这是我想到的。这是一个简单的贪婪的方法来产生一个分裂与平均总数。不完全是你所要求的，但希望它有帮助。

import bisect

class Cluster:
    def __init__(self, nums):
        self.nums = nums
        self.total = sum(nums)

def main():
    nums = [10, 20, 30, 40, 50, 60, 70, 80, 10, 20, 25, 20, 21, 26, 65, 95, 84, 65, 2, 3, 6, 198, 16, 651, 984, 651, 35, 61, 651, 16, 56, 651, 651, 651, 2, 32, 615, 651, 984, 615, 351, 651, 651, 3, 5]
    clusters = [Cluster([n]) for n in nums]
    while len(clusters) > 1:
        pairs = list(zip(clusters, clusters[1:]))
        best_pair_index = min(
            range(len(pairs)), key=lambda i: abs(pairs[i][0].total - pairs[i][1].total)
        )
        best_pair = pairs[best_pair_index]
        combined = Cluster(best_pair[0].nums + [-n for n in best_pair[1].nums])
        del clusters[best_pair_index : best_pair_index + 2]
        bisect.insort(clusters, combined, key=lambda c: c.total)
    [cluster] = clusters
    left = [n for n in cluster.nums if n > 0]
    right = [-n for n in cluster.nums if n < 0]
    print(sum(left))
    print(sum(right))
    print(len(left))
    print(len(right))

main()

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我将在这里发布我的尝试以供将来参考。我认为Alex Hall的答案是suberb，但有点复杂。这个函数可以很好地将列表与我想要的标准分开，并处理列表，如@RomanPerekhrest的评论。

def find_optimal_batch(numbers, ratio):

# Calculate the sum of the original list
total_sum = sum(numbers)

# Initialize variables to track the optimal batch and ratio
optimal_batch = []
optimal_difference = float('inf')  # Initialize to positive infinity

# Precompute the range of batch sizes
num_numbers = len(numbers)
batch_sizes = range(1, num_numbers + 1)

# Iterate through the batches
for batch_size in batch_sizes:

    # Iterate and calculate sums for each batch
    for i in range(num_numbers - batch_size + 1):

        batch_sum = sum(numbers[i:i+batch_size])
        batch_ratio = batch_sum / total_sum
        difference = abs(batch_ratio - ratio)

        # Check if the current batch has a closer ratio to the desired ratio
        if difference < optimal_difference:

            batch_1 = numbers[i:i+batch_size]
            batch_2 = numbers[0:i]+numbers[i+batch_size:]
            optimal_difference = difference
return batch_1, batch_2

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