Better way to shuffle two related lists

问题

Is there better ways to randomly shuffle two related lists without breaking their correspondence in the other list? I've found related questions in numpy.array and c# but not exactly the same one.

As a first try, a simple zip trick will do:

import random
a = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
b = [2, 4, 6, 8, 10]
c = zip(a, b)
random.shuffle(c)
a = [e[0] for e in c]
b = [e[1] for e in c]
print a
print b

It will get the output:

[[1, 2], [7, 8], [3, 4], [5, 6], [9, 10]]
[2, 8, 4, 6, 10]

Just find it a bit awkward. And it also need an additional list as well.

回答1:

Given the relationship demonstrated in the question, I'm going to assume the lists are the same length and that list1[i] corresponds to list2[i] for any index i. With that assumption in place, shuffling the lists is as simple as shuffling the indices:

 from random import shuffle
 # Given list1 and list2

 list1_shuf = []
 list2_shuf = []
 index_shuf = list(range(len(list1)))
 shuffle(index_shuf)
 for i in index_shuf:
     list1_shuf.append(list1[i])
     list2_shuf.append(list2[i])

回答2:

If you are willing to install a few more packages:

Req: NumPy (>= 1.6.1), SciPy (>= 0.9).

pip install -U scikit-learn

from sklearn.utils import shuffle
list_1, list_2 = shuffle(list_1, list_2)

回答3:

If you have to do this often, you could consider adding one level of indirection by shuffling a list of indexes.

Python 2.6.6 (r266:84297, Aug 24 2010, 18:13:38) [MSC v.1500 64 bit (AMD64)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import random
>>> a = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
>>> b = [2, 4, 6, 8, 10]
>>> indexes = range(len(a))
>>> indexes
[0, 1, 2, 3, 4]
>>> random.shuffle(indexes)
>>> indexes
[4, 1, 2, 0, 3]
>>> for index in indexes:
...     print a[index], b[index]
...
[9, 10] 10
[3, 4] 4
[5, 6] 6
[1, 2] 2
[7, 8] 8

回答4:

A fast answer using numpy please refer to here:
You can use

p = numpy.random.permutation(len(a))

To create a new list of indexes for both lists and use it to reorder them.

In your scenario:

In [61]: a = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
In [62]: b = [2, 4, 6, 8, 10]
In [63]: import numpy as np
In [64]: a_ar, b_ar = np.array(a), np.array(b)
In [65]: p = np.random.permutation(len(a))
In [66]: a, b = a_ar[p].tolist(), b_ar[p].tolist()
In [68]: a
Out[68]: [[3, 4], [7, 8], [5, 6], [1, 2], [9, 10]]
In [69]: b
Out[69]: [4, 8, 6, 2, 10]

回答5:

So far, all solutions created new lists in order to solve the problem. If the lists a and b are very long you may want to shuffle them in place. For that you would need a function like:

import random

def shuffle(a,b):
    assert len(a) == len(b)
    start_state = random.getstate()
    random.shuffle(a)
    random.setstate(start_state)
    random.shuffle(b)

a = [1,2,3,4,5,6,7,8,9]
b = [11,12,13,14,15,16,17,18,19]
shuffle(a,b)
print(a) # [9, 7, 3, 1, 2, 5, 4, 8, 6]
print(b) # [19, 17, 13, 11, 12, 15, 14, 18, 16]

回答6:

You can do an unzip at the end to limit the awkwardness a bit?

import numpy as np
list1 = [1,2,3]
list2 = [4,5,7]
list_zipped = list(zip(list1,list2))
np.random.shuffle(list_zipped)
list1,list2 = zip(*z) #unzipping

回答7:

I'm not sure if I'm missing something here, but it looks like you're just shuffling 1 of the lists and the other one is re-arranged to match the order of the first list. So what you have is the best way to do this without making it more complicated. If you want to go the complicated route you can just shuffle 1 list and use the unshuffled list to do a lookup in the shuffled list and rearrange it in that way. In the end you end up with the same result you started with. Why is creating a third list a problem? If you really want to recycle the lists then you can simply replace list b with what you're using for list c and then separate it later on back to a and b.

来源：https://stackoverflow.com/questions/11765061/better-way-to-shuffle-two-related-lists

标签

python

list

shuffle