如何用Python函数式编程实现刻度列表并优化现有代码？

需求说明 #

将以下C风格的Python代码重写为函数式范式，使其更符合Pythonic风格，最终得到指定的刻度列表输出。

原代码 #

X = [0, 1, 2, 3, 4, 3, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 2, 1, 1, 4, 3, 4, 4, 5, 1, 4, 2, 3, 3, 3, 2, 4, 4, 1, 3, 3, 2, 4, 1, 3, 3, 5, 5, 4, 4, 3, 2, 3, 4, 2, 1, 3, 4, 1, 2, 3, 5, 2, 3, 4, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 3, 1, 3, 3, 3, 2, 3, 5, 2, 1, 2, 3, 2, 1, 4, 1, 3, 2, 1, 3, 2, 3, 5, 2, 4, 1, 3, 4, 3, 3, 2, 4, 3, 4, 4, 3, 2, 1, 2, 3, 3, 4, 1, 4, 4, 3, 3, 3, 4, 3]

res = [0] * 121

for i in range(1, 31):
  k = 0
  for _ in range(0, 4):
    res[i] += X[i + k]
    k = k + 30

print(res)

期望输出 #

[0, 9, 12, 11, 12, 13, 7, 12, 11, 12, 12, 13, 13, 13, 13, 13, 9, 9, 11, 10, 8, 12, 15, 9, 8, 15, 11, 11, 11, 12, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

函数式风格实现 #

原代码逻辑是：对索引i从1到30，累加X[i]、X[i+30]、X[i+60]、X[i+90]四个值到res[i]，其余位置保持0。用函数式范式可以通过列表推导+生成器表达式一次性构建结果，避免嵌套循环和可变状态修改：

X = [0, 1, 2, 3, 4, 3, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 2, 1, 1, 4, 3, 4, 4, 5, 1, 4, 2, 3, 3, 3, 2, 4, 4, 1, 3, 3, 2, 4, 1, 3, 3, 5, 5, 4, 4, 3, 2, 3, 4, 2, 1, 3, 4, 1, 2, 3, 5, 2, 3, 4, 3, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 3, 1, 3, 3, 3, 2, 3, 5, 2, 1, 2, 3, 2, 1, 4, 1, 3, 2, 1, 3, 2, 3, 5, 2, 4, 1, 3, 4, 3, 3, 2, 4, 3, 4, 4, 3, 2, 1, 2, 3, 3, 4, 1, 4, 4, 3, 3, 3, 4, 3]

res = [
    sum(X[i + 30 * offset] for offset in range(4)) if 1 <= i <= 30 else 0
    for i in range(121)
]

print(res)

说明 #

• 用列表推导直接生成长度为121的结果列表，无需提前初始化再修改
• 利用sum()结合生成器表达式，简洁计算四个偏移位置的元素和，符合函数式"无副作用、用表达式替代语句"的特点
• 逻辑清晰直观，和原代码功能完全一致，输出与期望结果匹配

内容的提问来源于stack exchange，提问作者Ederson Corbari