将数字列表拆分为n个块，使这些块的总和（接近）相等，并保持原始顺序

Question

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]

我想要两块，然后分开就可以了

[[1, 6, 2, 3, 4, 1], [7, 6, 4]] 

每边的总和为17。对于三个块，结果将是

[[1, 6, 2, 3], [4, 1, 7], [6, 4]]

对于12、12和10的和。

编辑以获取其他说明

我目前用块的数量除以总和，并将其作为目标，然后迭代直到接近该目标。问题是某些数据集会搞乱算法，例如试图将以下内容分成3个：-

[95, 15, 75, 25, 85, 5]

总数是300，目标是100。第一个块总计为95，第二个块总计为90，第三个块总计为110，5是“剩余”。将它附加在应该是它的地方，将得到95，90，115，而一个更“合理”的解应该是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，我根据所有线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（最后一列明显更高）。

Answer 1

使用numpy的方式简洁明了。假设

import numpy.random as nr
import numpy as np

a = (nr.random(10000000)*1000).astype(int)

然后，假设您需要将列表分成总和大致相等的< code>p个部分

def equisum_partition(arr,p):
    ac = arr.cumsum()

    #sum of the entire array
    partsum = ac[-1]//p 

    #generates the cumulative sums of each part
    cumpartsums = np.array(range(1,p))*partsum

    #finds the indices where the cumulative sums are sandwiched
    inds = np.searchsorted(ac,cumpartsums) 

    #split into approximately equal-sum arrays
    parts = np.split(arr,inds)

    return parts

重要的是，这是矢量化的：

In [3]: %timeit parts = equisum_partition(a,20)
53.5 ms ± 962 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

你可以检查分裂的质量，

partsums = np.array([part.sum() for part in parts]).std()

分割不太好，但我怀疑它们是最佳的，因为顺序没有改变。

Answer 2

这是我目前得到的最好的O(n)贪婪算法。其思想是贪婪地将列表中的项目添加到一个块中，直到当前块的总和超过该点处块的平均预期总和。平均期望总和不断更新。这个解决方案并不完美，但是正如我所说的，它是O(n ),并且在我的测试中运行良好。我渴望听到反馈和改进建议。

我在代码中留下了调试打印语句以提供一些留档。请随意注释它们以查看每个步骤中发生的情况。

代码

def split_list(lst, chunks):
    #print(lst)
    #print()
    chunks_yielded = 0
    total_sum = sum(lst)
    avg_sum = total_sum/float(chunks)
    chunk = []
    chunksum = 0
    sum_of_seen = 0

    for i, item in enumerate(lst):
        #print('start of loop! chunk: {}, index: {}, item: {}, chunksum: {}'.format(chunk, i, item, chunksum))
        if chunks - chunks_yielded == 1:
            #print('must yield the rest of the list! chunks_yielded: {}'.format(chunks_yielded))
            yield chunk + lst[i:]
            raise StopIteration

        to_yield = chunks - chunks_yielded
        chunks_left = len(lst) - i
        if to_yield > chunks_left:
            #print('must yield remaining list in single item chunks! to_yield: {}, chunks_left: {}'.format(to_yield, chunks_left))
            if chunk:
                yield chunk
            yield from ([x] for x in lst[i:])
            raise StopIteration

        sum_of_seen += item
        if chunksum < avg_sum:
            #print('appending {} to chunk {}'.format(item, chunk))
            chunk.append(item)
            chunksum += item
        else:
            #print('yielding chunk {}'.format(chunk))
            yield chunk
            # update average expected sum, because the last yielded chunk was probably not perfect:
            avg_sum = (total_sum - sum_of_seen)/(to_yield - 1)
            chunks_yielded += 1
            chunksum = item
            chunk = [item]

测试代码

import random
lst = [1, 6, 2, 3, 4, 1, 7, 6, 4]
#lst = [random.choice(range(1,101)) for _ in range(100)]
chunks = 3
print('list: {}, avg sum: {}, chunks: {}\n'.format(lst, sum(lst)/float(chunks), chunks))
for chunk in split_list(lst, chunks):
    print('chunk: {}, sum: {}'.format(chunk, sum(chunk)))

使用您的列表进行测试：

list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 17.0, chunks: 2

chunk: [1, 6, 2, 3, 4, 1], sum: 17
chunk: [7, 6, 4], sum: 17

---

list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 11.33, chunks: 3

chunk: [1, 6, 2, 3], sum: 12
chunk: [4, 1, 7], sum: 12
chunk: [6, 4], sum: 10

---

list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 8.5, chunks: 4

chunk: [1, 6, 2], sum: 9
chunk: [3, 4, 1], sum: 8
chunk: [7], sum: 7
chunk: [6, 4], sum: 10

---

list: [1, 6, 2, 3, 4, 1, 7, 6, 4], avg sum: 6.8, chunks: 5

chunk: [1, 6], sum: 7
chunk: [2, 3, 4], sum: 9
chunk: [1, 7], sum: 8
chunk: [6], sum: 6
chunk: [4], sum: 4

