这不是标准的分区问题，因为我需要维护列表中元素的顺序。
例如，如果我有一个列表
[1, 6, 2, 3, 4, 1, 7, 6, 4]
我想要两个 block ，那么分割应该给
[[1, 6, 2, 3, 4, 1], [7, 6, 4]]
每边总和为 17。对于三个 block ，结果将是
[[1, 6, 2, 3], [4, 1, 7], [6, 4]]
对于 12、12 和 10 的总和。
编辑以获取更多解释
我目前将总和除以 block 数并将其用作目标，然后迭代直到接近该目标。问题是某些数据集可能会搞乱算法，例如试图将以下内容分成 3 份:-
[95, 15, 75, 25, 85, 5]
总和为 300，目标为 100。第一个 block 的总和为 95，第二个总和为 90，第三个总和为 110，5 为“剩余”。将它附加到它应该出现的位置会得到 95、90、115，其中更“合理”的解决方案是 110、100、90。
结束编辑
背景:
我有一个包含不同高度的文本(歌词)的列表，我想将文本分成任意数量的列。目前我根据所有行的总高度计算目标高度，但显然这是一个一致的低估，在某些情况下会导致次优解决方案(最后一列明显更高)。
最佳答案
此方法定义分区边界，将数组划分为大致相等数量的元素，然后反复搜索更好的分区，直到找不到更多为止。它不同于大多数其他已发布的解决方案，因为它希望通过尝试多个不同的分区来找到最佳解决方案。其他解决方案试图通过阵列单次传递创建良好的分区，但我想不出保证最优的单次传递算法。
此处的代码是此算法的有效实现，但可能难以理解，因此在末尾包含一个更具可读性的版本作为附录。
def partition_list(a, k):
if k <= 1: return [a]
if k >= len(a): return [[x] for x in a]
partition_between = [(i+1)*len(a)/k for i in range(k-1)]
average_height = float(sum(a))/k
best_score = None
best_partitions = None
count = 0
while True:
starts = [0]+partition_between
ends = partition_between+[len(a)]
partitions = [a[starts[i]:ends[i]] for i in range(k)]
heights = map(sum, partitions)
abs_height_diffs = map(lambda x: abs(average_height - x), heights)
worst_partition_index = abs_height_diffs.index(max(abs_height_diffs))
worst_height_diff = average_height - heights[worst_partition_index]
if best_score is None or abs(worst_height_diff) < best_score:
best_score = abs(worst_height_diff)
best_partitions = partitions
no_improvements_count = 0
else:
no_improvements_count += 1
if worst_height_diff == 0 or no_improvements_count > 5 or count > 100:
return best_partitions
count += 1
move = -1 if worst_height_diff < 0 else 1
bound_to_move = 0 if worst_partition_index == 0\
else k-2 if worst_partition_index == k-1\
else worst_partition_index-1 if (worst_height_diff < 0) ^ (heights[worst_partition_index-1] > heights[worst_partition_index+1])\
else worst_partition_index
direction = -1 if bound_to_move < worst_partition_index else 1
partition_between[bound_to_move] += move * direction
def print_best_partition(a, k):
print 'Partitioning {0} into {1} partitions'.format(a, k)
p = partition_list(a, k)
print 'The best partitioning is {0}\n With heights {1}\n'.format(p, map(sum, p))
a = [1, 6, 2, 3, 4, 1, 7, 6, 4]
print_best_partition(a, 1)
print_best_partition(a, 2)
print_best_partition(a, 3)
print_best_partition(a, 4)
b = [1, 10, 10, 1]
print_best_partition(b, 2)
import random
c = [random.randint(0,20) for x in range(100)]
print_best_partition(c, 10)
d = [95, 15, 75, 25, 85, 5]
print_best_partition(d, 3)
根据您的用途，可能需要进行一些修改。例如，为了确定是否找到了最佳分区，当分区之间没有高度差时，该算法停止，它没有找到比连续 5 次以上迭代看到的最好的东西更好的东西，或者在 100 次之后总迭代次数作为一个包罗万象的停止点。您可能需要调整这些常量或使用不同的方案。如果您的高度形成了一个复杂的值(value)观景观，知道何时停止可能会遇到试图逃避局部最大值等经典问题。
输出
Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 1 partitions
The best partitioning is [[1, 6, 2, 3, 4, 1, 7, 6, 4]]
With heights [34]
Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 2 partitions
The best partitioning is [[1, 6, 2, 3, 4, 1], [7, 6, 4]]
With heights [17, 17]
Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 3 partitions
The best partitioning is [[1, 6, 2, 3], [4, 1, 7], [6, 4]]
With heights [12, 12, 10]
