在Python中随机随机洗牌的键和值DIctionary

Question

有没有一种方法可以随机洗牌什么键对应什么值？我发现random.sample，但我想知道是否有更pythonic /更快的方式做到这一点。

例子：a = {"one":1,"two":2,"three":3}

抛去：a_shuffled = {"one":2,"two":3,"three":1}

Answer 1

对不起，使其更快的唯一方法是使用numpy的：/。 不管你做什么，它都会以某种方式争夺所有需要时间的指数 - 所以在C中这样做会有所帮助。此外，纯粹的随机和随机的区别在于你不能有重复的指数。

对不起这有点长了 - 所以你必须做一些滚动

E.g.                                    


# made for python 2.7 but should be able to work in python 3 
import random 
import numpy as np 
from time import time 


def given_seq(): 
#general example 
    start = time() 
    a = {"one":1,"two":2,"three":3} 
    keys = a.keys() 
    random.shuffle(keys) 
    a = dict(zip(keys, a.values())) 

#Large example 

a = dict(zip(range(0,100000), range(1,100001))) 

def random_shuffle(): 
    keys = a.keys() 
    random.shuffle(keys) 
    b = dict(zip(keys, a.values())) 

def np_random_shuffle(): 
    keys = a.keys() 
    np.random.shuffle(keys) 
    b = dict(zip(keys, a.values())) 

def np_random_permutation(): 
    #more concise and using numpy's permutation option 
    b = dict(zip(np.random.permutation(a.keys()), a.values())) 

#if you precompute the array key as a numpy array 

def np_random_keys_choice(): 
    akeys = np.array(a.keys()) 
    return dict(zip(akeys[np.random.permutation(len(akeys))],a.values())) 

def np_random_keys_shuffle(): 
    key_indexes = np.arange(len(a.keys())) 
    np.random.shuffle(key_indexes) 
    return dict(zip(np.array(a.keys())[key_indexes],a.values())) 

#fixed dictionary size 
key_indexes = np.arange(len(a.keys())) 
def np_random_fixed_keys_shuffle(): 
    np.random.shuffle(key_indexes) 
    return dict(zip(np.array(a.keys())[key_indexes],a.values())) 


#so dstack actually slows things down 
def np_random_shuffle_dstack(): 
    keys = a.keys() 
    np.random.shuffle(keys) 
    return dict(np.dstack((keys, a.values()))[0]) 

if __name__=='__main__': 
    import timeit 
    # i can use global namespace level introspection to automate the below line but it's not needed yet 
    for func in ['given_seq', 'random_shuffle', 'np_random_shuffle', 'np_random_permutation', 'np_random_keys_choice', 
      'np_random_keys_shuffle', 'np_random_fixed_keys_shuffle']: 
     print func, timeit.timeit("{}()".format(func), setup = "from __main__ import {}".format(''.join(func)), number = 200) 

given_seq 0.00103783607483 
random_shuffle 23.869166851 
np_random_shuffle 16.3060112 
np_random_permutation 21.9921720028 
np_random_keys_choice 21.8105020523 
np_random_keys_shuffle 22.4905178547 
np_random_fixed_keys_shuffle 21.8256559372 

使用选择/置换看起来更好 - 但它不以任何方式更快。不幸的是，复制通常是缓慢的，除非它是一个小尺寸 - 并且没有办法通过指针/引用，而不必占用额外的行 - 尽管我辩论如果这使得它'非pythonic'

即如果你看在Zen of Python或只是做import this在蟒蛇会议的行之一是：

尽管实用性胜过纯度。

所以它的开放当然:)

Answer 2

In [47]: import random 

In [48]: keys = a.keys() 

In [49]: values = a.values() 

In [50]: random.shuffle(values) 

In [51]: a_shuffled = dict(zip(keys, values)) 

In [52]: a_shuffled 
Out[52]: {'one': 2, 'three': 1, 'two': 3} 

，或者更简练是：

In [56]: dict(zip(a.keys(), random.sample(a.values(), len(a)))) 
Out[56]: {'one': 3, 'three': 2, 'two': 1} 

（但我想那就是你已经想出了解决方案）

---

注意，虽然使用random.sample是pithier，使用random.shuffle是快了一点：

import random 
import string 
def using_shuffle(a): 
    keys = a.keys() 
    values = a.values() 
    random.shuffle(values) 
    return dict(zip(keys, values)) 

def using_sample(a): 
    return dict(zip(a.keys(), random.sample(a.values(), len(a)))) 

N = 10000 
keys = [''.join(random.choice(string.letters) for j in range(4)) for i in xrange(N)] 
a = dict(zip(keys, range(N))) 

In [71]: %timeit using_shuffle(a) 
100 loops, best of 3: 5.14 ms per loop 

In [72]: %timeit using_sample(a) 
100 loops, best of 3: 5.78 ms per loop