带有长度为100的随机列表和从1到100的元素的测试（省略随机列表的打印）：

avg sum: 2776.0, chunks: 2

chunk: [25, 8, 71, 39, 5, 69, 29, 64, 31, 2, 90, 73, 72, 58, 52, 19, 64, 34, 16, 8, 16, 89, 70, 67, 63, 36, 9, 87, 38, 33, 22, 73, 66, 93, 46, 48, 65, 55, 81, 92, 69, 94, 43, 68, 98, 70, 28, 99, 92, 69, 24, 74], sum: 2806
chunk: [55, 55, 64, 93, 97, 53, 85, 100, 66, 61, 5, 98, 43, 74, 99, 56, 96, 74, 63, 6, 89, 82, 8, 25, 36, 68, 89, 84, 10, 46, 95, 41, 54, 39, 21, 24, 8, 82, 72, 51, 31, 48, 33, 77, 17, 69, 50, 54], sum: 2746

---

avg sum: 1047.6, chunks: 5

chunk: [19, 76, 96, 78, 12, 33, 94, 10, 38, 87, 44, 76, 28, 18, 26, 29, 44, 98, 44, 32, 80], sum: 1062
chunk: [48, 70, 42, 85, 87, 55, 44, 11, 50, 48, 47, 50, 1, 17, 93, 78, 25, 10, 89, 57, 85], sum: 1092
chunk: [30, 83, 99, 62, 48, 66, 65, 98, 94, 54, 14, 97, 58, 53, 3, 98], sum: 1022
chunk: [80, 34, 63, 20, 27, 36, 98, 97, 7, 6, 9, 65, 91, 93, 2, 27, 83, 35, 65, 17, 26, 41], sum: 1022
chunk: [80, 80, 42, 32, 44, 42, 94, 31, 50, 23, 34, 84, 47, 10, 54, 59, 72, 80, 6, 76], sum: 1040

---

avg sum: 474.6, chunks: 10

chunk: [4, 41, 47, 41, 32, 51, 81, 5, 3, 37, 40, 26, 10, 70], sum: 488
chunk: [54, 8, 91, 42, 35, 80, 13, 84, 14, 23, 59], sum: 503
chunk: [39, 4, 38, 40, 88, 69, 10, 19, 28, 97, 81], sum: 513
chunk: [19, 55, 21, 63, 99, 93, 39, 47, 29], sum: 465
chunk: [65, 88, 12, 94, 7, 47, 14, 55, 28, 9, 98], sum: 517
chunk: [19, 1, 98, 84, 92, 99, 11, 53], sum: 457
chunk: [85, 79, 69, 78, 44, 6, 19, 53], sum: 433
chunk: [59, 20, 64, 55, 2, 65, 44, 90, 37, 26], sum: 462
chunk: [78, 66, 32, 76, 59, 47, 82], sum: 440
chunk: [34, 56, 66, 27, 1, 100, 16, 5, 97, 33, 33], sum: 468

---

avg sum: 182.48, chunks: 25

chunk: [55, 6, 16, 42, 85], sum: 204
chunk: [30, 68, 3, 94], sum: 195
chunk: [68, 96, 23], sum: 187
chunk: [69, 19, 12, 97], sum: 197
chunk: [59, 88, 49], sum: 196
chunk: [1, 16, 13, 12, 61, 77], sum: 180
chunk: [49, 75, 44, 43], sum: 211
chunk: [34, 86, 9, 55], sum: 184
chunk: [25, 82, 12, 93], sum: 212
chunk: [32, 74, 53, 31], sum: 190
chunk: [13, 15, 26, 31, 35, 3, 14, 71], sum: 208
chunk: [81, 92], sum: 173
chunk: [94, 21, 34, 71], sum: 220
chunk: [1, 55, 70, 3, 92], sum: 221
chunk: [38, 59, 56, 57], sum: 210
chunk: [7, 20, 10, 81, 100], sum: 218
chunk: [5, 71, 19, 8, 82], sum: 185
chunk: [95, 14, 72], sum: 181
chunk: [2, 8, 4, 47, 75, 17], sum: 153
chunk: [56, 69, 42], sum: 167
chunk: [75, 45], sum: 120
chunk: [68, 60], sum: 128
chunk: [29, 25, 62, 3, 50], sum: 169
chunk: [54, 63], sum: 117
chunk: [57, 37, 42], sum: 136

如您所见，正如预期的那样，您想要生成的块越多，情况就越糟。我希望我能帮上忙。

edit：语法的收益需要Python 3.3或更高版本，如果您使用的是较旧的版本，只需将语句转换为正常的for循环即可。

Answer 3

这种方法定义了分区边界，将数组划分为大致相等数量的元素，然后重复搜索更好的分区，直到找不到更多的分区。它不同于大多数其他发布的解决方案，因为它试图通过尝试多个不同的分区来找到一个最佳解决方案。其他解决方案试图通过一次遍历数组来创建一个好的分区，但是我想不出一个保证最优的一次遍历算法。

这里的代码是该算法的有效实现，但可能很难理解，因此最后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)

根据您对此所做的操作，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，此算法会停止，它不会找到比连续5次以上迭代中看到的最佳结果更好的结果，或者在100次总迭代后作为一个catch-all停止点。您可能需要调整这些常数或使用不同的方案。如果你的身高构成了一个复杂的价值观景观，那么知道什么时候该停下来可能会遇到一些经典问题，比如试图逃避当地的最高标准等等。

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]

添加了新的测试用例[95、15、75、25、85、5]，该方法可以正确处理。

此版本的算法更易于阅读和理解，但由于对内置Python功能利用较少，因此时间更长。然而，它似乎在相当甚至稍微快一点的时间内执行。

#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)