生成具有指定行和列总和的随机矩阵

[0.1, 0.2, 0.7] = 1
[0.5, 0.3, 0.2] = 1
[0.4, 0.5, 0.1] = 1
  1    1    1

import numpy as np

def generate_matrix_incorrect(x, y, z):
    matrix = np.random.rand(x, y)
    # 尝试按行归一化
    row_sums = matrix.sum(axis=1, keepdims=True)
    matrix = matrix / row_sums * z

    # 此时行和满足条件，但列和通常不满足
    print("Row sums after first normalization:", matrix.sum(axis=1))
    print("Col sums after first normalization:", matrix.sum(axis=0))

    # 验证行和
    assert np.allclose(matrix.sum(axis=1), z), "Row sums are not equal to Z"
    # 验证列和（此处会失败）
    # assert np.allclose(matrix.sum(axis=0), z), "Col sums are not equal to Z" # This assertion will fail

    return matrix.round(2)

# x = 3, y = 3, z = 1
# result_matrix = generate_matrix_incorrect(x, y, z)
# print(result_matrix)

import numpy as np

def generate_matrix_with_constrained_sums(x, y, z, max_iters=100, tol=1e-6):
    """
    生成一个x行y列的随机矩阵，确保每行和每列的元素之和都等于z。

    参数:
    x (int): 矩阵的行数。
    y (int): 矩阵的列数。
    z (float): 目标行和与列和。
    max_iters (int): 最大迭代次数，防止无限循环。
    tol (float): 容忍度，用于判断行和列和是否已足够接近z。

    返回:
    numpy.ndarray: 满足条件的随机矩阵。
    """
    if x <= 0 or y <= 0:
        raise ValueError("矩阵的行数和列数必须为正数。")
    if z <= 0:
        raise ValueError("目标和Z必须为正数。")

    # 1. 初始化矩阵，元素为0到1之间的随机数
    matrix = np.random.rand(x, y)

    for i in range(max_iters):
        # 2.1 行归一化与缩放
        # 计算每行的和，keepdims=True 保持维度，以便广播
        row_sums = matrix.sum(axis=1, keepdims=True)
        # 避免除以零，对极小值进行处理
        row_sums[row_sums == 0] = 1e-10
        matrix = matrix / row_sums * z

        # 2.2 列归一化与缩放
        # 计算每列的和
        col_sums = matrix.sum(axis=0, keepdims=True)
        # 避免除以零
        col_sums[col_sums == 0] = 1e-10
        matrix = matrix / col_sums * z

        # 可选：检查收敛，提前退出
        if np.allclose(matrix.sum(axis=1), z, atol=tol) and \
           np.allclose(matrix.sum(axis=0), z, atol=tol):
            # print(f"矩阵在 {i+1} 次迭代后收敛。")
            break
    else:
        # print(f"警告：矩阵在 {max_iters} 次迭代后未能完全收敛到指定容忍度。")
        pass

    # 3. 最终验证
    # 使用np.allclose来处理浮点数精度问题
    assert np.allclose(matrix.sum(axis=1), z, atol=tol), "行和不等于Z"
    assert np.allclose(matrix.sum(axis=0), z, atol=tol), "列和不等于Z"

    # 返回四舍五入到两位小数的结果，方便查看
    return matrix.round(2)

# 示例用法
x_dim = 3
y_dim = 3
target_sum = 1

result_matrix = generate_matrix_with_constrained_sums(x_dim, y_dim, target_sum)
print("生成的矩阵:")
print(result_matrix)
print("\n行和:")
print(result_matrix.sum(axis=1))
print("\n列和:")
print(result_matrix.sum(axis=0))

# 另一个例子：非方阵，Z=10
x_dim_2 = 2
y_dim_2 = 4
target_sum_2 = 10
result_matrix_2 = generate_matrix_with_constrained_sums(x_dim_2, y_dim_2, target_sum_2)
print("\n--- 另一个例子 (2x4, Z=10) ---")
print("生成的矩阵:")
print(result_matrix_2)
print("\n行和:")
print(result_matrix_2.sum(axis=1))
print("\n列和:")
print(result_matrix_2.sum(axis=0))