按非零值的数量将稀疏矩阵划分为更小的块

问：

我有以下代码：

import io, sys, numpy, scipy, struct, os
import numpy as np
from scipy import io
from scipy import sparse
from numpy import inf
import random as rnd
import sys
from scipy import sparse

def myhorsplit(matrix,numPartitions):
    csr = sparse.csr_matrix(matrix)
    input_array = csr.getnnz(0)
    print(input_array)
        # Step 1: Calculate the total
    total = sum(input_array)

    # Step 3: Set number of compute units to 4
    num_compute_units = 4
 
    partitions = [[] for _ in range(num_compute_units)]
    current_sums = [0] * num_compute_units
    for element in input_array:
        # Find the partition with the smallest current sum
        min_partition = min(range(num_compute_units), key=lambda i: current_sums[i])

        # Add the element to the selected partition
        partitions[min_partition].append(element)
        current_sums[min_partition] += element

    for i, partition in enumerate(partitions):
        print(f"Partition {i}: {partition}")
    print("Rows Result:")
    print(csr.tolil().rows[0])
    print("Get Column Result:")
    print(csr.tolil().getcol(1))
    
    return partition

# Create an 8x8 adjacency matrix with the modified element
adjacency_matrix = [
    [1, 1, 1, 1, 0, 0, 0, 0],
    [1, 0, 1, 0, 0, 0, 0, 0],
    [1, 1, 0, 1, 0, 0, 0, 0],
    [1, 0, 1, 0, 0, 0, 0, 0],
    [0, 0, 1, 0, 0, 1, 0, 1],
    [0, 0, 0, 0, 1, 0, 0, 0],
    [0, 0, 0, 0, 1, 1, 0, 1],
    [0, 0, 1, 0, 1, 0, 1, 0]
]
csr_matrix = sparse.csr_matrix(adjacency_matrix)
# horizontalSplit(csr_matrix,4)
myhorsplit(csr_matrix,4)

打印出来：

[4 2 5 2 3 2 1 2]
Partition 0: [4, 1]
Partition 1: [2, 3]
Partition 2: [5]
Partition 3: [2, 2, 2]
Rows Result:
[0, 1, 2, 3]
Get Column Result:
  (0, 0)        1
  (2, 0)        1

我找不到一种方法将这个稀疏矩阵划分为更小的块，以便前 3 个块（假设我想要 4 个块）具有总非零值/4，最后一个块具有其余的非零值？