Partitioning [1, 6, 2, 3, 4, 1, 7, 6, 4] into 4 partitions
The best partitioning is [[1, 6], [2, 3, 4], [1, 7], [6, 4]]
With heights [7, 9, 8, 10]
Partitioning [1, 10, 10, 1] into 2 partitions
The best partitioning is [[1, 10], [10, 1]]
With heights [11, 11]
Partitioning [7, 17, 17, 1, 8, 8, 12, 0, 10, 20, 17, 13, 12, 4, 1, 1, 7, 11, 7, 13, 9, 12, 3, 18, 9, 6, 7, 19, 20, 17, 7, 4, 3, 16, 20, 6, 7, 12, 16, 3, 6, 12, 9, 4, 3, 2, 18, 1, 16, 14, 17, 7, 0, 14, 13, 3, 5, 3, 1, 5, 5, 13, 16, 0, 16, 7, 3, 8, 1, 20, 16, 11, 15, 3, 10, 10, 2, 0, 12, 12, 0, 18, 20, 3, 10, 9, 13, 12, 15, 6, 14, 16, 6, 12, 9, 9, 16, 14, 19, 1] into 10 partitions
The best partitioning is [[7, 17, 17, 1, 8, 8, 12, 0, 10, 20], [17, 13, 12, 4, 1, 1, 7, 11, 7, 13, 9], [12, 3, 18, 9, 6, 7, 19, 20], [17, 7, 4, 3, 16, 20, 6, 7, 12], [16, 3, 6, 12, 9, 4, 3, 2, 18, 1, 16], [14, 17, 7, 0, 14, 13, 3, 5, 3, 1, 5, 5], [13, 16, 0, 16, 7, 3, 8, 1, 20, 16], [11, 15, 3, 10, 10, 2, 0, 12, 12, 0, 18], [20, 3, 10, 9, 13, 12, 15, 6, 14], [16, 6, 12, 9, 9, 16, 14, 19, 1]]
With heights [100, 95, 94, 92, 90, 87, 100, 93, 102, 102]
Partitioning [95, 15, 75, 25, 85, 5] into 3 partitions
The best partitioning is [[95, 15], [75, 25], [85, 5]]
With heights [110, 100, 90]
编辑
添加了新的测试用例 [95, 15, 75, 25, 85, 5]，此方法可以正确处理。
附录
此版本的算法更易于阅读和理解，但由于较少利用内置的 Python 功能，因此有点长。然而，它的执行时间似乎相当，甚至稍快。
#partition list a into k partitions
def partition_list(a, k):
#check degenerate conditions
if k <= 1: return [a]
if k >= len(a): return [[x] for x in a]
#create a list of indexes to partition between, using the index on the
#left of the partition to indicate where to partition
#to start, roughly partition the array into equal groups of len(a)/k (note
#that the last group may be a different size)
partition_between = []
for i in range(k-1):
partition_between.append((i+1)*len(a)/k)
#the ideal size for all partitions is the total height of the list divided
#by the number of paritions
average_height = float(sum(a))/k
best_score = None
best_partitions = None
count = 0
no_improvements_count = 0
#loop over possible partitionings
while True:
#partition the list
partitions = []
index = 0
for div in partition_between:
#create partitions based on partition_between
partitions.append(a[index:div])
index = div
#append the last partition, which runs from the last partition divider
#to the end of the list
partitions.append(a[index:])
#evaluate the partitioning
worst_height_diff = 0
worst_partition_index = -1
for p in partitions:
#compare the partition height to the ideal partition height
height_diff = average_height - sum(p)
#if it's the worst partition we've seen, update the variables that
#track that
if abs(height_diff) > abs(worst_height_diff):
worst_height_diff = height_diff
worst_partition_index = partitions.index(p)
#if the worst partition from this run is still better than anything
#we saw in previous iterations, update our best-ever variables
if best_score is None or abs(worst_height_diff) < best_score:
best_score = abs(worst_height_diff)
best_partitions = partitions
no_improvements_count = 0
else:
no_improvements_count += 1
#decide if we're done: if all our partition heights are ideal, or if
