Question

我目前正在尝试获取随机排列的数字数组：

label_array = np.repeat(np.arange(6), 12)

唯一的限制是，随机播放的连续元素不得相同。为此，我目前正在使用以下代码：

# Check if there are any occurrences of two consecutive 
# elements being of the same category (same number)
num_occurrences = np.sum(np.diff(label_array) == 0)

# While there are any occurrences of this...
while num_occurrences != 0:
    # ...shuffle the array...
    np.random.shuffle(label_array)

    # ...create a flag for occurrences...
    flag = np.hstack(([False], np.diff(label_array) == 0))
    flag_array = label_array[flag]

    # ...and shuffle them.
    np.random.shuffle(flag_array)

    # Then re-assign them to the original array...
    label_array[flag] = flag_array

    # ...and check the number of occurrences again.
    num_occurrences = np.sum(np.diff(label_array) == 0)

尽管这适用于这种大小的数组，但我不知道它是否适用于更大的数组。即使这样，也可能会花费很多时间。

那么，有没有更好的方法呢？

Answer 1

从技术上讲，可能不是最佳答案，希望它能满足您的要求。

import numpy as np
def generate_random_array(block_length, block_count):
    for blocks in range(0, block_count):
        nums = np.arange(block_length)
        np.random.shuffle(nums)
        try:
            if nums[0] == randoms_array [-1]:
                nums[0], nums[-1] = nums[-1], nums[0]
        except NameError:
            randoms_array = []
        randoms_array.extend(nums)
    return randoms_array


generate_random_array(block_length=1000, block_count=1000)

Answer 2

在Python> = 3.6的情况下，这是一种使用random.choices的方法，该方法允许从具有权重的总体中进行选择。

想法是一一生成数字。每次生成新数字时，我们都会通过将其权重暂时设置为零来排除前一个数字。然后，我们减小所选对象的权重。

正如@roganjosh适当指出的那样，当我们剩下最后一个值的多个实例时，最终会遇到一个问题-这真的很常见，尤其是值少且重复次数多时

我使用的解决方案是使用简短的send_back函数将这些值重新插入不会产生冲突的列表中。

import random

def send_back(value, number, lst):
    idx = len(lst)-2
    for _ in range(number):
        while lst[idx] == value or lst[idx-1] == value:
            idx -= 1
        lst.insert(idx, value)


def shuffle_without_doubles(nb_values, repeats):
    population = list(range(nb_values))
    weights = [repeats] * nb_values
    out = []
    prev = None
    for i in range(nb_values * repeats):
        if prev is not None:
            # remove prev from the list of possible choices
            # by turning its weight temporarily to zero
            old_weight = weights[prev]
            weights[prev] = 0    

        try:
            chosen = random.choices(population, weights)[0]
        except IndexError:
            # We are here because all of our weights are 0,
            # which means that all is left to choose from
            # is old_weight times the previous value
            send_back(prev, old_weight, out)
            break

        out.append(chosen)
        weights[chosen] -= 1
        if prev is not None:
            # restore weight
            weights[prev] = old_weight
        prev = chosen
    return out

print(shuffle_without_doubles(6, 12))

[5, 1, 3, 4, 3, 2, 1, 5, 3, 5, 2, 0, 5, 4, 3, 4, 5,
 3, 4, 0, 4, 1, 0, 1, 5, 3, 0, 2, 3, 4, 1, 2, 4, 1,
 0, 2, 0, 2, 5, 0, 2, 1, 0, 5, 2, 0, 5, 0, 3, 2, 1,
 2, 1, 5, 1, 3, 5, 4, 2, 4, 0, 4, 2, 4, 0, 1, 3, 4,
 5, 3, 1, 3]

一些粗略的计时：生成(shuffle_without_doubles(600, 1200))大约需要30秒，因此有720000个值。

如何在不重复两个连续元素的情况下重新排列数字数组？

2 个答案: