随机化两个列表并在python中维护顺序

说我有两个简单的清单，

a = ['Spears', "Adele", "NDubz", "Nicole", "Cristina"]

b = [1,2,3,4,5]

len(a) == len(b)

我想做的是将a和a随机化，但要保持顺序。 因此，类似：

a = ["Adele", 'Spears', "Nicole", "Cristina", "NDubz"]

b = [2,1,4,5,3]

我知道我可以使用以下方法来随机排列一个列表：

import random

random.shuffle(a)

但这只是将a随机化，而我想将a随机化，并保持列表b中的“随机顺序”。

希望就如何实现这一目标提供任何指导。

JohnJ asked 2020-01-13T02:54:49Z

6个解决方案

84 votes

我将两个列表合并在一起，将结果列表随机排序，然后将它们拆分。 这利用了zip()

a = ["Spears", "Adele", "NDubz", "Nicole", "Cristina"]

b = [1, 2, 3, 4, 5]

combined = list(zip(a, b))

random.shuffle(combined)

a[:], b[:] = zip(*combined)

Tim answered 2020-01-13T02:55:10Z

16 votes

使用zip，它具有可以“两种”方式工作的出色功能。

import random

a = ['Spears', "Adele", "NDubz", "Nicole", "Cristina"]

b = [1,2,3,4,5]

z = zip(a, b)

# => [('Spears', 1), ('Adele', 2), ('NDubz', 3), ('Nicole', 4), ('Cristina', 5)]

random.shuffle(z)

a, b = zip(*z)

answered 2020-01-13T02:55:29Z

12 votes

为了避免重新发明轮子，请使用sklearn

from sklearn.utils import shuffle

a, b = shuffle(a, b)

Nimrod Morag answered 2020-01-13T02:55:49Z

8 votes

请注意，Tim的答案仅适用于Python 2，不适用于Python3。如果使用Python 3，则需要执行以下操作：

combined = list(zip(a, b))

random.shuffle(combined)

a[:], b[:] = zip(*combined)

否则会出现错误：

TypeError: object of type 'zip' has no len()

Adam_G answered 2020-01-13T02:56:14Z

2 votes

另一种方法可能是

a = ['Spears', "Adele", "NDubz", "Nicole", "Cristina"]

b = range(len(a)) # -> [0, 1, 2, 3, 4]

b_alternative = range(1, len(a) + 1) # -> [1, 2, 3, 4, 5]

random.shuffle(b)

a_shuffled = [a[i] for i in b] # or:

a_shuffled = [a[i - 1] for i in b_alternative]

这是相反的方法，但是仍然可以帮助您。

glglgl answered 2020-01-13T02:56:38Z

0 votes

这是我的风格：

import random

def shuffleTogether(A, B):

if len(A) != len(B):

raise Exception("Lengths don't match")

indexes = range(len(A))

random.shuffle(indexes)

A_shuffled = [A[i] for i in indexes]

B_shuffled = [B[i] for i in indexes]

return A_shuffled, B_shuffled

A = ['a', 'b', 'c', 'd']

B = ['1', '2', '3', '4']

A_shuffled, B_shuffled = shuffleTogether(A, B)

print A_shuffled

print B_shuffled

Nadav B answered 2020-01-13T02:56:58Z