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python - 如何约束具有多个随机选择位置的项目,使每个项目的平均位置在一定范围内

转载 作者:行者123 更新时间:2023-12-01 06:39:54 26 4
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问题:总共,我有 1-300 个职位需要填写,我有 50 个项目,每个项目有 6 个独特的职位可供选择(总共 300 个职位)。每个项目的平均排名需要在 145-155 范围内(越接近 150 越好)

约束:对于每个项目,其 6 个位置中的每一个都必须落在固定范围列表(称为运行)内,但不能在同一运行内,

例如运行次数为:[(1,36), (37,73), (74,110), (111,148), (149,186), (187,225), (226,262), (263, 300)] 。因此,第 1 项的位置可以为 1、38、158、198、238、271 - 因此不能在同一次运行中出现两次。

选择的位置应该是随机的 - 或尽可能随机

我遇到的问题:将随机选择的位置限制为平均值 150(或非常接近)似乎非常困难。我已经尝试了代码的各种不同实现,但大多数都会导致它卡在接近末尾的位置(由于没有足够的位置可供选择),或者不会非常接近。我最好的尝试只是涉及我放置的 if 语句,至少尝试限制范围,但它仍然只能得到大约 130-170 的范围。这感觉像是一个非常糟糕的方法,我想知道是否有一种方法可以通过算法来完成它,而不是仅仅粘贴 if 语句以希望某些东西能起作用。

我对随机创可贴限制的最佳尝试:https://pyfiddle.io/fiddle/dfc6dfd1-1e12-46df-957c-d7ae0e94fbe3/?i=true

^ 正如您所看到的,平均值有所不同,大多数情况都在可接受的范围内,有些只是偏离/非常偏离

我花了几个星期的时间来解决这个问题,但真的想不出什么办法可以将其限制在适当的平均值(145-155),并希望在这里得到任何帮助

最佳答案

这种方法使用动态规划来随机提取尽可能好的组 50 次。然后,它将不完美的东西与一些好的东西一起放回原处,并尝试同样的事情。它不断这样做,同时放宽完美的定义,直到获得 50 个可接受的组。

速度很慢。 (在我的笔记本电脑上大约 20 秒。)但它经常使所有组的平均值完全相同。在多次运行中,我没有看到任何组超出范围[150.0, 151.0]

#! /usr/bin/env python3

import math
import collections
import random

BaseNode = collections.namedtuple('BaseNode', 'sum value ways run prev_node')
class Node(BaseNode):
def merge(self, other):
if self.sum != other.sum:
print(self)
print(other)
raise Exception('Can only merge nodes with the same sum.')
ways = self.ways + other.ways
if self.ways <= random.randint(1, ways):
return Node(sum=self.sum, value=self.value, ways=ways,
run=self.run, prev_node=self.prev_node)
else:
return Node(sum=other.sum, value=other.value, ways=ways,
run=other.run, prev_node=other.prev_node)

def extract_group(self):
values = []
node = self
while node.value is not None:
node.run.remove(node.value)
values.append(node.value)
node = node.prev_node
return sorted(values)


def random_next_group_from_runs (runs):
runs_by_count = {}
# organize by count
for run in runs:
count = len(run)
if count in runs_by_count:
runs_by_count[count].append(run)
else:
runs_by_count[count] = [run]


required_runs = []
optional_runs = []
largest = max(runs_by_count.keys())
if 6 < len(runs_by_count[largest]):
largest = largest + 1
else:
required_runs = runs_by_count[largest]

for count, these_runs in runs_by_count.items():
if count < largest:
optional_runs.extend(these_runs)

# We start with the empty sum.
node_by_sum_by_count = {0: {0: Node(sum=0, value=None, ways=1, run=None, prev_node=None)}}
# We update to use a value from each required run.
for run in required_runs:
new_node_by_sum_by_count = {}
for count, node_by_sum in node_by_sum_by_count.items():
if count+1 not in new_node_by_sum_by_count:
new_node_by_sum_by_count[count+1] = {}
new_node_by_sum = new_node_by_sum_by_count[count+1]
for s, node in node_by_sum.items():
for i in run:
new_node = Node(sum=s+i, value=i, ways=node.ways, run=run, prev_node=node)
if s+i not in new_node_by_sum:
new_node_by_sum[s+i] = new_node
else:
# This merge hides the random choice of which one to take.
new_node_by_sum[s+i] = new_node_by_sum[s+i].merge(new_node)
node_by_sum_by_count = new_node_by_sum_by_count
# We may use a value from each optional run.
for run in optional_runs:
new_node_by_sum_by_count = {}
for count, node_by_sum in node_by_sum_by_count.items():
# The options where we do not use this run.
if count not in new_node_by_sum_by_count:
new_node_by_sum_by_count[count] = {}
new_node_by_sum = new_node_by_sum_by_count[count]
for s, node in node_by_sum.items():
if s not in new_node_by_sum:
new_node_by_sum[s] = node
else:
new_node_by_sum[s] = new_node_by_sum[s].merge(node)


# The options where we do use this run.
if count+1 not in new_node_by_sum_by_count:
new_node_by_sum_by_count[count+1] = {}
new_node_by_sum = new_node_by_sum_by_count[count+1]
for s, node in node_by_sum.items():
for i in run:
new_node = Node(sum=s+i, value=i, ways=node.ways, run=run, prev_node=node)
if s+i not in new_node_by_sum:
new_node_by_sum[s+i] = new_node
else:
# This merge hides the random choice of which one to take.
new_node_by_sum[s+i] = new_node_by_sum[s+i].merge(new_node)
node_by_sum_by_count = new_node_by_sum_by_count

# Widening sums close to 903
avg = 903
def try_sum():
yield avg
i = 1
while True:
yield avg - i
yield avg + i
i = i+1

# We only want groups with 6 elements.
node_by_sum = node_by_sum_by_count[6]
for i in try_sum():
if i in node_by_sum:
return node_by_sum[i].extract_group()


runs = [
set(range(1, 37)),
set(range(37, 74)),
set(range(74, 111)),
set(range(111, 149)),
set(range(149, 187)),
set(range(187, 226)),
set(range(226, 263)),
set(range(263, 301)),
]
in_run = {}
for i in range(len(runs)):
for j in runs[i]:
in_run[j] = i
#runs = [ set(range(i*36+1, i*36+37)) for i in range(8) ]
groups = []
bad_groups = []
attempt = 0
while attempt == 0 or 0 < len(bad_groups):
attempt = attempt + 1
# We add a few groups to bad groups in principle.
for i in range(attempt):
if 20 < len(groups):
bad_groups.append(groups.pop())
for group in bad_groups:
for i in group:
runs[in_run[i]].add(i)
bad_groups = []
while len(groups) + len(bad_groups) < 50:
group = random_next_group_from_runs(runs)
if abs(sum(group) - 903) <= math.floor(attempt/5.0):
groups.append(group)
else:
bad_groups.append(group)
random.shuffle(groups)

for group in groups:
print([sum(group), group])

关于python - 如何约束具有多个随机选择位置的项目,使每个项目的平均位置在一定范围内,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59505520/

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