元素和下一个元素与条件python之和

Question

我有一个脚本，它生成0,60之间的随机数字列表，并按升序排序。

本质上，我希望检查每个元素与其旁边的元素之间的差异是否高于3，如果不是，我希望重新生成列表，直到条件适用为止。

例如：

my_list = [1, 2, 3, 4, 5]
# this would not pass

my_list = [1, 5, 9, 13, 20]
# this would pass as the difference between each element and the next is more than 3

到目前为止我的代码是：

def generateList():
    timeSlots = list(range(0, 60)) # generate random list
    random.shuffle(timeSlots) # shuffle list

    timeSlots = timeSlots[:9] # shorten list to 9 elements
    timeSlots.sort() # sort list in ascending order

    for cur, nxt in zip(timeSlots, timeSlots[1:]):
        diff = (nxt - cur) # check difference
        if diff < 3:
            # HELP HERE
            # regenerate timeSlots until the sum of each element and the next element is bigger than 3
     
    return timeSlots

Answer 1

你想要使用all()

def generateList():
    while True:
        timeSlots = list(range(0, 60)) # generate random list
        random.shuffle(timeSlots) # shuffle list

        timeSlots = timeSlots[:9] # shorten list to 9 elements
        timeSlots.sort() # sort list in ascending order
        if all(nxt - cur > 3 for cur, nxt in zip(timeSlots, timeSlots[1:])):
            return timeSlots

注意，如果只想选择9个元素，则可以使用randome.sample()。

import random
def generate_list():
    while True:
        time_slots = random.sample(range(60), 9) # note this will not include 60 in the population
        time_slots.sort() # sort list in ascending order

        # or combine the above 2 lines as
        # time_slots = sorted(random.sample(range(60), 9))

        if all(nxt - cur > 3 for cur, nxt in zip(time_slots, time_slots[1:])):
            return time_slots

Answer 2

我希望检查每个元素与其旁边的元素之间的差异是否高于3，如果不是，我希望重新生成列表，直到条件适用为止。

答案已经显示了如何检查列表，但是根据您的数字，生成有效列表的可能性很低，或者根本就没有有效的列表。在这种情况下，循环会运行很长时间，或者无限长。

相反，您可以只生成一个有效的列表。假设K=10元素必须小于N=60，且差值大于M=3。然后，您知道M*(K-1)必须为空白“保留”，您可以从其余的random.sample数字，然后应用累积的差距之后。

import random

N, M, K = 60, 3, 10

nums = sorted(random.sample(range(N - (K-1)*M), K))
# [2, 5, 13, 16, 21, 23, 27, 28, 31, 32]  (random)

res = [x + i*M for i, x in enumerate(nums)]
# [2, 8, 19, 25, 33, 38, 45, 49, 55, 59]  (random)

作为一个副作用，如果没有这样的列表，这将立即引起一个例外。

因此，您的generateList函数可以如下所示，不需要循环：

def generateList(n=60, m=3, k=10):
    nums = sorted(random.sample(range(n - (k-1)*m), k))
    return [x + i*m for i, x in enumerate(nums)]

Answer 3

import random
def generateList():
    while True:
        timeSlots = list(range(0, 60))  # generate random list
        random.shuffle(timeSlots)  # shuffle list
        timeSlots = timeSlots[:9]  # shorten list to 9 elements
        timeSlots.sort()  # sort list in ascending order

        for cur, nxt in zip(timeSlots, timeSlots[1:]):
          diff = (nxt - cur)  # check difference
          if diff < 3:
              break
        else:
            return timeSlots


print(generateList())