random
random.sample(list(range(10)), 10)

outputs: [2, 9, 7, 8, 3, 0, 4, 1, 6, 5]


回答 10

当使用’foo’调用时,’print func(foo)’将输出’func’的返回值。但是,’shuffle’的返回类型为None,因为该列表将被修改,因此不打印任何内容。解决方法:

# shuffle the list in place 
random.shuffle(b)

# print it
print(b)

如果您更喜欢函数式编程风格,则可能需要创建以下包装函数:

def myshuffle(ls):
    random.shuffle(ls)
    return ls

‘print func(foo)’ will print the return value of ‘func’ when called with ‘foo’. ‘shuffle’ however has None as its return type, as the list will be modified in place, hence it prints nothing. Workaround:

# shuffle the list in place 
random.shuffle(b)

# print it
print(b)

If you’re more into functional programming style you might want to make the following wrapper function:

def myshuffle(ls):
    random.shuffle(ls)
    return ls

回答 11

可以定义一个函数shuffledsort与vs 相同sorted

def shuffled(x):
    import random
    y = x[:]
    random.shuffle(y)
    return y

x = shuffled([1, 2, 3, 4])
print x

One can define a function called shuffled (in the same sense of sort vs sorted)

def shuffled(x):
    import random
    y = x[:]
    random.shuffle(y)
    return y

x = shuffled([1, 2, 3, 4])
print x

回答 12

import random

class a:
    foo = "bar"

a1 = a()
a2 = a()
a3 = a()
a4 = a()
b = [a1,a2,a3,a4]

random.shuffle(b)
print(b)

shuffle 到位,因此不要打印结果None,而是列表。

import random

class a:
    foo = "bar"

a1 = a()
a2 = a()
a3 = a()
a4 = a()
b = [a1,a2,a3,a4]

random.shuffle(b)
print(b)

shuffle is in place, so do not print result, which is None, but the list.


回答 13

您可以这样做:

>>> 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']

如果要返回到两个列表,则可以将此长列表分成两部分。

You can go for this:

>>> 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']

if you want to go back to two lists, you then split this long list into two.


回答 14

您可以构建一个将列表作为参数并返回列表的随机版本的函数:

from random import *

def listshuffler(inputlist):
    for i in range(len(inputlist)):
        swap = randint(0,len(inputlist)-1)
        temp = inputlist[swap]
        inputlist[swap] = inputlist[i]
        inputlist[i] = temp
    return inputlist

you could build a function that takes a list as a parameter and returns a shuffled version of the list:

from random import *

def listshuffler(inputlist):
    for i in range(len(inputlist)):
        swap = randint(0,len(inputlist)-1)
        temp = inputlist[swap]
        inputlist[swap] = inputlist[i]
        inputlist[i] = temp
    return inputlist

回答 15

""" 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
""" 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

回答 16

您可以使用随机播放或采样。两者均来自随机模块。

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

要么

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

you can either use shuffle or sample . both of which come from random module.

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

回答 17

确保您没有命名源文件random.py,并且工作目录中没有名为random.pyc ..的文件,这可能会导致程序尝试导入本地random.py文件而不是pythons random模块。

Make sure you are not naming your source file random.py, and that there is not a file in your working directory called random.pyc.. either could cause your program to try and import your local random.py file instead of pythons random module.


回答 18

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
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

回答 19

import random
class a:
    foo = "bar"

a1 = a()
a2 = a()
b = [a1.foo,a2.foo]
random.shuffle(b)
import random
class a:
    foo = "bar"

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

回答 20

改组过程是“有替换的”,因此每个项目的出现可能会改变!至少当列表中的项目也同时列出时。

例如,

ml = [[0], [1]] * 10

后,

random.shuffle(ml)

[0]的数目可以是9或8,但不完全是10。

The shuffling process is “with replacement”, so the occurrence of each item may change! At least when when items in your list is also list.

E.g.,

ml = [[0], [1]] * 10

After,

random.shuffle(ml)

The number of [0] may be 9 or 8, but not exactly 10.


回答 21

计划:无需依赖库就可以完成改组工作。示例:从元素0的开头开始浏览列表;找到一个新的随机位置,例如6,将0的值放在6中,将6的值放在0中。移到元素1并重复此过程,以此类推。

import random
iteration = random.randint(2, 100)
temp_var = 0
while iteration > 0:

    for i in range(1, len(my_list)): # have to use range with len()
        for j in range(1, len(my_list) - i):
            # Using temp_var as my place holder so I don't lose values
            temp_var = my_list[i]
            my_list[i] = my_list[j]
            my_list[j] = temp_var

        iteration -= 1

Plan: Write out the shuffle without relying on a library to do the heavy lifting. Example: Go through the list from the beginning starting with element 0; find a new random position for it, say 6, put 0’s value in 6 and 6’s value in 0. Move on to element 1 and repeat this process, and so on through the rest of the list

import random
iteration = random.randint(2, 100)
temp_var = 0
while iteration > 0:

    for i in range(1, len(my_list)): # have to use range with len()
        for j in range(1, len(my_list) - i):
            # Using temp_var as my place holder so I don't lose values
            temp_var = my_list[i]
            my_list[i] = my_list[j]
            my_list[j] = temp_var

        iteration -= 1

回答 22

它工作正常。我在这里尝试使用功能作为列表对象:

    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()

它打印出来:foo1 foo2 foo3 foo2 foo3 foo1(最后一行中的foos具有随机​​顺序)

It works fine. I am trying it here with functions as list objects:

    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()

It prints out: foo1 foo2 foo3 foo2 foo3 foo1 (the foos in the last row have a random order)