다음은 각 하위 목록에 대해 2-ples의 조각 인덱스를 반환하는 코드입니다.

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)

출력은 다음과 같습니다.

[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)

여기에 가장 좋은 O (n) greedy 알고리즘이 있습니다. 아이디어는 현재 청크의 합계 가 그 시점의 청크에 대한 예상 평균 합계 를 초과 할 때까지 목록의 항목을 청크에 추가하는 것입니다. 예상 평균 금액은 지속적으로 업데이트됩니다. 이 솔루션은 완벽하지는 않지만, 내가 말했듯이, 그것은 O (n)이며 테스트 결과 나쁘지 않습니다. 개선을위한 의견 및 제안을 듣고 자합니다.

몇 가지 문서를 제공하기 위해 디버그 인쇄 문을 코드에 남겼습니다. 각 단계에서 어떤 일이 벌어지고 있는지 보려면 언제든지 댓글을 달아주세요.

암호

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]

코드 검사

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)))

귀하의 목록과 함께 시험 :

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

길이 100의 무작위 목록과 1에서 100까지의 요소가있는 시험 (무작위 목록 인쇄가 생략 됨) :

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

보시다시피 예상대로 생성하려는 청크가 많을수록 악화됩니다. 나는 조금 도울 수 있었으면 좋겠다.

편집 : 구문 yield from 는 Python 3.3 이상이 필요합니다. 이전 버전을 사용하는 경우 명령문을 일반 for 루프로 바꾸십시오.

이 방법은 대략 동일한 수의 요소로 배열을 나누는 파티션 경계를 정의한 다음 더 이상 찾을 수 없을 때까지 더 나은 파티션을 반복적으로 검색합니다. 다른 게시 된 솔루션의 대부분과 다른 점은 여러 가지 다른 파티션을 시도하여 최적의 솔루션을 찾는 것입니다. 다른 솔루션은 배열을 통해 단일 패스에서 좋은 파티션을 만들려고 시도하지만 최적의 보장을받는 단일 패스 알고리즘은 생각할 수 없습니다.

여기에있는 코드는이 알고리즘을 효율적으로 구현 한 것으로 이해하기는 어려울 수 있으므로 더 읽기 쉬운 버전이 끝에 추가 정보로 포함됩니다.

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)

이 작업으로 수행하는 작업에 따라 약간의 수정이있을 수 있습니다. 예를 들어, 최상의 파티셔닝이 발견되었는지 여부를 결정하기 위해 파티션간에 높이 차이가 없을 때이 알고리즘이 중지되고, 연속적으로 5 회 이상 또는 100 회 이후에 가장 좋은 것보다 나은 것을 찾지 못합니다 전체 반복은 포괄적 인 중지 지점으로 간주됩니다. 이러한 상수를 조정하거나 다른 구성표를 사용해야 할 수도 있습니다. 당신의 고소가 복잡한 가치관을 형성한다면, 멈출 때를 알면 로컬 맥시마와 같은 것들을 피하려고하는 고전적인 문제에 빠질 수 있습니다.

산출

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]

편집하다

이 메서드가 올바르게 처리하는 새 테스트 사례 [95, 15, 75, 25, 85, 5]를 추가했습니다.

추가

이 버전의 알고리즘은 읽기 쉽고 이해하기 쉽지만 내장 된 Python 기능의 이점을 덜 사용하기 때문에 조금 더 길어집니다. 그러나 비슷한 시간 또는 약간 더 빠른 시간 내에 실행되는 것으로 보입니다.

#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)

좋은 방법은 입력 목록을 정렬하는 것입니다. 그런 다음 최소 목록과 최대 목록을 하나의 목록에 추가하십시오. 목록에 모든 요소가 추가 될 때까지 다음 목록에서 두 번째로 큰 목록과 두 번째로 큰 목록 등등.

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)

산출

([1, 7, 2, 6], [1, 6, 3, 4, 4])
([1, 10], [1, 10])