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我有一个概念问题，我有几个包，每个包里面都包含许多元素。元素是 typeA或 type B。我想以这样一种方式将包分配在有限数量的箱中，以便箱之间的分布A和B箱之间的分布差异不大。

这个问题非常复杂，因此我将尝试用硬约束和概念示例来解释它。

约束

A package can only be used once
A package must be used entirely
The bins should have relatively equal distributions between `A` and `B` (max 5% deviation from the original ratio)
A package can be spread across all the bins in the given batch
I want to end up with as little as batches (size <= 3 bins) as possible

示例（概念）

Plate 1: 92 * `A`
Plate 2: 92 * `A`
Plate 3: 64 * `A`
Plate 4: 42 * `A`, 50 * `B`
Plate 5: 12 * `A`, 49 * `B`
Plate 6: 92 * `B`

这样的总分布是 302 *A和 191 *B总共产生 493 个样本，得到的比率是 61.25%A和 38.75%B

期望的结果：

一组最小化的批次，其中每个批次最多包含 3 个箱（长度 <= 92），假设每个箱的类型在 52 到 60 之间，类型A在 32 到 40 之间B（总和不超过 92）。

问题

建议使用什么算法或方法来解决这个问题，一个简单的建议方案就可以了（考虑到我到目前为止一直在尝试的东西（见下文）并没有走得太远）

迄今为止我的尝试背后的想法

data = ... # This is a list containg all values in a tuple format of `(unit, [(type, location)])` format
while len(data) > 0:
   batch = []
   counter1 = 0
   counter2 = 0
   for i in data:
      for j in i[1]:
         if j[0] == 'A':
            counter1 += 1
         else:
            counter2 += 1
   ratio1 = counter1/(counter1+counter2)
   ratio2 = counter2/(counter1+counter2)
   # Now we know the maximum number of A and B per batch
   counter3 = 0 # This keeps track of howmany type `A` we have in current batch
   counter4 = 0 # This keeps track of howmany type `B` we have in current batch
   while counter3 < ratio1:
      for i in data:
         for j in i[1]:
            if Counter(elem[0] for elem in j)['A'] < (ratio1 - counter3) and Counter(elem[0] for elem in j)['B'] < (ratio2 - counter4):
               # Add this unit (from data) to the batch
               batch.append(i)
               counter3 += Counter(elem[0] for elem in j)['A']
               counter4 += Counter(elem[0] for elem in j)['B']
               # Remove the used unit from data

这也是我被卡住的地方，目前这并没有尝试最小化垃圾箱的数量，也没有检查比率。此外，我有一个挥之不去的想法，即我尝试这样做的方式与解决此类问题的智能方式相去甚远。

4

1 回答 1

1 
#quick generator to rotate bin numbers
def getBin(maxBin):
    number = -1
    while True:
        number +=1 
        if number >= maxBin:
            number = 0
        yield number

batches = []
data = ....

#calculate the minimum number of bins we need
numberOfBins = (len(data))/ 92 + 1 

aBinPlacement = getBin(numberOfBins)
bBinPlacement = getBin(numberOfBins)

bins = numberOfBins * [[]]

#the ratio will be maintained because we rotate bins by type
for datum in data:
    if datum[0] == 'A':
        bins[aBinPlacement.next()].append(datum)
    else:
        bins[bBinPlacement.next()].append(datum)

batches.extend(bins)
于 2015-09-19T03:53:10.063 回答