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我想知道是否有一种“pythonic”方法可以将数学 argmin/argmax 用于成员函数，而不使用像numpy这样的库。

我有一个类，里面有一个成员函数，它返回一个整数。我实例化了这个类的几个对象。我想知道哪个对象的该方法的返回值较低。

请在下面找到我的源代码。我想改进的部分就在我想改进的标记代码之后。这段代码运行良好，但我很确定有更好的方法来做同样的事情。

#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""argmin example"""


class People(object):
    """People class"""
    ret_ages = {"Half": 60, "AlmostFull": 65, "Full": 71}  # years

    def __init__(self, name, age, ret_mode):
        super(People, self).__init__()

        if ret_mode not in self.ret_ages.keys():
            raise KeyError(ret_mode + " not in " + str(self.ret_ages.keys()))

        self.name = name
        self.age = age
        self.ret_mode = ret_mode

    def get_remaining_years(self):
        """
                Return how many years People have still to work before earning
                <rate> retirement.
                <rate> could be "Half", "Middle" or "Full".
        """
        try:
            return self.ret_ages[self.ret_mode] - self.age
        except KeyError:
            raise KeyError("rate has to be in " + str(self.ret_ages.keys()))


def main():
    """Main function"""
    people_list = [
        People("Juliette", 35, "Full"),
        People("Coralie", 26, "Half"),
        People("Laura", 27, "AlmostFull")
    ]

    # Debugging print
    for people in people_list:
        print people.name, "has still to work",\
            people.get_remaining_years(), "years."
    print
    # End of debugging print

    ############################
    # Code I'd like to improve #
    ############################

    people_closer_to_ret = people_list[0]
    minimum_remainining_years = people_closer_to_ret.get_remaining_years()

    for people in people_list:
        if people.get_remaining_years() < minimum_remainining_years:
            people_closer_to_ret = people
            minimum_remainining_years = people.get_remaining_years()

            minimum_remainining_years = people.get_remaining_years()

    ###################################
    # End of code I'd like to improve #
    ###################################

    print people_closer_to_ret.name, "will be retired soon !"


if __name__ == '__main__':
    main()

这是此脚本的输出：

Juliette has still to work 36 years.
Coralie has still to work 34 years.
Laura has still to work 38 years.

Coralie will be retired soon !
4

1 回答 1

2

以更Python的方式编写此代码的一个好方法是使用min函数，由于它的参数key，它实际上可以像argmin函数一样使用。

如果我们将我想改进的代码代码和我想改进的代码结束之间的代码替换为：

people_closer_to_ret = min(people_list,
                           key=lambda people: people.get_remaining_years()),

它运行良好。key参数对于告诉min函数它必须最小化哪个标准很有用。

所以你的完整代码如下：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""argmin example"""


class People(object):
    """People class"""
    ret_ages = {"Half": 60, "AlmostFull": 65, "Full": 71}  # years

    def __init__(self, name, age, ret_mode):
        super(People, self).__init__()

        if ret_mode not in self.ret_ages.keys():
            raise KeyError(ret_mode + " not in " + str(self.ret_ages.keys()))

        self.name = name
        self.age = age
        self.ret_mode = ret_mode

    def get_remaining_years(self):
        """
                Return how many years People have still to work before earning
                <rate> retirement.
                <rate> could be "Half", "Middle" or "Full".
        """
        try:
            return self.ret_ages[self.ret_mode] - self.age
        except KeyError:
            raise KeyError("rate has to be in " + str(self.ret_ages.keys()))


def main():
    """Main function"""
    people_list = [
        People("Juliette", 35, "Full"),
        People("Coralie", 26, "Half"),
        People("Laura", 27, "AlmostFull")
    ]

    # Debugging print
    for people in people_list:
        print people.name, "has still to work",\
            people.get_remaining_years(), "years."
    print
    # End of debugging print

    people_closer_to_ret = min(people_list,
                               key=lambda people: people.get_remaining_years())

    print people_closer_to_ret.name, "will be retired soon !"


if __name__ == '__main__':
    main()
于 2016-06-05T23:33:49.587 回答