带有条件语句的 For-Loop 输出问题 - python

Question

这个函数只返回 1 的值。有人可以帮忙吗？这些参数描述了我正在尝试实现的内容，但我不确定为什么会发生这种情况。

def clamp(alist,min,max):
    """
    MODIFIES the list so that every element is between min and max.

    Any number in the list, less than min is replaced with min.  Any 
    number in the list greater than max is replaced with max. Any number 
    between min and max is left unchanged.

    This is a PROCEDURE. It modifies alist, but does not return a new 
    list.

    Example: if alist is [-1, 1, 3, 5], then clamp(thelist,0,4) changes
    alist to have [0,1,3,4] as its contents.

    Parameter alist: the list to modify
    Precondition: alist is a list of numbers (float or int)

    Parameter min: the minimum value for the list
    Precondition: min <= max is a number

    Parameter max: the maximum value for the list
    Precondition: max >= min is a number
    """
    alist.append(min)
    alist.append(max)
    for a in alist:
        if max >= a >= min:
            return a
    alist.sort()
    return alist

返回：1，无论输入。

Answer 1

代码的三个问题

1. 只要元素在范围内，return a 就会导致返回
2. alist.append(min)、alist.append(max) 和 alist.sort() 与声明的所需功能无关
3. 不应使用内置函数名作为变量名（即min & max）

代码

def clamp(alist, min_, max_):
    for i, a in enumerate(alist):
        if a > max_:
          alist[i] = max_     # limit to max
        elif a < min_:
          alist[i] = min_     # limit to min

    return alist

print((clamp([-1, 1, 3, 5], 0, 4)))
# Output: [0, 1, 3, 4]

替代代码（使用列表理解）

def clamp(alist, min_, max_):
  return [min_ if a < min_ else max_ if a > max_ else a for a in alist]