如何在python上降低嵌套if的复杂度

def is_valid_input(input):
    return input > 0

def process_positive_input(input):
    # 处理正数输入的逻辑
    return input * 2

def process_negative_input(input):
    # 处理负数输入的逻辑
    return input / 2

def main(input):
    if not is_valid_input(input):
        return "Invalid input"
    
    if input > 0:
        return process_positive_input(input)
    else:
        return process_negative_input(input)

# 示例调用
result = main(5)
print(result)  # 输出: 10

def main(input):
    if input <= 0:
        return "Invalid input"
    
    if input > 0:
        return input * 2
    else:
        return input / 2

# 示例调用
result = main(5)
print(result)  # 输出: 10

def process_positive(input):
    return input * 2

def process_negative(input):
    return input / 2

def main(input):
    action = {
        True: process_positive,
        False: process_negative
    }.get(input > 0, lambda x: "Invalid input")
    
    return action(input)

# 示例调用
result = main(5)
print(result)  # 输出: 10

class InputProcessor:
    def process(self, input):
        raise NotImplementedError

class PositiveInputProcessor(InputProcessor):
    def process(self, input):
        return input * 2

class NegativeInputProcessor(InputProcessor):
    def process(self, input):
        return input / 2

def get_processor(input):
    if input > 0:
        return PositiveInputProcessor()
    elif input < 0:
        return NegativeInputProcessor()
    else:
        return None

def main(input):
    processor = get_processor(input)
    if processor is None:
        return "Invalid input"
    return processor.process(input)

# 示例调用
result = main(5)
print(result)  # 输出: 10