Python类调用max_profit_assignment方法报错：参数数量不匹配

问题重现 #

你定义了Profile类的max_profit_assignment方法，通过实例调用时触发报错：

class JOB:
    def __init__(self, job, profit):
        self.job = job
        self.profit = profit

class Profile:
    def max_profit_assignment(difficulty, profit, worker):
        jobs = [JOB(difficulty[i], profit[i]) for i in range(len(difficulty))]

        n = len(worker)
        i = 0
        current_profit = 0
        max_profit = 0

        for able in worker:
            while i < n:
                if jobs[i].job <= able:
                    current_profit = max(current_profit, jobs[i].profit)
                i += 1
            i = 0

            print(current_profit)
            max_profit += current_profit
            current_profit = 0

        return max_profit

if __name__ == "__main__":
    difficulty = [5, 50, 92, 21, 24, 70, 17, 63, 30, 53]
    profit = [68, 100, 3, 99, 56, 43, 26, 93, 55, 25]
    worker = [96, 3, 55, 30, 11, 58, 68, 36, 26, 1]
    profile = Profile()
    print(dir(profile))
    print(profile.max_profit_assignment(difficulty, profit, worker))

执行后抛出错误：

File "C:\Users\jerry\Downloads\GOLong\PythonCode\Profile.py", line 34, in <module>
    print(profile.max_profit_assignment(difficulty, profit, worker))
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
TypeError: Profile.max_profit_assignment() takes 3 positional arguments but 4 were given

报错原因 #

在Python中，实例方法的第一个参数必须是self（参数名可自定义，但行业约定用self），它代表调用该方法的实例本身。

你定义max_profit_assignment时只声明了3个参数，但通过profile实例调用方法时，Python会自动把实例profile作为第一个参数传递给方法，导致实际传递了4个参数（实例本身 + difficulty + profit + worker），而方法仅接受3个，因此触发参数不匹配的TypeError。

修复方案 #

有两种可行的修复方式：

1. 改为标准实例方法，添加self作为第一个参数：

class Profile:
    def max_profit_assignment(self, difficulty, profit, worker):
        # 原有代码逻辑保持不变
        jobs = [JOB(difficulty[i], profit[i]) for i in range(len(difficulty))]
        # ... 后续代码省略

2. 如果该方法不需要依赖实例的任何属性，可定义为静态方法，用@staticmethod装饰器省略self参数：

class Profile:
    @staticmethod
    def max_profit_assignment(difficulty, profit, worker):
        # 原有代码逻辑保持不变
        jobs = [JOB(difficulty[i], profit[i]) for i in range(len(difficulty))]
        # ... 后续代码省略

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