Python实现二维列表矩阵减法及结果元素求和的最优方法

分两种场景选择最优实现方式：

1. 无第三方依赖场景（纯Python实现） #

如果矩阵规模小、不想额外安装依赖库，优先用生成器表达式直接边减边求和，不需要存储完整的中间减法矩阵，内存开销最低：

a = [
    [2, 3, 4],
    [1, 5, 2]
]
b = [
    [1, 4, 2],
    [0, 1, 3]
]

# 直接计算差值总和，无额外中间存储
total = sum(a[i][j] - b[i][j] for i in range(len(a)) for j in range(len(a[0])))
print(total) # 输出6

如果需要保留减法后的矩阵结果，可分开实现：

# 先计算矩阵减法结果
diff = [[a[i][j] - b[i][j] for j in range(len(a[0]))] for i in range(len(a))]
# 再对所有元素求和
total = sum(num for row in diff for num in row)

print(diff) # 输出[[1, -1, 2], [1, 4, -1]]
print(total) # 输出6

2. 大规模数值计算场景（numpy实现） #

如果处理的矩阵规模大，优先用numpy实现，底层为C语言优化的运算逻辑，执行效率远高于纯Python，语法也更简洁：

import numpy as np

a = np.array([
    [2, 3, 4],
    [1, 5, 2]
])
b = np.array([
    [1, 4, 2],
    [0, 1, 3]
])

# 矩阵减法
diff = a - b
# 所有元素求和
total = diff.sum()

print(diff)
# 输出：
# [[ 1 -1  2]
#  [ 1  4 -1]]
print(total) # 输出6

内容的提问来源于stack exchange，提问作者Carlos Andrés del Valle