如何将列表随机化以满足条件 [英] How to randomize a list to meet conditions

问题描述

我想生成一个列表的n个随机版本，以使每次随机化的顺序都不同于之前的顺序，而且每个元素的位置都必须与之前列表中的位置不同.我已经生成了列表的所有可能排列的列表，但是我在如何选择符合我的条件的子列表上遇到了麻烦.我在想也许列表理解可以奏效，但不确定如何完成.

I would like to generate n randomized versions of a list such that with each randomization the ordering is different from the one before and also each element must have a different position than in the list before. I have generated a list of all possible permutations of the list but I am stuck on how to select the sub-lists that match my conditions. I am thinking maybe a list comprehension could work but not sure how to complete it.

# constraints: n <= 12

lst = ['John', 'William', 'Michael', 'Victor', 'Tom', 'Charley', 'Patrick', 'David']

permutations = list(itertools.permutations(lst))

randomized_lists = [i for i in permutations if <conditions>]

有什么想法可以做到这一点吗?另外，是否有更好(更有效)的方法来解决该问题?

Any ideas how this could be done? Also, is there a better (more efficient) approach to the problem?

推荐答案

这可以通过修改

This can be done by modifying the Fisher–Yates shuffle algorithm to avoid swapping one item with itself. That is, for each item at k (where k starts at 0), instead of choosing a random item in [0, k] or [k, n - 1] (including k), choose a random item in [0, k) or (k, n - 1] (excluding k), and swap the item at k with the random item.

以下方法实现了这个想法:

The following method implements this idea:

import random

def shuffle_diff_pos(list):
  """ Returns a shuffled list in which 
      each item moves to a different position. """
  list=[x for x in list]
  if len(list)>=2:
    i=len(list)-1
    while i>0:
      k=random.randint(0, i-1)
      tmp=list[i];list[i]=list[k];list[k]=tmp
      i-=1
  return list

lst = ['John', 'William', 'Michael', 'Victor', 'Tom', 'Charley', 'Patrick', 'David']
randomized_lists = [shuffle_diff_pos(lst) for _ in range(12)]

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