#we haven't seen improvement in >5 iterations, or we've tried 100
#different partitionings
#the criteria to exit are important for getting a good result with
#complex data, and changing them is a good way to experiment with getting
#improved results
if worst_height_diff == 0 or no_improvements_count > 5 or count > 100:
return best_partitions
count += 1
#adjust the partitioning of the worst partition to move it closer to the
#ideal size. the overall goal is to take the worst partition and adjust
#its size to try and make its height closer to the ideal. generally, if
#the worst partition is too big, we want to shrink the worst partition
#by moving one of its ends into the smaller of the two neighboring
#partitions. if the worst partition is too small, we want to grow the
#partition by expanding the partition towards the larger of the two
#neighboring partitions
if worst_partition_index == 0: #the worst partition is the first one
if worst_height_diff < 0: partition_between[0] -= 1 #partition too big, so make it smaller
else: partition_between[0] += 1 #partition too small, so make it bigger
elif worst_partition_index == len(partitions)-1: #the worst partition is the last one
if worst_height_diff < 0: partition_between[-1] += 1 #partition too small, so make it bigger
else: partition_between[-1] -= 1 #partition too big, so make it smaller
else: #the worst partition is in the middle somewhere
left_bound = worst_partition_index - 1 #the divider before the partition
right_bound = worst_partition_index #the divider after the partition
if worst_height_diff < 0: #partition too big, so make it smaller
if sum(partitions[worst_partition_index-1]) > sum(partitions[worst_partition_index+1]): #the partition on the left is bigger than the one on the right, so make the one on the right bigger
partition_between[right_bound] -= 1
else: #the partition on the left is smaller than the one on the right, so make the one on the left bigger
partition_between[left_bound] += 1
else: #partition too small, make it bigger
if sum(partitions[worst_partition_index-1]) > sum(partitions[worst_partition_index+1]): #the partition on the left is bigger than the one on the right, so make the one on the left smaller
partition_between[left_bound] -= 1
else: #the partition on the left is smaller than the one on the right, so make the one on the right smaller
partition_between[right_bound] += 1
def print_best_partition(a, k):
#simple function to partition a list and print info
print ' Partitioning {0} into {1} partitions'.format(a, k)
p = partition_list(a, k)
print ' The best partitioning is {0}\n With heights {1}\n'.format(p, map(sum, p))
#tests
a = [1, 6, 2, 3, 4, 1, 7, 6, 4]
print_best_partition(a, 1)
print_best_partition(a, 2)
print_best_partition(a, 3)
print_best_partition(a, 4)
print_best_partition(a, 5)
b = [1, 10, 10, 1]
print_best_partition(b, 2)
import random
c = [random.randint(0,20) for x in range(100)]
print_best_partition(c, 10)
d = [95, 15, 75, 25, 85, 5]
print_best_partition(d, 3)
关于python - 将数字列表拆分为 n 个 block ，使这些 block 具有(接近)相等的总和并保持原始顺序，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56086075/