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]
我想要两个块，然后拆分应该给出
[[1, 6, 2, 3, 4, 1], [7, 6, 4]]
每边17的总和。对于三个块，结果将是
[[1, 6, 2, 3], [4, 1, 7], [6, 4]]
为12,12和10的总和。
修改以获取其他说明
我目前将总和除以块数并将其用作目标，然后迭代直到接近该目标。问题是某些数据集会使算法混乱，例如试图将以下内容分成3： -
[95, 15, 75, 25, 85, 5]
总和为300，目标为100.第一个块总和为95，第二个总和为90，第三个总和为110，而5将为“剩余”。将它添加到它应该的位置将会给出95,90,115，其中一个更合理的＆＃39;解决方案将是110,100,90。
结束修改
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，但显然这是一致的低估，在某些情况下会导致次优解（最后一列明显更高）。
答案 0 :(得分：8)
这种方法定义了分区边界，它将数组分成大致相等数量的元素，然后重复搜索更好的分区，直到找不到它为止。它与大多数其他发布的解决方案的不同之处在于，它希望通过尝试多个不同的分区来找到最佳解决方案。其他解决方案试图在一次通过阵列时创建一个好的分区，但我不能想到一个保证最优的单通道算法。
这里的代码是这个算法的有效实现，但是它很难理解，因此最后可以包含一个更易读的版本作为附录。
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)
根据您对此的操作，可能会进行一些修改。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异时，该算法停止，它没有找到比它在5次迭代中看到的最好的东西更好的东西。一行，或总共100次迭代后作为一个全能停止点。您可能需要调整这些常量或使用不同的方案。如果你的高度形成一个复杂的价值观，那么知道什么时候停止可以进入试图逃避局部最大值等经典问题。
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]
添加了新测试用例，[95,15,75,25,85,5]，此方法正确处理。
此版本的算法更易于阅读和理解，但由于内置Python功能的优势较少，因此更长一些。然而，它似乎在可比较的或甚至更快的时间内执行。
#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)
答案 1 :(得分：4)
这是我现在获得的最佳O（n）贪婪算法。 我们的想法是贪婪地将列表中的项目附加到一个块，直到当前块的总和超过该点处的块的平均预期总和。平均预期金额不断更新。这个解决方案并不完美，但正如我所说，它是O（n）并且对我的测试没有坏处。我渴望听到反馈和改进建议。
我将调试打印语句留在代码中以提供一些文档。随意评论他们，看看每一步中发生了什么。
<强> CODE
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)))
测试代码
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
TESTS ：
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
TESTS ，长度为100的随机列表和1到100之间的元素（省略随机列表的打印）：
yield from正如您所看到的，正如预期的那样，您想要生成的块越多越好。我希望我有所帮助。
编辑：tr .selectAll("td")
.data(function(d) {
//your function
}).enter()
语法需要Python 3.3或更高版本，如果您使用的是旧版本，只需将该语句转换为正常for循环。
答案 2 :(得分：1)
我认为一个好的方法是对输入列表进行排序。然后将最小和最大的列表添加到一个列表中。第二个最小，第二个到下一个列表，依此类推，直到所有元素都添加到列表中。
[ProjectName] $ /usr/bin/security unlock-keychain -p ******** /Users/Shared/Jenkins/Home/jobs/JobName/workspace/build.keychain
security: SecKeychainUnlock /Users/Shared/Jenkins/Home/jobs/JobName/workspace/build.keychain: The user name or passphrase you entered is not correct.
FATAL: Unable to unlock the keychain.
def divide_list(A):
A.sort()
l = 0
r = len(A) - 1
l1,l2= [],[]
i = 0
while l < r:
ends = [A[l], A[r]]
if i %2 ==0:
l1.extend(ends)
else:
l2.extend(ends)
i +=1
l +=1
r -=1
if r == l:
smaller = l1 if sum(l1) < sum(l2) else l2
smaller.append(A[r])
return l1, l2
myList = [1, 6, 2, 3, 4, 1, 7, 6, 4]
print divide_list(myList)
myList = [1,10,10,1]
print divide_list(myList)
答案 3 :(得分：1)
这种情况有点迟了但是我想出了一个能够满足您需要的功能需要第二个参数来告诉它应该如何分割列表
import math
my_list = [1, 6, 2, 3, 4, 1, 7, 6, 4]
def partition(my_list, split):
solution = []
total = sum(my_list)
div = total / split
div = math.ceil(div)
criteria = [div] * (total // div)
criteria.append(total - sum(criteria)) if sum(criteria) != total else criteria
temp = []
pivot = 0
for crit in criteria:
for count in range(len(my_list) + 1):
if sum(my_list[pivot:count]) == crit:
solution.append(my_list[pivot:count])
pivot = count
break
return solution
print(partition(my_list, 2)) # Outputs [[1, 6, 2, 3, 4, 1], [7, 6, 4]]
print(partition(my_list, 3)) # Outputs [[1, 6, 2, 3], [4, 1, 7], [6, 4]]
它将失败4个部门，因为你明显在你的问题中说明你想维持秩序
4 divisions = [9, 9, 9, 7]
并且您的序列无法匹配
答案 4 :(得分：1)
以下是一些代码，为每个子列表返回2个切片索引。
weights = [1, 6, 2, 3, 4, 1, 7, 6, 4]
def balance_partitions(weights:list, n:int=2) -> tuple:
if n < 1:
raise ValueError("Parameter 'n' must be 2+")
target = sum(weights) // n
results = []
cost = 0
start = 0
for i, w in enumerate(weights):
delta = target - cost
cost += w
if cost >= target:
if i == 0 or cost - target <= delta:
results.append( (start, i+1) )
start = i+1
elif cost - target > delta:
