python，洗牌直到列表完全不同

Question

我想问一下如何洗牌，也许是在一个 while 循环中，直到所有列表都完全不同（就像在数独中一样）？

#lists you want to shuffle
s1 = [1, 2, 3, 4]
s2 = [1, 2, 3, 4]
s3 = [1, 2, 3, 3]
s4 = [1, 2, 3, 4]

def mid_generator():

    while True:
        random.shuffle(s1)
        random.shuffle(s2)
        random.shuffle(s3)
        random.shuffle(s4)

        # if ... all lists are different...:
            #break

    return s1, s2, s3, s4

使得第i行第j列的次数只有一次：

s1 = [3, 1, 2, 4]
s2 = [4, 2, 1, 3]
s3 = [2, 4, 3, 1]
s4 = [1, 3, 4, 2]

如果我尝试使用 if s1[0].= s2[0] 的长 if 语句...... output 是错误的。 也许你可以帮助我。

Answer 1

这仍然是一种低效的算法，因为它基于原始代码，但是您可以将每个位置的元素放入集合中并检查它们的长度； len({1, 2, 3, 3})将为3 ，因为只有3个唯一元素：

import random

#lists you want to shuffle
s1 = [1, 2, 3, 4]
s2 = [1, 2, 3, 4]
s3 = [1, 2, 3, 4]
s4 = [1, 2, 3, 4]

def mid_generator():

    while True:
        random.shuffle(s1)
        random.shuffle(s2)
        random.shuffle(s3)
        random.shuffle(s4)

        test0 = {s1[0], s2[0], s3[0], s4[0]}
        test1 = {s1[1], s2[1], s3[1], s4[1]}
        test2 = {s1[2], s2[2], s3[2], s4[2]}
        test3 = {s1[3], s2[3], s3[3], s4[3]}

        if len(test0) == len(test1) == len(test2) == len(test3) == 4:
            break

    return s1, s2, s3, s4

交互式输出示例：

>>> mid_generator()
([3, 4, 2, 1], [1, 2, 4, 3], [4, 3, 1, 2], [2, 1, 3, 4])

Answer 2

这很丑陋，但是如果您仔细地针对每个列表正确地将每个列彼此进行比较，则and语句的长列表and可以正常工作：

def mid_generator():

    while True:
        random.shuffle(s1)
        random.shuffle(s2)
        random.shuffle(s3)
        random.shuffle(s4)

        if (    s1[0] != s2[0] and s1[0] != s3[0] and s1[0] != s4[0]
            and s2[0] != s3[0] and s2[0] != s4[0] and s3[0] != s4[0]
            and s1[1] != s2[1] and s1[1] != s3[1] and s1[1] != s4[1]
            and s2[1] != s3[1] and s2[1] != s4[1] and s3[1] != s4[1]                
            and s1[2] != s2[2] and s1[2] != s3[2] and s1[2] != s4[2]
            and s2[2] != s3[2] and s2[2] != s4[2] and s3[2] != s4[2]                
            and s1[3] != s2[3] and s1[3] != s3[3] and s1[3] != s4[3]
            and s2[3] != s3[3] and s2[3] != s4[3] and s3[3] != s4[3]):
            break
    return s1, s2, s3, s4

Answer 3

仅出于完整性考虑，以下是任意列表长度的通用版本：

def mid_generator(n):
    s = [list(range(1, n+1)) for x in range(n)]
    while True:
        for x in s:
            random.shuffle(x) 
        for i in range(n):
            test = {s[j][i] for j in range(n)}
            if len(test) != n:
                break
        else:
            return s
    return None # never hit

互动：

>>> mid_generator(1)
[[1]]
>>> mid_generator(2)
[[1, 2], [2, 1]]
>>> mid_generator(3)
[[1, 3, 2], [3, 2, 1], [2, 1, 3]]
>>> mid_generator(4)
[[2, 4, 1, 3], [4, 2, 3, 1], [1, 3, 2, 4], [3, 1, 4, 2]]
>>> mid_generator(5)
[[4, 5, 2, 1, 3], [1, 2, 3, 5, 4], [3, 4, 5, 2, 1], [5, 1, 4, 3, 2], [2, 3, 1, 4, 5]]
>>> mid_generator(6)

（仍在等待...）

Answer 4

使用基于geegksforgeeks代码的Fisher-Yates shuffle。 修改以下代码以确保该值永远不会放在原始索引处。

def fisher_yates_shuffle_diff(orig_list):
    result_list = orig_list.copy()
    list_len = len(orig_list)
    indc_list = list(range(0, list_len))
    for i in range(list_len-1, 0, -1):
     
        # Pick a random index from 0 to i 
        indc = indc_list.copy()
        # Remove current index
        indc.remove(i)
        j = indc[random.randint(0, i)]

        # Swap arr[i] with the element at random index
        result_list[i], result_list[j] = result_list[j], result_list[i]
    return result_list

      

Answer 5

这将起作用-但不是最佳选择-但是if检查将起作用

def mid_generator():

    while True:
        random.shuffle(s1)
        random.shuffle(s2)
        random.shuffle(s3)
        random.shuffle(s4)

        if not (s1 == s2 or s1 == s3 or s1 == s4 or s2 == s3 or s2 == s4 or s3 == s4):
            break

    return s1, s2, s3, s4