# Better if we didn't include this one.
results.append( (start, i) )
start = i
cost -= target
if len(results) == n-1:
results.append( (start, len(weights)) )
break
return tuple(results)
def print_parts(w, n):
result = balance_partitions(w, n)
print("Suggested partition indices: ", result)
for t in result:
start,end = t
sublist = w[start:end]
print(" - ", sublist, "(sum: {})".format(sum(sublist)))
print(weights, '=', sum(weights))
for i in range(2, len(weights)+1):
print_parts(weights, i)
输出是：
[1, 6, 2, 3, 4, 1, 7, 6, 4] = 34
Suggested partition indices: ((0, 6), (6, 9))
- [1, 6, 2, 3, 4, 1] (sum: 17)
- [7, 6, 4] (sum: 17)
Suggested partition indices: ((0, 4), (4, 7), (7, 9))
- [1, 6, 2, 3] (sum: 12)
- [4, 1, 7] (sum: 12)
- [6, 4] (sum: 10)
Suggested partition indices: ((0, 3), (3, 5), (5, 7), (7, 9))
- [1, 6, 2] (sum: 9)
- [3, 4] (sum: 7)
- [1, 7] (sum: 8)
- [6, 4] (sum: 10)
Suggested partition indices: ((0, 2), (2, 4), (4, 6), (6, 7), (7, 9))
- [1, 6] (sum: 7)
- [2, 3] (sum: 5)
- [4, 1] (sum: 5)
- [7] (sum: 7)
- [6, 4] (sum: 10)
Suggested partition indices: ((0, 2), (2, 3), (3, 5), (5, 6), (6, 7), (7, 9))
- [1, 6] (sum: 7)
- [2] (sum: 2)
- [3, 4] (sum: 7)
- [1] (sum: 1)
- [7] (sum: 7)
- [6, 4] (sum: 10)
Suggested partition indices: ((0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 9))
- [1, 6] (sum: 7)
- [2] (sum: 2)
- [3] (sum: 3)
- [4] (sum: 4)
- [1] (sum: 1)
- [7] (sum: 7)
- [6, 4] (sum: 10)
Suggested partition indices: ((0, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9))
- [1, 6] (sum: 7)
- [2] (sum: 2)
- [3] (sum: 3)
- [4] (sum: 4)
- [1] (sum: 1)
- [7] (sum: 7)
- [6] (sum: 6)
- [4] (sum: 4)
Suggested partition indices: ((0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9))
- [1] (sum: 1)
- [6] (sum: 6)
- [2] (sum: 2)
- [3] (sum: 3)
- [4] (sum: 4)
- [1] (sum: 1)
- [7] (sum: 7)
- [6] (sum: 6)
- [4] (sum: 4)
答案 5 :(得分：0)
以下是我可能会针对两个所需子列表的情况来解决此问题。它可能没那么高效，但它是第一次削减。
def divide(l):
total = sum(l)
half = total / 2
l1 = []
l2 = []
for e in l:
if half - e >= 0 or half > abs(half - e):
l1.append(e)
half -= e
else:
l2.append(e)
return (l1, l2)
你可以在这里看到它：
(l1, l2) = divide([1, 6, 2, 3, 4, 1, 7, 6, 4])
print(l1)
# [1, 6, 2, 3, 4, 1]
print(l2)
#[7, 6, 4]
(l1, l2) = divide([1,1,10,10])
print(l1)
# [1, 1, 10]
print(l2)
#[10]
我会把其他案例留给你作为练习。 ：）
答案 6 :(得分：0)
使用numpy的简单方法。假设
import numpy.random as nr
import numpy as np
a = (nr.random(10000000)*1000).astype(int)
然后，假设您需要将列表划分为p个部分，它们的和大约相等。
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)
您可以检查拆分的质量，
partsums = np.array([part.sum() for part in parts]).std()
这些分割不是很好，但是鉴于顺序没有变化，我怀疑它们是最佳的。
答案 7 :(得分：0)
这是@Milind R的numpy方法的次要编辑版本（长官，非常感谢，先生）。即，我意识到，在现实生活中，如果元素的值不是“均匀地”分布在数组中，则脚本建议的分区可能最终不是最优的。为了解决这个问题，我通过重新排列“最小”，“最大”，“第二最小”，“第二最大”等元素来“统一”数组。下面的部分是，这使脚本相当大（〜5倍）慢点。
import numpy.random as nr
import numpy as np
a = (nr.random(10000000)*1000).astype(int)
编辑后的分区算法：
def equisum_partition(arr,p, uniformify=True):
#uniformify: rearrange to ['smallest', 'largest', 'second smallest', 'second largest', etc..]
if uniformify:
l = len(arr)
odd = l%2!=0
arr = np.sort(arr)
#add a dummy element if odd length
if odd:
arr = np.append(np.min(arr)-1, arr)
l = l+1
idx = np.arange(l)
idx = np.multiply(idx,
np.subtract(1,
np.multiply(
np.mod(idx, 2),
2))
)
arr = arr[idx]
#remove the dummy element
if odd:
arr = arr[1:]
#cumulative summation
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
在原始答案的示例中，由于示例数组的随机性，这没有太大作用。
使用统一：
%%time
parts = equisum_partition(a,20)
partsums = np.array([part.sum() for part in parts])#
partsums.std()
Wall time: 624 ms
266.6111212984185
不统一：
%%time
parts = equisum_partition(a,20, uniformify=False)
partsums = np.array([part.sum() for part in parts])#
partsums.std()
Wall time: 105 ms
331.19071544957296